Class 12

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 2 Inverse Trigonometric Functions will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 2 Inverse Trigonometric Functions

Definitions:
Let us first consider the function
Sin : R → [-1, 1]
Let y = Sin x, x ∈ R. Look at the graph of sin x. For y ∈ [-1, 1], there is a unique number x in each of the intervals…, \(\left[-\frac{3 \pi}{2},-\frac{\pi}{2}\right],\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right], \ldots\) such that y = sin x.

Hence any one of these intervals can be chosen to make sine function bijective.

We usually choose \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) as the domain of sine function. Thus sin: \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) → [-1, 1] is bijective and hence admits of an inverse function with range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) denoted by sin^-1 or arcsin (see footnote).

Each of the above-mentioned intervals as range gives rise to different branches of sin^-1 function. The function sin^-1 with the range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is called the principal branch which is defined below.

sin^-1 : [-1, 1] → \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) defined by y = sin^-1 x ⇔ x = sin y.

The values of y (=sin^-1 x) in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) are called principal values of sin^-1.

Similar considerations for other trigonometric functions give rise to respective inverse functions. We define below the principal branches of cos^-1, tan^-1 and cot^-1.

cos^-1 : [-1, 1] → [0, π] defined by
y = cos^-1 x ⇔ x = cos y
tan^-1 : R → \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) defined by
y = tan^-1 x ⇔ x = tan y
cot^-1 : R → (0, π) defined by
y = cot^-1 x ⇔ x = cot y

Important Properties

Property – I
We know that when
f : X → Y is invertible then fof^-1 = I_y and f^-1of = I_x.
Applying this we have.
(i) sin (sin^-1 x) = x, x ∈ [-1, 1]
cos (cos^-1 x) = x, x ∈ [-1, 1]
tan (tan^-1 x) = x, x ∈ R
cot (cot^-1 x) = x, x ∈ R
sec (sec^-1 x) = x, x ∈ R -(-1, 1)
cosec (cosec^-1 x) = x, x ∈ R -(-1, 1)

Property – II
(i) sin^-1(-x) = -sin^-1 x, x ∈ [-1, 1]
(ii) cosec^-1(-x) = -cosec^-1 x , |x| ≥ 1
(iii) tan^-1(-x) = -tan^-1 x, x ∈ R

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 1 Relation and Function will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 1 Relation and Function

Ordered Pair
It is a pair of numbers of functions listed in a specific order, eg. (a, b) is an ordered pair.

Cartesian Product:
Let A and B are two non empty sets. The cartesian product of A and B = A × B = {(a, b) : a ∈ A, b ∈ B}.

Relation
A relation R from A to B is a subset of A × B i.e. R ⊆ A × B.

Relation on a set:
R is a relation on A if R ⊆ A × A.

Domain, Range and Co-domain of a relation:
Let R is a relation from A to B.
Dom R = {x ∈ A: (x, y) ∈ R for y ∈ B }
Rng R = {y ∈ B : (x, y) ∈ R for x ∈ A}
Co-domain of R = B

Types of Relation
(a) Empty or void relation:
As Φ ⊂ A × B we have Φ is a relation known as empty or void relation.

(b) Universal Relation:
As A × B ⊆ A × B, we have A × B is a relation, known as universal relation.

(c) Identity Relation:
A relation R on A is an identity relation of (a, a) ∈ R, For a ∈ A.

(d) Reflexive Relation:
A relation R on A is reflexive if (a, a) ∈ R for all a ∈ A.

(e) Symmetric Relation:
A relation R on A is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R, where a, b ∈ A.

(f) Transitive Relation:
A relation R on A is transitive if (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R where a, b, c ∈ A.

(g) Anti Symmetric Relation:
A relation R on A is anti-symmetric if (a, b), (b, a) ∈ R ⇒ a = b.

(h) Equivalence Relation:
A relation R on A is an equivalence relation if it is reflexive, symmetric as well as Transitive.

(i) Partial ordering:
A relation R on A is a partial ordering if it is reflexive, transitive and antisymmetric.

(j) Total ordering:
A relation R on A is a total ordering. If it is a partial ordering and either (a, b) or (b, a) ∈ R for a, b ∈ A.

Equivalence Class:
Let R is an equivalence relation on X for any x ∈ X the equivalence class of x.
= [x] = {y ∈ X : (x, y) ∈ R}
(a) For all x ∈ X, [x] ≠ Φ
(b) If (x, y) ∈ R, then [x] = [y]
(c) If (x, y) ∉ R then [x] ∩ [y] = Φ
(d) An equivalence relation partitions a set into disjoint equivalence classes.

Function
Function as a Rule:
Let A and B are two non empty sets. If each element of A is mapped to a unique element of B by rule ‘f’ then f is a function.
Function as a relation
A relation f from A to B is a function if
(i) Dom f = A
(ii) (x, y), (x, z) ∈ f ⇒ y = z

Domain, Co-domain and Range of a function

Domain of a Real Function:
Let f : A → B, defined as y = f(x)
Dom f = {x ∈ A: = f(x) for y ∈ B}

Range of a function:
Let f : A → B, defined as y = f(x)
Rng f = {y ∈ B : y = f(x) for all x ∈ A}

Co-Domain of a function:
Let f : A → B defined as y = f(x)
Co dom f = B
Note: Rng f ⊆ Co dom f

Types of Functions

Injective (one-one) function:
f : A→ B is one one if f(x₁) = f(x₂) ⇒ x₁ = x₂
Step-1
Methods of check Injective function:
Consider any two arbitrary x₁, x₂ ∈ A.
Step-2
Put f(x₁) = f(x₂) and simplify.
Step-3
If we get x₁ = x₂ then f is one-one, otherwise f is many one.

Surjective (onto) function:
A function f : A → B is onto if Rng f = Co-domain f = B.

Methods of Check for onto
Method-1:
Find Range of f
If Rng f = B then f is onto.

Method-2:
Step-1
Consider any arbitrary y ∈ B
Step-2
Write y = f(x) and simplify to express x in terms of y.
Step-3
If x ∈ A, then find f(x)
Step-4
If f(x) = y, then f is onto
In case x ∉ A or f(x) ≠ y then f is not onto, it is into function.

Bijective function (one-one and onto):
f : A → B is a bijective function if its both one-one and onto.

Composition of functions:
Let f : X → Y and g : Y → Z, the composition of f and g denoted by gof, is defined as
gof : X → Z defined by
gof(x) = g(f (x)) for all x ∈ X.

Note:

1. gof is defined when Rngf ⊆ Dom g.
2. Dom gof = Dom f.
3. As gof(x) = g(f(x)), first f rule is applied then g rule is applied.
4. gof ≠ fog
5. If f : R → R and g : R→ R then gof and fog exist.
6. ho(gof) = (hog)of
7. If f : X → Y and g : Y → Z are one-one, then gof : X → Z is also one-one.
8. If f : X → Y and g : Y → Z are onto then go f : X → Z is also onto.
9. Let f : X → Y and g : Y → Z then
   (i) gof : X → Z is onto ⇒ g is onto.
   (ii) gof is one-one ⇒ f is one-one.
   (iii) gof is on to and g is one-one then f is onto.
   (iv) gof is one-one and f is onto ⇒ g is one-one.

Invertible functions:
A function f : X → Y is invertible if there exists a function g : Y → X, such that gof = idx and fog = idy.
The function g is called the inverse of f denoted by g = f^-1.

Note:

1. Not all functions are invertible.
2. Only bijective functions are invertible.
3. Inverse of a function may not be a function.
4. Dom f^-1 = Y
5. \(\left(f^{-1}\right)^{-1}\)
6. (gof)^-1 = f^-1og^-1

Methods to check invertibility of a function and to find f^-1:

Method-1:
Step-1
Write f : X→ Y, defined as y = f(x) = an expression in x.
Step-2
Take any arbitrary y ∈ Y. and write y = f(x).
Step-3
Express x in terms of y and check that x ∈ X.
Step-4
Define g : Y → X as x = g(y) = Value of x in terms of y.
Step-5
find gof (x) and fog of (y).
Step-6
If gof (x) = x and fog (y) = y then f is invertible with f^-1.

Method-2:
Step-1
Show that ‘f’ is one-one.
Step-2
Show that ‘f’ is on to.
Step-3
Either from step-2 or otherwise, write x in terms of y and write f^-1 : Y → X defined as f^-1(y) = x (in terms of y).

Even and odd function:
A function f is even if f(-x) and odd if f(-x) = -f(x).

Note:

1. fi(x) + f(-x) is always even.
2. f(x) – f(-x) is always odd.
3. Every function can be expressed as the sum of one even and one odd function as
   f(x) = \(\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}\)

Binary operations:
Let A is a nonempty set. Then a binary operation * on a set A is a function *:
A × A → A

Note :

1. Closure property: An operation * on a non-empty set A is a said to satisfy closure property if for every a, b ∈ A
   ⇒ a * b ∈ A.
2. If * is a binary operation on a non-empty set A, then must satisfy the closure property.
3. Number of binary operations on A where |A| = n is \(n^{n^2}\).

Properties of a Binary operation
Let * is a binary operation on A.
1. Commutative Law:
* is commutative if for all a, b ∈ A. a * b = b * a

2. Associative Law:
* is a associative if for all a, b, c ∈ A.
(a * b) * c = a * (b * c)

3. Distributive Law:
Let ‘*’ and ‘a’ are two binary operations on A.
‘*’ is said to be distributive over ‘o’ if for all a, b, ∈ A.
a * (b o c) = (a * b) o (a * c)

4. Existence of identity element:
e ∈ A is said to be the identity element for the binary operation if for all a ∈ A.
a * e = e * a = a

5. Existence of inverse of an element:
a^-1 ∈ A is inverse of a ∈ A if a * a^-1 = a^-1 * a = e, where e is the identity.

Operation or composition table for a binary operation.
Let * is any operation on a finite set A = {a₁, a₂ …… a_n}
The table containing the results of the operation * is known as the operation table.
We can study different properties of binary operation * from the table.

Operation table for * on A:

Study of properties from operation table:
1 . If all the entries of the table belong to A, then A is a binary operation.
2. If each row coincides with the corresponding column, then * is commutative.
3. If elements of a row are identical to the top row, then the leftmost element of that row is the identity element.
4. Mark the position of identity elements in the table. The leading elements in the corresponding row and column of that cell are inverse of each other.

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 13 Three Dimensional Geometry Additional Exercise Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 13 Three Dimensional Geometry Additional Exercise

Question 1.
Find the equation in vector and Cartesian form of the plane passing through the point (3, -3, 1) and normal to the line joining the points (3, 4, -1) and (2, -1, 5)
Solution:

Changing the vector equation to cartesian form we have
(xi + yj + zk) . (-i – 5j + 6k) = 18
⇒ -x – 5y + 6z = 18
⇒ x + 5y – 6z + 18 = 0

Question 2.
Find the vector equation of the plane whose Cartesian form of equation is 3x – 4y + 2z = 5
Solution:
The cartesian equation is 3x – 4y + 2z = 5
The vector equation is \(\vec{r}\). (3i – 4j + 2k) = 5

Question 3.
Show that the normals to the planes \(\vec{r}\) . (î – ĵ + k̂) = 3 and \(\vec{r}\) . (3î + 2ĵ – k̂) = 0 are perpendicular to each other.
Solution:

Question 4.
Find the angle between the planes \(\vec{r} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})=6 \text { and } \vec{r} \cdot(3 \hat{i}+6 \hat{j}-2 \hat{k})=9\).
Solution:

Question 5.
Find the angle between the line \(\vec{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and the \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 4.
Solution:

Question 6.
Prove that the acute angle between the lines whose direction cosines are given by the relations l + m + n = 0 and l² + m² – n² = 0 is \(\frac{\pi}{3}\).
Solution:
l + m + n = 0 ⇒ n = -(l + m)
l² + m² – n² = 0
⇒ l² + m² – (l + m)² = 0
⇒ 2lm = 0
Now l = 0 ⇒ m + n = 0 ⇒ m = -n

Question 7.
Prove that the three lines drawn from origin with direction cosines l₁, m₁, n₁ ; l₂, m₂, n₂ ; l₃, m₃, n₃ are coplanar if \(\left|\begin{array}{lll}
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2 \\
l_3 & m_3 & n_3
\end{array}\right|\) = 0.
Solution:

Question 8.
Prove that three lines drawn from origin with direction cosines proportional to (1, -1, 1), (2, -3, 0), (1, 0, 3) lie on one plane.
Solution:

= 1 (-9) + 1 (6) + 1(3) = 0
∴ The lines are co-planar.

Question 9.
Determine k so that the lines joining the points P₁ (k, 1, -1) and P₂ (2k, 0, 2) shall be perpendicular to the line from P₂ to P₃ (2 + 2k, k, 1).
Solution:

Question 10.
Find the angle between the lines whose direction ratios are proportional to a, b, c and b-c, c-a, a-b.
Solution:

Question 11.
O is the origin and A is the point (a, b, c). Find the equation of the plane through A at right angles to \(\overrightarrow{OA}\).
Solution:

Question 12.
Find the equation of the plane through (6, 3, 1) and (8, -5, 3) parallel to x-axis.
Solution:
Given points are A (6, 3, 1) and B (8, -5, 3) D.C.S of x-axis are < 1, 0, 0 >
Let P (x, y, z) is any point on the plane.

⇒ 2 (y – 3) + 8 (z – 1) = 0
⇒ 2y – 6 + 8z – 8 = 0
⇒ y + 4z – 7 = 0

(A) Multiple Choice Questions (Mcqs) With Answers

Question 1.
Write the value of k such that the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies on the plane 2x – 4y + z = 7.
(a) k = 7
(b) k = 2
(c) k = 5
(d) k = 3
Solution:
(a) k = 7

Question 2.
If \((\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2\) = 144, write the value of ab.
(a) 4
(b) 12
(c) 3
(d) 24
Solution:
(b) 12

Question 3.
If the vectors \(\vec{a}, \vec{b}\text { and }\vec{c}\) from the sides \(\overline{BC}, \overline{CA} \text { and } \overline{AB}\) respectively of a triangle ABC, then write the value of \(\vec{a} \times \vec{c}+\vec{b} \times \vec{c}\).
(a) 1
(b) -1
(c) 2
(d) 0
Solution:
(d) 0

Question 4.
If \(|\vec{a}|=3|\vec{b}|=2 \text { and } \vec{a} \cdot \vec{b}=0\), then write the value of \(|\vec{a} \times \vec{b}|\).
(a) 6
(b) 0
(c) 2
(d) 3
Solution:
(a) 6

Question 5.
If \(|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2\) = 144 and \(|\vec{a}|\) = 4 then \(|\vec{b}|\) is equal to:
(a) 3
(b) 8
(c) 12
(d) 16
Solution:
(a) 3

Question 6.
Write down the equation to the plane perpendicular to the y-axis at the point (0, -2, 0).
(a) y – 2 = 0
(b) y + 2 = 0
(c) y = 0
(d) 2y + 1 = 0
Solution:
(b) y + 2 = 0

Question 7.
How many straight lines in space through the origin are equally inclined to the coordinate axes?
(a) 2
(b) 1
(c) 0
(d) -1
Solution:
(a) 2

Question 8.
Write the value of a if the vectors \(\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k} \text { and } \vec{b}=\alpha \hat{i}-\hat{j}+2 \hat{k}\) are parallel.
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{-3}{2}\)
(d) \(\frac{-2}{3}\)
Solution:
(d) \(\frac{-2}{3}\)

Question 9.
Write the equation of the line passing through the point (4, -6, 1) and parallel to the line \(\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-1}{-1}\).
(a) \(\frac{x+4}{1}=\frac{y+6}{3}=\frac{z+1}{-1}\)
(b) \(\frac{x-4}{1}=\frac{y+6}{3}=\frac{z-1}{-1}\)
(c) \(\frac{x-4}{1}=\frac{y-6}{3}=\frac{z-1}{-2}\)
(d) \(\frac{x+4}{-1}=\frac{y+6}{3}=\frac{z-1}{-1}\)
Solution:
(b) \(\frac{x-4}{1}=\frac{y+6}{3}=\frac{z-1}{-1}\)

Question 10.
What is the image of the point (-2, 3, -5) with respect to the zx-plane?
(a) (-2, -3, -5)
(b) (2, -3, -5)
(c) (2, 3, 5)
(d) (2, 3, -5)
Solution:
(a) (-2, -3, -5)

Question 11.
If \(|\vec{a} \times \vec{b}|^2=k|\vec{a} \cdot \vec{b}|^2+|\vec{a}|^2|\vec{b}|^2\), then what is the value of k?
(a) 1
(b) 2
(c) 0
(d) -1
Solution:
(d) -1

Question 12.
Write the value of m and n for which the vectors (m – 1) î + (n + 2) ĵ + 4k̂ and (m + 1) î + (n – 2) ĵ + 8k̂ will be parallel.
(a) m = 3 and n = 6
(b) m = -3 and n = 6
(c) m = -3 and n = -6
(d) m = 3 and n = -6
Solution:
(d) m = 3 and n = -6

Question 13.
For what value of λ the vectors λî + 3ĵ + λk̂ and λî – 2ĵ +k̂ are perpendicular to each other.
(a) (-2, -3)
(b) (2, 3)
(c) (2, -3)
(d) (-2, 3)
Solution:
(c) (2, -3)

Question 14.
If |x| = 1, |y| = 2 and |z| = 3, then how many points in R³ are there having coordinates (x, y, z)?
(a) 8
(b) 2
(c) 6
(d) 4
Solution:
(a) 8

Question 15.
Write the equation of the plane passing through the point (1, -2, 3) and perpendicular to the y-axis.
(a) y – 2 = 0
(b) y = -2
(c) y + 2 = 0
(d) 2y – 2 = 0
Solution:
(c) y + 2 = 0

Question 16.
What is the projection of î + ĵ – k̂ upon the vector î?
(a) 0
(b) 1
(c) 2
(d) None of the above
Solution:
(b) 1

Question 17.
If \(\vec{a}=2 \hat{i}+\hat{j}, \vec{b}=\hat{k}\) what is \(\vec{a} \cdot \vec{b}\)?
(a) 0
(b) 1
(c) 2
(d) None of the above
Solution:
(a) 0

Question 18.
What is the work done by a force \(\overrightarrow{\mathrm{F}}\) = 4i + 2j + 3k in displacing a particle from A (1, 2, 0) to B (2, -1, 3)?
(a) 5
(b) 6
(c) 7
(d) 8
Solution:
(c) 7

Question 19.
If \(\vec{a} \times \vec{b}=\hat{n}\) then what is the angle between \(\vec{a} \text { and } \vec{b}\)?
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{3}\)
(d) None of the above
Solution:
(b) \(\frac{\pi}{2}\)

Question 20.
The plane 2x + 3z = 5 is parallel to:
(a) x-axis
(b) y-axis
(c) z-axis
(d) line x = y = z
Solution:
(b) y-axis

Question 21.
The equation of the plane containing the points (1, 0, 0), (0, 2, 0) and (0, 0, 3) is given by:
(a) x + 2y + 3z = 1
(b) 3x + 2y + z = 2
(c) 6x + 3y + 2z = 6
(d) 6x + 3y + 2z = 8
Solution:
(c) 6x + 3y + 2z = 6

Question 22.
If on action of force f = 2i + j – k, a particle displaced from A (0, 1, 2) to B (-2, 3, 0) then what is the work done by the force?
(a) 1
(b) 2
(c) 0
(d) None of the above
Solution:
(c) 0

Question 23.
If \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors and \(\vec{a}+\vec{b}+\vec{c}=0\) then evaluate \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\).
(a) \(\frac{3}{2}\)
(b) –\(\frac{3}{2}\)
(c) \(\frac{2}{3}\)
(d) –\(\frac{2}{3}\)
Solution:
(b) –\(\frac{3}{2}\)

Question 24.
What is the value of (î + ĵ) × (ĵ + k̂) × (k̂ + î)?
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(c) 2

Question 25.
Write the distance of the point of intersection of the plane, ax + by + cz + d = 0 and the z-axis from the origin.
(a) \(\left|\frac{d}{c}\right|\)
(b) \(\left|\frac{d}{a}\right|\)
(c) \(\left|\frac{a}{c}\right|\)
(d) \(\left|\frac{a}{d}\right|\)
Solution:
(a) \(\left|\frac{d}{c}\right|\)

Question 26.
Write down the equation of the plane through (0, 0, 0) perpendicular to the line joining (0, 0, 1) and (0, 0, -1).
(a) x = 0
(b) z = 0
(c) y = 0
(d) None of the above
Solution:
(b) z = 0

Question 27.
What is the distance of the point (1, 1, 1) from the plane y = x?
(a) 0
(b) 1
(c) -1
(d) None of the above
Solution:
(a) 0

Question 28.
What is the angle between the planes y + x = 0 and z = 0?
(a) 45°
(b) 60°
(c) 75°
(d) 90°
Solution:
(d) 90°

Question 29.
For what ‘k’ the line \(\frac{x-3}{2}=\frac{y+k}{-1}=\frac{z+1}{-5}\) lies on the plane 2x – y + z – 7 = 0.
(a) 0
(b) 1
(c) -1
(d) 2
Solution:
(d) 2

Question 30.
Projection of the line segment joining (1, 3, -1) and (3, 2, 4) on z-axis is
(a) 4
(b) 5
(c) 3
(d) 2
Solution:
(b) 5

Question 31.
The image of the point (6, 3, -4) with respect to yz-plane is ______.
(a) (-6, 3, 4)
(b) (6, 3, -4)
(c) (-6, 3, -4)
(d) (-6, -3, -4)
Solution:
(c) (-6, 3, -4)

Question 32.
Find the equation of a plane through (1, 1, 2) and parallel to x + y + z – 1 = 0
(a) x + y – z + 4 = 0
(b) x + y – z – 4 = 0
(c) x + y + z – 4 = 0
(d) x + y + z + 4 = 0
Solution:
(c) x + y + z – 4 = 0

Question 33.
The distance between the parallel planes 2x – 3y + 6z + 1 = 0 and 4x – 6y + 12z – 5 = 0 is ______.
(a) \(\frac{1}{4}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{2}\)
(d) None
Solution:
(c) \(\frac{1}{2}\)

Question 34.
Find k if the normal to the plane parallel to the line joining (-1, 1, -4) and (5, 6, -2) has d.rs (3, -2, k).
(a) 1
(b) -1
(c) 2
(d) -2
Solution:
(b) -1

Question 35.
If a line makes angles 35° and 55° with x-axis and y-axis respectively, then the angle which this line subtends with z-axis is:-
(a) 35°
(b) 45°
(c) 55°
(d) 90°
Solution:
(d) 90°

Question 36.
Write the equation of the plane passing through (3, -6, -9) and parallel to xz-plane.
(a) z = -5
(b) z = -9
(c) z = 9
(d) z = -7
Solution:
(b) z = -9

Question 37.
In which condition x + y + z = α + β + γ will contain the line \(\frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}\).
(a) l + m – n = 0
(b) l – m – n = 0
(c) l – m + n = 0
(d) l + m + n = 0
Solution:
(d) l + m + n = 0

Question 38.
The angle between the planes x + y + 1 = 0 and y + z + 1 = 0 is ______.
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Solution:
(c) 60°

Question 39.
Find the number of points (x, y, z) in space other than the point (1, -2, 3) such that |x| = 1, |y|= 2, |z| = 3.
(a) 2
(b) 3
(c) 5
(d) 7
Solution:
(d) 7

Question 40.
Write the ratio in which the line segment joining the points (1, 2, -2) and (4, 3, 4) is divided by the xy-plane.
(a) 1:2
(b) 3:4
(c) 2:3
(d) 2:5
Solution:
(a) 1:2

(B) Very Short Type Questions With Answers

Question 1.
Write the value of k such that the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies on the plane 2x – 4y + z = 7.
Solution:
The line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies on the plane 2x – 4y + z = 7.

Question 2.
Write the equations of the line 2x + z – 4 = 0 = 2y + z in the symmetrical form.
Solution:
Given line is 2x + z – 4 = 0 = 2y + z
⇒ z = -(2x – 4) = -2(x – 2) and z = -2y
∴ -2(x – 2) = -2y = z
∴ The equation of the line in symmetrical form is \(\frac{x-2}{1}=\frac{y}{1}=\frac{z}{-2}\).

Question 3.
Write the distance between parallel planes 2x – y + 3z = 4 and 2x – y + 3z = 18
Solution:
Distance between the given parallel planes
= \(\frac{|18-4|}{\sqrt{4+1+9}}=\frac{14}{\sqrt{14}}=\sqrt{14}\)

Question 4.
Write down the equation to the plane perpendicular to the y-axis at the point (0, -2, 0).
Solution:
The equation of the plane perpendicular to y-axis at (0, -2, 0) is
(x – 0) . 0 + (y + 2) . 1 + (z – 0) . 0 = 0
⇒ y + 2 = 0

Question 5.
Under which conditions the straight line \(\frac{\mathrm{x}-\mathrm{a}}{l}=\frac{\mathrm{y}-\mathrm{b}}{\mathrm{m}}=\frac{\mathrm{z}-\mathrm{c}}{\mathrm{n}}\) intersects the plane Ax + By + Cz = 0 at a point other than (a, b, c)?
Solution:
The line \(\frac{\mathrm{x}-\mathrm{a}}{l}=\frac{\mathrm{y}-\mathrm{b}}{\mathrm{m}}=\frac{\mathrm{z}-\mathrm{c}}{\mathrm{n}}\) will intersect the plane Ax + By + Cz + D = 0 at a point other than (a, b, c) if Al + Bm + Cn ≠ 0 and Aa + Bb + Cc + D ≠ 0.

Question 6.
How many straight lines in space through the origin are equally inclined to the coordinate axes?
Solution:
There are two lines in space through origin which are equally inclined to coordinate axes.

Question 7.
Write the value of a if the vectors \(\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k} \text { and } \vec{b}=\alpha \hat{i}-\hat{j}+2 \hat{k}\) are parallel.
Solution:

Question 8.
Write the equation of the line passing through the point (4, -6, 1) and parallel to the line \(\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-1}{-1}\).
Solution:
Equation of the line passing through (4, -6, 1) and parallel to
\(\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-1}{-1} \text { is } \frac{x-4}{1}=\frac{y+6}{3}=\frac{z-1}{-1}\).

Question 9.
What is the image of the point (-2, 3, -5) with respect to the zx-plane?
Solution:
Image of the point (-2, 3, -5) w.r.t. zx-plane is (-2, -3, -5)

Question 10.
What is the point of intersection of the line x = y = z with the plane x + 2y + 3z = 6?
Solution:
Given line is x = y = z = λ (say)
Any point on this line has coordinates (λ, λ, λ).
Putting x = y = z = λ in x + 2y + 3z = 6
We get 6λ = 6 ⇒ λ = 1
∴ The point of interesetion is (1, 1, 1).

Question 11.
How many directions a null vector has?
Solution:
A null vector has infinitely many directions (arbitrary direction).

Question 12.
For what value of λ the vectors λî + 3ĵ + λk̂ and λî – 2ĵ + k̂ are perpendicular to each other?
Solution:
Given vectors are perpendicular to each other iff λ² – 6 + λ = 0.
⇒ λ² + 3λ – 2λ- 6 = 0
⇒ λ (λ + 3) – 2 (λ + 3) = 0
⇒ (λ – 2) (λ + 3) = 0
⇒ λ = 2, λ = (-3)

Question 13.
If |x| = 1, |y| = 2 and |z| = 3, then how many points in R³ are there having coordinates (x, y, z)?
Solution:
Required number of points are ‘8’.

Question 14.
Write the equation of the plane passing through the point (1, -2, 3) and perpendicular to the y-axis.
Solution:
D.rs. of any line parallel to y-axis are < 0, 1, 0 >.
∴ The equation of required plane is
(x – 1) 0 + (y + 2) . 1 + (z – 3) . 0 = 0
⇒ y + 2 = 0

(C) Short Type Questions With Answers

Question 1.
Find the point where the line \(\frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{2}\) meets the plane 2x + y + z = 2.
Solution:
Equation of the line is \(\frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{2}\)
Co-ordinates of any point on the line are (k + 2, -k, 2k + 1)
This point lies on the plane
⇒ 2 (k + 2) + (-k) + (2k + 1) = 2
⇒ 3k + 5 = 2 ⇒ k = -1
⇒ The required point of intersection is (1, 2, -1)

Question 2.
If the sum of two unit vectors is a unit vector, show that the magnitude of their difference is √3.
Solution:

Question 3.
The position vectors of two points A and B are 3î + ĵ + 2k̂ and î – 2ĵ – 4k̂ respectively. Find the equation of the plane passing through B and perpendicular to \(\overrightarrow{AB}\).
Solution:

Question 4.
Find the equation of the plane through the point (2, 1, 0) and passing through the intersection of the planes 3x – 2y + z – 1 = 0 and x – 2y + 3z = 1.
Solution:
Equation of any plane passing through the intersection of given two planes is:
(3x – y + z – 1) + λ (x – 2y + 3z – 1) = 0
⇒ x (3 + 2λ) + y (-1 – 2λ) + z (1 + 3λ) – 1 – λ = 0
This plane passes through A (2, 1, 0)
⇒ 2 (3 + λ) – 1 – 2λ – 1 – λ = 0
⇒ 6 + 2λ – 2 – 3λ = 0
⇒ λ = 4
∴ Equation of the required plane is 7x – 9y + 13z – 5 = 0

Question 5.
Prove that the vectors 2î – ĵ + k̂, î – 3ĵ – 5k̂, 3î – 4ĵ – 4k̂ are the sides of a right angled triangle.
Solution:

Question 6.
Prove that \(|\overrightarrow{a}+\overrightarrow{b}| \leq|\overrightarrow{a}|+|\overrightarrow{b}|\). Write when equality will hold
Solution:

Question 7.
The projections of a line segment \(\overline{OP}\), through the origin O on the coordinate axes are 6, 2, 3. Find the length of the line segment \(\overline{OP}\) and its direction cosines.
Solution:

Question 8.
Prove that the lines \(\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}\) and 3x – 2y + z + 5 = 0 = 2x + 3y + 4z – 4 are coplanar.
Solution:
Given lines are
\(\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}\) = r (say) … (1)
and 3x – 2y + z + 5 = 0
2x + 3y + 4z – 4 = 0 … (2)
Two lines are coplanar if either they are parallel or intersecting coordinates of any point on the line (1) are P (3r – 4, -5r – 6, -2r + 1) putting in the equations (2) we get.
3 (3r – 4) – 2 (5r – 6) + (-2r + 1) + 5 = 0
and 2 (3r – 4) + 3 (5r – 6) + 4 (-2r + 1) -4 = 0
⇒ -3r + 6 = 0 and 13r – 26 = 0
⇒ r = 2 and r = 2
Thus two lines are intersecting.
⇒ The lines are co-planar.

Question 9.
If the sun of two unit vectors is a unit vector then find the magnitude of the difference.
Solution:

Question 10.
Find the equation of a plane parallel to the plane 2x – y + 3z + 1 = 0 and at a distance of 3 units away from it.
Solution:
Any plane parallel to 2x – y + 3z + 1 = 0 … (1)
has equation 2x – y + 3z + λ = 0.
Distance between two parallel planes (1) and (2) is 3 units.
Equation of required plane is.

Question 11.
Show that the poiints (3, -2, 4), (1, 1, 1) and (-1, 4, -2) are collinear.
Solution:
Let the given points are A (3, -2, 4), B (1, 1, 1) and C (-1, 4, -2).

The points A, B and C are collinear.

Question 12.
Determine the value of m for which the following vectors are orthogonal:
(m + 1) j + m²ĵ – mk̂, (m² – m + 1) î – mĵ + k̂
Solution:

Question 13.
Find the equation of the plane passing through the line x = y = z and the point (3, 2, 1).
Solution:
Equation of any plane through the line is (x – y) + λ (y – z) = 0 … (1)
Since the point (3, 2, 1) is on the plane, so it satisfies (1)
i.e., ( 3 – 2) + λ ( 2 – 1) = 0 ⇒ 1 + λ = 0
⇒ λ = -1
Using the value of λ in (1) we get
x – y + (-y) – z = 0 ⇒ x – 2y + z = 0 is the required equation of the plane.

Question 14.
Find the scalar projection of the vector \(\overrightarrow{a}=3 \hat{i}+6 \hat{j}+9 \hat{k} \text { on } \overrightarrow{b}=2 \hat{i}+2 \hat{j}-\hat{k}\).
Solution:

Question 15.
Prove that the straight line \(\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{1}\) lies on the plane 7x + 5y + z = 0.
Solution:
Given line is \(\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{1}\) lies on the plane 7x + 5y + z = 0.
As 7 × 1 + 5 × (-2) + 3 = 0 the point (1,-2, 3) lies on the plane.
D.rs. of the given line = < 2, -3, 1 > and
d.rs. of the normal to the plane are < 7, 5, 1 >
As 2 × 7 + (-3) × 5 + 1 × 1 = 0
The given line is parallel to the given plane.
Hence the given line lies in the plane.

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 13 Three Dimensional Geometry Ex 13(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 13 Three Dimensional Geometry Exercise 13(c)

Question 1.
State which of the following statements are true (T) or false (F):
(a) The line \(\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-1}{2}\) pass through the origin.
Solution:
True

(b) The lines \(\frac{x+2}{-k}=\frac{y-3}{k}=\frac{z+4}{k}\) and \(\frac{x-4}{-4}=\frac{y-3}{k}=\frac{z+1}{2}\) are perpendicular every value of k.
Solution:
True

(c) The line \(\frac{x+5}{-2}=\frac{y-3}{1}=\frac{z-2}{3}\) lies on the plane x – y + z + 1 = 0
Solution:
False

(d) The line \(\frac{x-2}{3}=\frac{1-y}{4}=\frac{5-z}{1}\) is parallel to the plane 2x – y – 2z = 0
Solution:
False

(e) The line \(\frac{x+3}{-1}=\frac{y-2}{3}=\frac{z-1}{4}\) is perpendicular to the plane 3x – 3y + 3z – 1 = 0
Solution:
False

Question 2.
Fill in the blanks by choosing the correct alternative from the given ones:
(a)

[parallel, perpendicular, coincident]
Solution:
perpendicular.

(b) The line passing through (-1, 0, 1) and perpendicular to the plane x + 2y + 1 = 0 is _____.

Solution:
\(\frac{x+1}{1}=\frac{y}{2}=\frac{z-1}{0}\)

(c) The line \(\frac{x+1}{2}=\frac{y-6}{1}=\frac{z-4}{0}\) is _____. [parallel to x-axis, perpendicular to y-axis, perpendicular to z-axis]
Solution:
perpendicular to. z-axis

(d) If the line \(\frac{x-3}{2}=\frac{y+k}{-1}=\frac{z+1}{-5}\) lies on the plane 2x – y + z – 7 = 0; then k = -(2, -1, -2)
Solution:
2

(e) If l, m, n be d.cs. of a line, then the line is perpendicular to the plane x – 3y + 2z + 1 = 0 if _____. [(i) l = 1, m = -3, n = 2, (ii) \(\frac{l}{1}=\frac{m}{-3}=\frac{n}{2}\) , (iii) l – 3m + 2n = 0]
Solution:
\(\frac{l}{1}=\frac{m}{-3}=\frac{n}{2}\)

Question 3.
Find the equation of lines joining the points.
(i) (4, -6, 1) and (0, 3, -1)
Solution:
Equation of the line joining the points (4, -6, 1) and (0, 3, -1) is

(ii) (a, a, a) and (a, 0, a)
Solution:
Equation of the line joining the points (a, a, a) and (a, 0, a) is

(iii) (2, 1, 3) and (4, -2, 5).
Solution:
Equation of the line joining the points (2, 1, 3) and (4, -2, 5) is

Question 4.
Write the symmetric form of equation of the following lines:
(i) x-axis
Solution:
D.cs. of x-axis are < 1, 0, 0 >.
x-axis passes through the origin.
So the equation of x-axis in symmetrical form is
\(\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}\)

(ii) y = b, z = c
Solution:
Given line in unsymmetrical from is y = b, z = c
⇒ y – b = 0 and z – c = 0.
The straight line is parallel to x-axis.
D.rs. of the straight line are < 0, 0, k >.
So the equation of the line is
\(\frac{y-b}{0}=\frac{z-c}{0}=\frac{x}{k}\)

(iii) ax + by + d = 0, 5z = 0
Solution:
Given lines ax + by + d = 0, 5z = 0

(iv) x – 2y = 3, 2x + y – 5z = 0;
Solution:
Given straight line is x – 2y – 3 = 0 … (1)
and 2x + y – 5z = 0 … (2)
Putting z = 0 in (1) and (2) we get x – 2y – 3 = 0 and 2x + y = 0.
Solving we get y = -2x and x + 4x – 3 = 0

(v) 4x + 4y – 5z – 12 = 0, 8x + 12y – 13z = 32;
Solution:
Given straight line in unsymmetrical form is
4x + 4y – 5z – 12 = 0 … (1)
8x + 12y – 13z – 32 = 0 … (2)
Putting z = 0 in (1) and (2) we get
4x + 4y – 12 = 0 ⇒ x + y – 3 =0
8x + 12y – 32 = 0 ⇒ 2x + 3y – 8 = 0
Solving we get

(vi) 3x – 2y + z = 1, 5x + 4y – 6z = 2
Solution:
Given straight line is
3x – 2y + z – 1 = 0 … (1)
and 5x – 4y + 6z – 2 = 0 … (2)
Putting x = 0 in (1) and (2) we get

Question 5.
(a) Obtain the equation of the line through the point (1, 2, 3) and parallel to the line x – y + 2z – 5 = 0, 3x + y + z = -6
Solution:
Equation of the straight line through

(b) Find the equation of the line through the point (3, -1, 2) and parallel to the planes x + y + 2z – 4 = 0 and 2x – 3y + z + 3 = 0
Solution:
Equation of the straight line through the point (3, -1, 2) is

Question 7.
(a) Show that the line passing through the points (a₁, b₁, c₁) and (a₂, b₂, c₂) passes through the origin, if a₁a₂ + b₁b₂ + c₁c₂ = p₁p₂, where p₁ and p₂ are distances of the points from origin.
Solution:
The equation of the line passing through the points (a₁, b₁, c₁) and (a₂, b₂, c₂) is

(b) Prove that the lines x = az + b, y = cz + d and x = a₁z + b₁, y = c₁z + d₁ are perpendicular if aa₁ + cc₁ + 1 = 0.
Solution:
Given lines are
x – az – b = 0 = y – cz – d … (1)
and x – a₁z – b₁ = 0 and < l₂, m₂, n₂ > … (2)
Let < l₁, m₁, n₁ > and < l₂, m₂, n₂ > be the d.cs. of the lines (1) and(2)

D.rs. of the two lines are < a, c, 1 > and < a₁, c₁, 1 >.
If the lines are perpendicular then the sum of product of d.rs. is zero.
So aa₁ + cc₁ + 1 = 0 (Proved)

Question 8.
Find the points of intersection of the line \(\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-1}{-1}\) and the plane 2x + y + a = 9.
Solution:
Given line is \(\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-1}{-1}\) = r (say)
∴ Any point on (1) is (r + 1, 3r – 2, -r + 1)
If it lies on the plane 2x + y + z – 9 = 0 then
2 (r + 1) + 3r – 2 + (-r + 1) – 9 = 0
or, 2r + 3r – r + 2 – 2 + 1 – 9 = 0
or, 4r = 8, or, r = 2
∴ The point of intersection is (3, 4, -1).

Question 9.
Find the coordinates of the point where the line joining (3, 4, -5) and (2, -3, 1), meets the plane 2x + y + z – 7 = 0.
Solution:
The straight line joining the points (3, 4, -5) and (2, -3, 1) is
\(\frac{x-3}{2-3}=\frac{y-4}{-3-4}=\frac{z+5}{1+5}\)
or, \(\frac{x-3}{-1}=\frac{y-4}{-7}=\frac{z+5}{6}\) = r (say)
Any point on the line is
(-r + 3, -7r + 4, 6r – 5)
If this point is the point of intersection of the line with the plane 2x + y + z – 7 = 0 then
2 (-r + 3) + (-7r + 4) + 6r – 5 – 7 = 0
or, -2r + 6 – 7r + 4 + 6r – 12 = 0
or, -3r = 2 or, r = \(\frac{-2}{3}\)
∴ The point of intersection is
\(\left(\frac{11}{3}, \frac{26}{3}, \frac{-27}{3}\right) \text { i.e., }\left(\frac{11}{3}, \frac{26}{3},-9\right)\).

Question 10.
(a) Find the distance of the point (-1, -5, -10) from the point of intersection of the line \(\frac{x-2}{2}=\frac{y+1}{4}=\frac{z-2}{12}\) and the plane x – y + z = 5.
Solution:
Given line is
\(\frac{x-2}{2}=\frac{y+1}{4}=\frac{z-2}{12}\) = z
Any point on (1) is (2r + 2, 4r – 1, 12r + 2).
If this point is the point of intersection of the line with the plane
x – y + z = 5
then 2r + 2 – 4r + 1 + 12r + 2 = 5
or, 10r = 0, or, r = 0
The point of intersection is (2, -1, 2).
Distance between the points (-1, -5, -10) and the point of intersection of the given line and plane.
= \(\sqrt{9+16+144}=\sqrt{169}\) = 13

(b) Find the image of the point (2, -1, 3) in the plane 3x – 2y + z – 9 = 0
Solution:

Let A = (2, -1, 3)
Let B be the image point of A with respect to the plane
3x – 2y + z – 9 = 0 … (1)
Then AB is normal to the plane.
Again let C be the point of intersection of the line with the plane (1).
Again as B is the image point of A then C must be the mid-point of AB.
Now d.rs. of AB are < 3, -2, 1 > because AB is perpendicular to the plane.
Eqn. of the line AB is

Question 11.
Prove that the lines, \(\frac{x+3}{2}=\frac{y+5}{3}=\frac{z-7}{-3}\) and \(\frac{x+1}{4}=\frac{y+1}{5}=\frac{z+1}{-1}\) are coplanar.
Find the equation of the plane containing them.
Solution:
Given lines are

Two lines are co-planar if either they are parallel or intersecting. Now the line (1) and (2) are not parallel because their d.rs. are not proportional. So we shall show that they are intersecting.
Any point on (1) is (2r₁ – 3, 3r₁ – 5, -3r₁ + 7)
Any point on (2) is (4r₂ – 1, 5r₂ – 1, -r₂ – 1)
If the lines are intersecting then for some values of r₁ and r₂.
2r₁ – 3 = 4r₂ – 1 … (3)
3r₁ – 5 = 5r₂ – 1 … (4)
-3r₁ + 7 = -r₂ – 1 … (5)
From (3) we get r₁ = \(\) = 2r₂ + 1
Putting it in (4) we get 6r₂ + 3 – 5 = 5r₂ – 1
or, r₂ = 1. Again r₁ = 3
With these values r₁ = 3, r₂ = 1
We see that eqn. (5) is satisfied. So the straight lines are intersecting. Hence they are coplanar.
The equation of the plane containing the line (1) is:
a (x + 3) + b (y + 5) + c (z – 7) = 0 … (6)
where 2a + 3b – 3c = 0 … (6)
If the plane (6) contains the line (2) then
4a + 5b – c = 0 … (8)
Solving (7) and (8) we get

Equation of the plane is
6 (x + 3) – 5 (y + 5) – 1 (z – 7) = 0
or, 6x – 5y – z = 0

Question 12.
Prove that the lines \(\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}\) and 3x – 2y + z + 5 = 0 = 2x + 3y + 4z – 4 are co-planar.
Solution:
Given lines are
\(\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}\) = r (say) … (1)
and 3x – 2y + z + 5 = 0
= 2x + 3y + 4z – 4 … (2)
Any point on (1) and (2) are coplanar if they are either parallel or intersecting. If the line are intersecting then
3 (3r – 4) – 2 (5r – 6) + (-2r + 1) + 5 = 0 … (3)
and 2 (3r – 4) + 3 (5r – 6) + 4 (-2r + 1) – 4 = 0 … (4) are consistent.
Solving (3) we get
9r – 12 – 10r + 12 – 2r + 1 + 5 = 0
or, -3r + 6 = 0 or, r = 2
For r = 2, eqn (4) is satisfied. Thus the lines are intersecting and hence they are co-planar. (Proved)

Question 13.
Show that the lines 7x – 4y + 7z + 16 = 0 = 4x + 3y – 2z + 3 and x – 3y + 4z + 6 = z – y + z + 1 intersect. Find the coordinates of their point of intersection and equation of the plane containing them.
Solution:
Let < l₁, m₁, n₁ > and < l₂, m₂, n₂ > be the d.cs. of the lines.
7x – 4y + 7z + 16 = 0 = 4x + 3y – 2z + 3 … (1)
and x – 3y + 4z + 6 = 0 = x – y + z + 1 … (2)
Then 7l₁ – 4m₁ + 7n₁ = 0
4l₁ + 3m₁ – 2n₁ = 0

Question 14.
Show that the line joining the points (0, 2, -4) and (-1 , 1, -2) and the lines joining the points (-2, 3, 3) and (-3, -2, 1) are co-planar. Find their point of intersection.
Solution:
The eqn. of the line joining the points (0, 2, -4) and (-1, 1, -2) is

The lines (1) and (2) are coplanar if either they are parallel or intersecting. These lines are not parallel. So we have to prove that they are intersecting.
Any point on (1) is (-r₁, -r₁ + 2, 2r₁ – 4)
Any point on (2) is (-r₂ – 2, -5r₂ + 3, -2r₂ + 3)
If two lines are intersecting then for some r₁ and r₂
-r₁ = -r₂ – 2 … (1)
-r₁ + 2 = -5r₂ + 3 … (2)
2r₁ – 4 = -2r₁ + 3 … (3)
From (3), r₁ = r₂ + 2

Question 15.
Show that the lines x – mz – a = 0 = y – nz – b and x – m’z’ – a’ = 0 = y – n’z’ – b’ intersect, if (a – a’) (n – n’) = (b – b’) (m – m’).
Solution:
Given lines are
x – mz – a = 0 = y – nz – b … (1)
and x – m’z’ – a’ = 0 = y – n’z’ – b … (2)
Putting z = 0 in (1) we get x = a and y = b.
So (a, b, 0) is a point on (1).
Again putting z = 0 in (2) we get x = a’, y = b’.
So (a’, b’, 0) is a point on (2).
Let < L₁, M₁, N₁ > and < L₂, M₂, N₂ > be the d.cs. of the lines (1) and (2).
Then L₁ – mN₁ = 0, M₁ – nN₁ = 0

or, n (a’ – a) + b (m – m’)
= n’ (a’ – a) + b (m – m’)
⇒ (a – a’) (n – n’) = (b – b’) (m – m’) (Proved)

Question 16.
Proved that the line \(\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{1}\) lies on the plane 7x + 5y + z = 0
Solution:
Given line is
\(\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{1}\) = r (say) … (1)
Any point on (1) is (2r + 1, -3r – 2, r + 3)
The straight line lies on the plane 7x + 5y + z = 0 … (2)
if every point of the line lies on (2).
Now 7 × (2r + 1) + 5 (-3r -2) + (r + 3)
= 14r + 7 – 15r – 10 + r + 3 = 0
Thus the point (2r + 1, -3r – 2, r + 3) lies on (2).
Hence the straight line lies on the plane. (Proved)

Question 17.
(a) Find the angle between the plane x + y + 4 = 0 and the line \(\frac{x+3}{2}=\frac{y-1}{1}=\frac{z+4}{-2}\).
Solution:

(b) Find the angle between the plane 4x + 3y + 5z – 1 = 0 and the line \(\frac{x+3}{2}=\frac{y-1}{3}=\frac{z+4}{6}\).
Solution:
Given plane and the line have equations
4x + 3y + 5z – 1 = 0 … (1)

Question 18.
(a) Find the equation of the line passing through the point (1, 0, -1) and intersecting the lines x = 2y = 2z; 3x + 4y – 1 = 0 = 4x + 5z – 2.
Solution:
Given lines are x = 2y = 2z … (1)
and 3x + 4y – 1 = 0 = 4x + 5z – 2 … (2)
Any plane containing the line (1)
i.e., x – 2y = 0 and y – z = 0 is
x – 2y + k₁ (y – z)= 0 = 0
or, x + (k₁ – 2) y – k₁z = 0
If it passes through the point (1, 0, -1) then 1 + k₁ = 0 or, k₁ = -1
The plane containing the line (1) and passing through the point (1, 0, -1) is x – 3y + z = 0
Again any plane containing the line (2) is
3x + 4y – 1 + k₂ (4x + 5z – 2) = 0
or, (3 + 4k₂) x + 4y + 5k₂z – (2k₂ + 1) = 0
If it passes through the point (1, 0, -1) then
3 + 4k₂ – 5k₂ – 2k₂ – 1 = 0
or, 3k₂ = 2 or, k₂ = \(\frac{2}{3}\)
The equation of the plane through the line (2) and passing through (1, 0, -1) is

(b) A line with direction ratios < 2, 1, 2 > meets each of the lines x = y + a = z and x + a = 2y = 2z. Find the coordinates of the points of intersection.
Solution:
Given lines are
x = y + a = z and x + a = 2y = 2z
These can be written in symmetrical form as
\(\frac{x}{1}=\frac{y+a}{1}=\frac{z}{1}\) = r₁ (say)
and \(\frac{x+a}{2}=\frac{y}{1}=\frac{z}{1}\) = r₂ (say)
Any point on (1) is (r₁, r₁ – a, r₁)
Any point on (2) is (2r₂ – a, r₂, r₂)
Suppose that the line meets the lines (1) and (2) at P and Q respectively.
Let P = (r₁, r₁ – a, r₁), Q = (2r₂ – a, r₂, r₂)
D.rs. of PQ are
< r₁ – 2r₂ + a, r₁ – r₂ – a, r₁ – r₂ >
But given that d.rs. are < 2, 1, 2 >.
So \(\frac{r_1-2 r_2+a}{2}=\frac{r_1-r_2-a}{1}=\frac{r_1-r_2}{2}\)
From the 1st two ratios we get
r₁ – 2r₂ + a = 2r₁ – 2r₂ – 2a ⇒ r₁ = 3a
From the last two ratios we get
2r₁ – 2r₂ – 2a = r₁ – r₂
⇒ r₁ = r₂ + 2a ⇒ 3a = r₂ + 2a ⇒ r₂ = a
∴ P (3a, 2a, 3a), Q = (a, a, a)
∴ The co-prdinates of the points of intersection and (3a, 2a, 3a) and (a, a, a).

Question 19.
Obtain the co-ordinates of the foot of the perpendicular drawn from the point (3, -1, 11) to the line \(\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}\). Obtain the equation of the perpendicular also.
Solution:

Let P be the given point (3, -1, 11).
Draw PM perpendicular from P onto the straight line.
Any point on (1) is (2λ, 3λ + 2, 4λ + 3)
Let M = (2λ, 3λ + 2, 4λ + 3)
D.rs. of PM are 2λ – 2, 3λ + 3, 4λ – 8
As PM is perpendicular to the line then
(2λ – 3) . 2 +3 (3λ + 3) + (4λ – 8) . 4 = 0
or, 4λ – 6 + 9λ + 9 + 16λ – 32 = 0
or, 29λ – 29 = 0 or, λ = 1
∴ The foot of the perpendicular is (2, 5, 7)
Equation of the perpendicular line is
\(\frac{x-3}{2-3}=\frac{y+1}{5+1}=\frac{z-11}{7-11}\)
or, \(\frac{x-3}{1}=\frac{y+1}{-6}=\frac{z-11}{4}\)

Question 20.
Find the perpendicular distance of the point (-1, 3, 9) from the line \(\frac{x-13}{5}=\frac{y+8}{-8}=\frac{z-31}{1}\).
Solution:
Let P be the point (-1, 3, 9).
Suppose that M is the foot of the perpendicular drawn from P onto the straight line.
\(\frac{x-13}{5}=\frac{y+8}{-8}=\frac{z-31}{1}\) = λ (say)
Let M = (5λ + 13, -8λ – 8, λ + 31)
D.rs. of PM are
< 5λ + 14, -8λ – 11, λ + 22 >
As PM is perpendicular to the line (1) then
5 (5λ + 14) – 8 (-8λ – 11) + λ + 22 = 0
or, 25λ + 64λ + λ + 70 + 88 + 22 = 0
or, 90λ = -180 or, λ = -2
Thus M = (3, 8, 29)
Distance PM = \(\sqrt{(3+1)^2+(8-3)^2+(29-9)^2}\)
= \(\sqrt{16+25+400}\) = \(\sqrt{441}\) = 21

Question 21.
Find the distance of the point (1, -2, 3) from the plane x – y + z = 5, measured parallel to the line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}\).
Solution:
Let P be the point (1, 2, 3). Draw the straight line PM parallel to the line

Question 22.
Find the distance of the point (1, -1, -10) from the line \(\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\) measured parallel to the line \(\frac{x+2}{2}=\frac{y-3}{-3}=\frac{z-4}{8}\).
Ans.
Let P be the point (1, -1, -10). Equation of the line through P parallel to the line

∴ Distance of the point (1, -1, -10) from the given line is √308.

Question 23.
Find the equation of plane through the point (2, 0, -3) and containing the line 3x + y + z – 5 = 0 = x – 2y + 4z + 4
Solution:
Any plane containing the line
3x + y + z – 5 = 0 = x – 2y + 4z + 4
3x + y + z – 5 + k (x – 2y + 4z + 4) = 0
or, (3 + k) x + (1 – 2k) y + (1 + 4k) z + (4k – 5) = 0 … (1)
If this plane contains the point (2, 0, -3) then
2 (3 + k)² + (1 – 2k) . 0 + (1 + 4k) . (-3) + 4k – 5 = 0
⇒ 6 + 2k – 3 – 12k + 4k – 5 = 0
⇒ -6k – 2 = 0 ⇒ k = –\(\frac{1}{3}\)
Required plane is
\(\left(3-\frac{1}{3}\right) x+\left(1+\frac{2}{3}\right) y+\left(1-\frac{4}{3}\right) z+\left(-\frac{4}{3}-5\right)\) = 0
⇒ 8x + 5y – z – 19 = 0

Question 24.
Find the equation of the plane containing the line x + 2 = 2y – 1 = 3z and parallel to the line x = 1 – 5y = 2z – 7. Also find the shortest distance between the two lines.
Solution:
Given line is x + 2 = 2y – 1 = 3z
[x – 2y + 3 = 0
2y – 3z – 1 = 0] … (1)
Any plane containing the line (1) is
(x – 2y + 3) + k (2y – 3z – 1) = 0
or, x + (2k – 2) y – 3kz + (3 – k) = 0… (2)
Again given that the plane (2) is parallel to the line
x = 1 – 5y = 2z – 7
⇒ \(\frac{x}{10}=\frac{y-\frac{1}{5}}{-2}=\frac{z-\frac{7}{2}}{5}\) … (3)
D.rs. of the line (3) are < 10, -2, 5 >
If the plane (2) is parallel to the line (3) then the normal plane (2) is perpendicular to the line (3). D.rs. of the normal of the plane(2) are
< 1, 2k – 2, -3k >.
Thus 10 – 2 (2k – 2) – 15 k = 0
⇒ 10 – 4k + 4 – 15k = 0

We have to find the shortest distance between the lines (3) and (4). The shortest distance is the line segment perpendicular to both the lines. Let < l, m, n > be the d.cs. of the shortest distance.
Then 6l + 3m + 2n = 0
10l – 2m+ 5n = 0
Solving we get

Question 25.
Find the equation of the two planes through the origin and parallel to the line \(\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}\) and at a distance \(\frac{5}{3}\) from it.
Solution:
Equation of the plane through the origin is ax + by + cz = 0 … (1)
If the plane (1) is parallel to the line

⇒ 9b² = 4a² + 4b² + 4c²
⇒ 4a² – 5b² + 4c² = 0 … (4)
From (3) we get b = 2a – 3c.
Putting it in (4) we get
4a² – 5 × 4 (a – c)² + 4c² = 0
⇒ a² – 5a² – 5c² + 10ac + c² = 0
⇒ -4a² – 4c² + 10ac = 0
⇒ 2a² + 2c² – 5ac = 0
⇒ 2a² + 2c² – 4ac – ac = 0
⇒ 2a (a – 2c) – c (a – 2c) = 0
⇒ (a – 2c) (2a – c) = 0

∴ The plane is x – 2y + 2z = 0
Hence the planes are
2x + 2y + z = 0 and x – 2y + 2z = 0.

Question 26.
Find the equation of the straight line perpendicular to the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{7}\) and lying in the plane x – 2y + 4z – 51 = 0.
Solution:
Let < l, m, n > be the d.cs. of the straight line perpendicular to the line.

D.rs. of the line is < -6, 1, 2 >.
Again let A be the point of intersection of the line with the line (1).
Let A = (3r + 2, 4r – 1, 7r + 6)
Then this point a lies also on the plane (2).
So 3r + 2 – 8r + 2 + 28r + 24 – 51 =0
or, 23r – 23 = 0 or, r = 1
∴ A = (5, 3, 13)
Hence the equation of the required line is
\(\frac{x-5}{-6}=\frac{y-3}{1}=\frac{z-13}{2}\)

Question 27.
Find the shortest distance between the lines \(\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}\) and \(\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}\). Find also the equation of the line of shortest distance.
Solution:
Given lines are

Any point on the line (1) and (2) are P (3 + 3α, 8 – α, 3 + α) and Q (-3 – 3β, -7 + 2β, 6 + 4β) respectively.
D.r.s. of PQ are < 6 + 3α + 3β, 15 – α – 4β >
D.r.s. of the lines are < 3, -1, 1 > and < -3, 2, 4 > respectively.
PQ is perpendicular to the given lines.
∴ 3 (6 + 3α + 3β) – (15 – α – 2β) + 1 (-3 + α – 4β) = 0
and -3 (6 + 3α + 3β) + 2 (15 – α – 2β) + 1 (-3 + α – 4β)
⇒ 18 + 9α + 9β – 15 + α + 2β – 3 + α – 4β = 0
and -18 – 9α – 9β + 30 – 2α – 4β – 12 + 4α – 16β = 0
⇒ 11a + 7b = 0
and -7a – 29b = 0
⇒ a = b = 0
co-ordinates P and Q are (3, 8, 3) and (-3, -7, 6) respectively.
The shortest distance

Question 28.
Show that the shortest distance between the lines x + a = 2y = -12z and x = y + 2a = 6z – 6a is 2a.
Solution:
Given lines are
x + a = 2y = -12z
and x = y + 2a – 6z – 6a

Question 29.
Find the length and equation of the line of shortest distance between the lines 3x – 9y + 5z = 0 = x + y – z and 6x + 8y + 3z – 13 = 0 = x + 2y + z – 3
Solution:
Given lines are
3x – 9y + 5z = 0
x + y – z = 0 … (1)
and 6x + 8y + 3z – 13 = 0
x + 2y + z – 3 = 0 … (2)
Let us consider the line (1)
Now z = x + y
∴ 3x – 9y + 5 ( x + y) = 0
⇒ 8x – 4y = 0 ⇒ 2x = y
Again y = z – x
∴ 3x – 9 (z – x) + 5z = 0
⇒ 12x – 4z = 0
⇒ 3x = z
∴ 6x = 3y = 2z

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 13 Three Dimensional Geometry Ex 13(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 13 Three Dimensional Geometry Exercise 13(b)

Question 1.
State, which of the following statements are true (T) or false (F).
(a) Through any four points one and only one plane can pass.
Solution:
False

(b) The equation of xy-plane is x + y = 0.
Solution:
False

(c) The plane ax + by + c = 0 is perpendicular to z-axis.
Solution:
False

(d) The equation of the plane parallel to xz-plane and passing through (2, -4, 0) is y + 4 = 0.
Solution:
True

(e) The planes 2x – y + z – 1 = 0 and 6x – 3y + 3z = 1 are coincident.
Solution:
False

(f) The planes 2x + 4y – z + 1 = 0 and x – 2y – 6z + 3 = 0 are perpendicular to each other.
Solution:
True

(g) The distance of a point from a plane is same as the distance of the point from any line lying in that plane.
Solution:
False

Question 2.
Fill in the blanks by choosing the appropriate answer from the given ones:
(a) The equation of a plane passing through (1, 1, 2) and parallel to x + y + z – 1 = 0 is ______. [x + y + z = 0, x + y + 2z – 1 = 0, x + y + z = 0, x + y + z = 4]
Solution:
x + y + z = 4

(b) The equation of plane perpendicular to z-axis and passing through (1, -2, 4) is ______. [x = 1, y + 2 = 0, z – 4 = 0, x + y + z – 3 = 0]
Solution:
z – 4 = 0

(c) The distance between the parallel planes 2x – 3y + 6z + 1 = 0 and 4x – 6y + 12z – 5 = 0 is ______. \(\left[\frac{1}{2}, \frac{1}{7}, \frac{4}{7}, \frac{6}{7}\right]\)
Solution:
\(\frac{1}{2}\)

(d) The plane y – z + 1 = 0 is ______. [parallel to x-axis, perpendicular to x-axis, parallel xy-plane, perpendicular to yz-plane].
Solution:
parallel to x-axis

(e) A plane whose normal has direction ratios < 3, -2, k > is parallel to the line joining (-1, 1, -4) and (5, 6, -2). Then the value of k = ______. [6, -4, -1, 0]
Solution:
k = -4

Question 3.
Find the equation of planes passing through the points.
(a) (6, -1, 1), (5, 1, 2) and (1, -5, -4)
Solution:
Equation of the plane through the points (6, -1, 1), (5, 1, 2) and (1, -5, -4) is

⇒ (x – 6) (-10 + 4) – (y + 1) (5 + 5) + (z – 1) (4 + 10) = 0
⇒ -6 (x – 6) – 10 (y + 1) + 14 (z – 1) = 0
⇒ -6x + 36 – 10y – 10 + 14z – 14 = 0
⇒ -6x – 10y + 14z + 12 = 0
⇒ 3x + 5y – 7z – 6 = 0

(b) (2, 1, 3), (3, 2, 1) and (1, 0, -1)
Solution:
Equation of the plane though the given points is

⇒ (x – 2) (-4 – 2) – (y – 1) (-4 – 2) + (z – 3) (-1 + 1) = 0
⇒ (x – 2) (-6) – (y – 1) (-6) = 0
⇒ x – y – 1 = 0

(c) (-1, 0, 1), (-1, 4, 2) and (2, 4, 1)
Solution:
Equation of the plane through the given points is

⇒ (x + 1) (0 – 4) – y ( 0 – 3) + (z – 1) (0 – 12) = 0
⇒ -4x – 4 + 3y – 12z + 12 = 0
⇒ -4x + 3y – 12z + 8 = 0
⇒ 4x – 3y + 1 2z – 8 = 0

(d) (-1, 5, 4), (2, 3, 4) and (2, 3, -1)
Solution:
Equation of the plane through the given points is

⇒ (x + 1) (10 – 0) – (y – 5) (-15 – 0) + (z – 4) (-6 + 6) = 0
⇒ 10x + 10 + 15y – 75 = 0
⇒ 2x + 3y – 13 = 0

(e) (1, 2, 3), (1, -4, 3) and (-1, 3, 2)
Solution:
Equation of the plane through the given points is

⇒ (x – 1) (6 – 0) – (y – 2) (0 – 0) + ( z – 3) ( 0 – 12) = 0
⇒ (x – 1) 6 + (z – 3) (-12) = 0
x – 1 – 2z + 6 = 0
⇒ x – 2z + 5 = 0

Question 4.
Find the equation of plane in each of the following cases:
(a) Passing through the point (2, 3, -1) and parallel to the plane 3x – 4y + 7z = 0.
Solution:
Given plane is 3x – 4y + 7z = 0 … (1)
Any plane parallel to the plane (1) is given by 3x – 4y + 7z + d = 0.
If it passes through the point (2, 3, -1) then 6 – 12 – 7 + d = 0 or, d = 13
Equation of the plane is 3x – 4y + 7z + 13 = 0

(b) Passing through the point (2, -3, 1) and (-1, 1, -7) and perpendicular to the plane x – 2y + 5z + 1 = 0.
Solution:
Equation of the plane passing through the point (2, -3, 1) is
a (x – 2)+ b (y + 3) + c (z – 1) = 0 … (1)
If it passes through the point (-1, 1, -7)
then a (-1 – 2) + b (1 + 3) + c (-7 – 1) = 0
or 3a – 4b + 8c = 0 … (2)
Again given that plane (1) is perpendicular to the plane
x – 2y + 5z + 1 = 0
so a – 2b + 5c = 0 … (3)
Solving (2) and (3) we get
\(\frac{a}{-20+16}=\frac{b}{8-15}=\frac{c}{-6+4}\)
or, \(\frac{a}{-4}=\frac{b}{-7}=\frac{c}{-2} \text { or, } \frac{a}{4}=\frac{b}{7}=\frac{c}{2}\)
Equation of the required plane is
4 (x – 2) + 7 (y + 3) + 2 (z – 1) = 0
or, 4x + 7y + 2z + 11 = 0

(c) Passing through the foot of the perpendiculars drawn from P (a, b, c) on the coordinate planes.
Solution:
Let A, B, C be the feet of the perpendiculars drawn from point P(a, b, c) on to the coordinate axes. Then
A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)
Equation of the plane through A, B, C is
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1

(d) Passing through the point (-1, 3, 2) perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.
Solution:
Equation of the plane through the point (-1, 3, 2) is
a (x + 1) + b (y – 3) + c (z – 2) = 0
If this plane is perpendicular to the plane
x + 2y + 2z = 5 and 3x + 3y + 2z = 8
then a + 2b +2c = 0
3a + 3b + 2c = 0
Solving we get \(\frac{a}{4-6}=\frac{b}{6-2}=\frac{c}{3-6}\)
or \(\frac{a}{-2}=\frac{b}{4}=\frac{c}{-3} \text { or, } \frac{a}{2}=\frac{b}{-4}=\frac{c}{3}\)
Equation of the required plane is
2 (x + 1) – 4 (y – 3) + 3 (z – 2) = 0
or, 2x – b – 3z – 8 = 0

(e) Bisecting the line segment joining (-1, 4, 3) and (5, -2, -1) at right angles.
Solution:

Let A = (-1, 4, 3) and B (5, -2, -1).
Suppose that a plane bisects AB at right angle. Let C be the mid-point of AB. Then the plane passes through C and the line AB is perpendicular to the plane
C = \(\left(\frac{-1+5}{2}, \frac{4-2}{2}, \frac{3-1}{2}\right)\) = (2, 1, 1)
D.rs. of AB are < 6, -6, -4 >
Equation of the plane is
6 (x – 2) – 6 (y – 1) – 4 (z – 1) = 0
or, 3x – 6 – 3y + 3 – 2z + 2 = 0
or, 3x – 3y – 2z – 1 = 0

(f) Parallel to the plane 2x – y + 3z + 1 = 0 and at a distance 3 units away from it.
Solution:
Given plane is 2x – y + 3z + 1 = 0 … (1)
Equation of the plane parallel to the plane (1) is 2x – y + 3z +k = 0 … (2)
Given that the distance between two planes (1) and (2) is 3 units.
We see that (0, 1, 0) is a point on (1).
The length of the perpendicular from (0, 1, 0) onto the plane (2).

Question 5.
(a) Write the equation of the plane 3x – 4y + 6z – 12 = 0 in intercept form and hence obtain the coordinates of the point where it meets the co-ordinate axes.
Solution:
Given plane is 3x – 4y + 6z – 12 = 0
⇒ 3x – 4y + 6z = 12
⇒ \(\frac{x}{4}+\frac{y}{-3}+\frac{z}{2}\) = 1
This is in intercept form. The coordinates of the points where the plane meets the coordinate axes are (4, 0, 0), (0, -3, 0), (0, 0, 2).

(b) Write the equation of the plane 2x – 3y + 5z + 1 = 0 in normal form and find its distance from the origin. Find also the distance from the point (3, 1, 2).
Solution:
Given plane is 2x – 3y + 5z + 1 = 0
2x – 3y + 5z = -1
-2x + 3y – 5z = 1
Its normal form is

(c) Find the distance between the parallel planes 2x – 2y + z + 1 = 0 and 4x – 4y + 2z + 3 = 0.
Solution:
The parallel planes are
2x – 2y + z + 1 = 0 … (1)
and 4x – 4y + 2z + 3 = 0 … (2)
We see that (0, 0, -1) lies on the plane (1).
Distance between two planes = Length of the perpendicular from (0, 0, -1) onto the plane (2)
\(=\left|\frac{-2+3}{\sqrt{16+16+4}}\right|=\frac{1}{6}\)

Question 6.
In each of the following cases, verify whether the four given points are coplanar or not.
(a) (1, 2, 3), (-1, 1, 0), (2, 1, 3), (1, 1, 2)
Solution:
As

The given points are coplanar.

(b) (1, 1, 1), (3, 1, 2), (1, 4, 0), (-1, 1, 0)
Solution:
As

The given points are coplanar.

(c) (0, -1, -1), (4, 5, 1), (3, 9, 4), (-4, 4, 4)
Solution:
As

= 100 – 210 + 110
= 0
The given points are coplanar.

(d) (-6, 3, 2), (3, -2, 4), (5, 7, 3) and (-13, 17, -1)
Solution:
As

= 14 × (-26) + 2 × 182 = 0
The given points are coplanar.

Question 7.
Find the equation of plane in each of the following cases:
(a) Passing through the intersection of planes 2x + 3y – 4z + 1 = 0, 2x – y + z + 2 = 0 and passing through the point (3, 2, 1).
Solution:
Given planes are
2x + 3y – 4z + 1 = 0 … (1)
and 3x – y + z + 2 = 0 … (2)
Any plane through the line of intersection of the planes (1) and (2) is
(2x + 3y – 4z + 1) + k (3x – y + z + 2) = 0
or (2 + 3k) x + (3 – k) y + (k – 4) z + (2k + 1) = 0 … (3)
If the plane (3) passes through the point (3, 2, 1)
then (2 + 3k) 3 + (3 – k) 2 + (k – 4) + 2k + 1 = 0
⇒ 6 + 9k + 6 – 2k + k – 4 + 2k + 1 = 0

or, -7x + 39y – 49z – 8 = 0
or, 7x – 39y + 49z + 8 = 0

(b) Which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and perpendicular to the plane 5x + 3y + 6z + 8 = 0.
Solution:
Given planes are
x + 2y + 3z – 4 = 0 … (1)
and 2x + y – z + 5 = 0 … (2)
Any plane containing the line of intersection of the planes (1) and (2) is
x + 2y + 3z – 4 + k (2x + y – z + 5) = 0
⇒ (2k + 1) x + (k + 2) y + (3 – k) z + (5k – 4) = 0 … (3)
Given that the plane (3) is perpendicular to the plane
5x + 3y + 6z + 8 = 0 … (4)
Two planes are perpendicular if their normals are perpendicular.
D.rs. of the normal of the plane (3) are < 2k + 1, k + 2, 3 – k >.
D.rs. of the normal of the plane (4) are < 5, 3, 6 >.
If the normal are perpendicular then
5 (2k + 1) + 3 (k + 2) + 6 (3 – k) = 0
⇒ 10k + 3k – 6k + 5 + 6 + 18 = 0
⇒ 7k = -29

or, -51x – 15y + 50z – 173 = 0
or, 51x + 15y – 50z + 173 = 0

(c) Passing through the intersection of ax + by + cz + d = 0 and a₁x + b₁y + c₁z + d₁ = 0 and perpendicular to xy-plane.
Solution:
Given planes are
ax + by + cz + d = 0 … (1)
and a₁x + b₁y + c₁z + d₁ = 0 … (2)
Any plane passing through the line of intersection of the planes (1) and (2) is given by:
ax + by + cz + d + k (a₁x + b₁y + c₁z + d₁) = 0
or, (a + ka₁) x + (b + kb₁) y + (c + kc₁) z + (d + kd₁) = 0 … (3)
D.rs. of the normal of the plane (3) are < a + ka₁, b + kb₁, c + kc₁).
D.rs. of the normal of the xy-plane i.e., z = 0 are < 0, 0, 1 >
If the plane (3) is perpendicular to the xy-plane then
(a + ka₁) × 0 + (b + kb₁) × 0 + (c + kc₁) × 1 = 0

(d) Passing through the intersection of the planes x + 3y – z + 1 = 0 and 3x – y+ 5z + 3 = 0 and is at a distance 2/3 units from origin.
Solution:
Given planes are
x + 3y – z + 1 = 0 … (1)
and 3x – y + 5z + 3 = 0 … (2)
Any plane through the line of intersection of the planes (1 ) and (2) is
(x + 3y – z + 1)+ k (3x – y + 5z + 3) = 0
⇒ (3k + 1) x + (3 – k) y + (5k – 1) z + (3k + 1) = 0
The length of the perpendicular from origin onto the plane (3)
\(\frac{3 k+1}{\sqrt{(3 k+1)^2(3-k)^2+(5 k-1)^2}}=\frac{2}{3}\) (Given)
⇒ 9 (3k + 1)²
⇒ 4 {(3k + 1)² + (3 – k)² + (5k – 1)²}
⇒ 81k² + 54k + 9
= 4 {9k² + 6k + 1 + 9 + k² – 6k + 25k² + 1 – 10k}
= 4 {35k² – 10k + 11}
= 140k² – 40k + 44
⇒ 59k² – 94k + 35 = 0
⇒ 59k² – 59k – 35k + 35 = 0
⇒ 59k (k – 1) – 35 (k – 1) = 0
⇒ (k – 1) (59k – 35) = 0
⇒ k – 1 = 0 or 59k – 35 = 0
⇒ k = 1 or, k = \(\frac{35}{59}\)
∴ The planes are 4x + 2y + 4z + 4 = 0 and 82x + 71y + 58z + 82 = 0

Question 8.
Find the angle between the following pairs of planes.
(a) x + 3y – 5z + 1 = 0 and 2x + y – z+ 3 = 0
Solution:
Given planes are
x + 3y – 5z + 1 = 0 … (1)
and 2x + y – z + 3 = 0 … (2)
Angle between two planes is equal to the angle between their normals.
D.rs. of the normal to the plane (1) are < 1, 3, -5 >
D.rs. of the normal to the plane (2) are < 2, 1, -1 >.
If θ is the angle between the planes then

(b) x + 2y + 2z – 3 = 0 and 3x + 4y + 5z + 1 = 0
Solution:
Given planes are
x + 2y + 2z – 3 = 0 and 3x + 4y + 5z + 1 = 0
D.rs. of normals of two planes are < 1, 2, 2 > and < 3, 4, 5 >.
If θ is the angle between the planes then

(c) x + 2y + 2z – 7 = 0 and 2x – y +z = 6
Solution:
Two planes are
x + 2y + 2z – 7 = 0 and 2x – y + z = 6
The d.rs. of the normals are < 1, 2, 2 > and <2, -1, 1 >.
If θ is the angle between two planes then

Question 9.
(a) Find the equation of the bisector of the angles between the following pairs of planes and specify the ones which bisects the acute angles;
(i) 3x – 6y + 2z + 5 = 0 and 4x – 12y + 3z – 3 = 0
Solution:
Given planes are
3x – 6y + 2z + 5 = 0 … (1)
and 4x – 12y + 3z – 3 = 0 … (2)
The equations of the bisectors are given by

So the plane (3) is a bisector for obtuse angle and the plane (4) is the bisector for acute angle.

(ii) 2x + y – 2z – 1 = 0 and 4x – 12y + 3z + 3 = 0
Solution:
Given planes are
2x + y – 2z – 1 = 0 … (1)
and 4x – 12y + 3z + 3 = 0 … (2)
The equation of bisector planes are

∴ θ > 45°
∴ Equation (3) is the bisector of the use angle and equation (4) is the bisector of acute angle.

(b) Show that the origin lies in the interior of the acute angle between planes.
x + 2y + 2z + 9 and 4x – 3y + 12z + 13 = 0;
Find the equation of bisector of the acute angle.
Solution:
Given planes are x + 2y + 2z – 9 = 0
and 4x – 3y + 12z + 13 = 0
These equations can be written as
-x – 2y – 2z+ 9 = 0 … (1)
4x – 3y+ 12z + 13 = 0 … (2)
Here a₁a₂ + b₁b₂ + c₁c₂
= (-1) 4 + (-2) (-3) + (-2) . 12
= -4 + 6 – 24 = -22 < 0
So the origin lies in the acute angle between the given planes.
The equations of the planes bisecting the angle between the given planes are

⇒ tan θ < 1 ⇒ θ < 45°
Hence the plane (3) bisects the acute angle between the given planes.

Question 10.
(a) Prove that the line joining (1, 2, 3), (2, 1, -1) intersects the line joining (-1, 3, 1) and (3, 1, 5)
Solution:
Let A = (1, 2, 3), B = (2, 1, -1), C = (-1, 3, 1), D = (3, 1, 5)
The line AB intersects CD if A, B, C, D are coplanar.
Equation of the plane

or, (x – 1) (2 + 4) – (y – 2) (-2 – 8) + (z – 3) (1 – 2) = 0
or, 6 (x – 1) + 10 (y – 2) – (z – 3) = 0
or, 6x+ 10y – z – 23 = 0
The point (3, 1, 5) lies on the plane (1) because
6x + 10y – z – 23 = 18 + 10 – 5 – 23 = 0 … (1)
As (1) is satisfied for the point (3, 1, 5), the point lies on the plane (1).
Hence the four points are coplanar. So the line AB is either parallel to the line CD.
or, AB is intersecting to CD.
D.rs. of AB are < 1, -1, -4 >.
D.rs. of CD are < 4, -2, 4 >.
So d.rs. of AB are not proportional to d.rs. of CD. So they are not parallel. Hence they are intersecting.

(b) Show that the point (-\(\frac{1}{2}\), 2, 0) is the circumcentre of the triangle formed by the points (1, 1, 0), (1, 2, 1) and (-2, 2, -1).
Solution:

Thus PA = PB = PC
Hence P is the circumcentre of the ΔABC.

Question 11.
Show that plane ax + by + cz + d = 0 divides the line segment joining
(x₁, y₁, z₁) and (x₂, y₂, z₂) in a ratio – \(\frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}\).
Solution:
Let A = (x₁, y₁, z₁) , B = (x₂, y₂, z₂)
Suppose that the line AB intersects the plane
ax + by + cz + d = 0 at P. … (1)
Suppose the P divides AB into the ratio m:n.
Then the coordinates of P are

Question 12.
A variable plane is at a constant distance p from the origin and meets the axes at A, B, C. Through A, B, C planes are drawn parallel to the co-ordinate planes. Show that the locus of their points of intersection is \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}\).
Solution:

The planes through the points A, B, C parallel to the coordinate planes are
x = a, y = b and z = c respectively.
Let P be the point of intersection of these planes.
Then p = (a, b, c)
Hence the locus of P is obtained by putting a = x, b = y, c = z in (2).
Hence the required locus is
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}\). (Proved)

Question 13.
A variable plane passes through a fixed point (a, b, c) and meets the co-ordinate axes at A, B, C. Show that the locus of the point common to the planes drawn through A, B and C parallel to the coordinate planes is \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\) = 1.
Solution:
Let the variable plane be
\(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\) = 1 … (1)
It meets the axes at A (α, 0, 0), B (0, β, 0) and C (0, 0, γ).
If P is the point of intersection of the planes drawn through A, B and C parallel to the coordinate planes then P = (α, β, γ).
Again the plane (1) passes through a fixed point (a, b, c).
So \(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\) = 1
∴ The locus of P is \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\) = 1. (Proved)

Question 14.
The plane 4x + 7y + 4z + 81 = 0 is rotated through a right angle about its line of intersection with the plane 5x + 3y + 10z – 25 = 0. Find the equation of the plane in new position.
Solution:
Given that the plane
4x + 7y + 4z + 81 = 0 … (1)
is rotated through a right angle about the line of intersection with the plane
5x + 3y + 10z – 25 = 0 … (2)
Let the equation of the plane in new position be
4x + 7y + 4z + 81 + λ (5x + 3y -10z – 25) = 0
or, (4 + 5λ) x + (7 + 3λ) y + (4 + 10λ) z + (81 – 25λ) = 0
Then the angle between the planes (1 ) and (3) is 90° i.e., they are perpendicular.
So 4 (4 + 5λ) + 7 (7 + 3λ) + 4 (4 + 10λ) = 0
16 + 20λ + 49 + 21λ + 16 + 40λ = 0
81λ = -81
⇒ λ = -1
∴ The equation of plane in new position is
-x + 4y – 6z + 106 = 0
or, x – 4y + 6z – 106 = 0

Question 15.
The plane lx + my = 0 is rotated about its line of intersection with the plane z = 0 through angle measure α. Prove that the equation of the plane in new position is lx + my ± z \(\sqrt{l^2+m^2}\) tan α = 0.
Solution:
Given plane is lx + my = 0 … (1)
The plane (1) is rotated through an angle α about line of intersection of the plane (1) and the plane z = 0 … (2)
Let the equation of the plane in it’s new position be lx + my + kz = 0. … (3)
D.rs. of the normal to the plane (1) are < l, m, 0 >.
D.rs. of the normal to the plane (3) are < l, m, k >.
If α is the angle between the planes (1) and (3) then

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 13 Three Dimensional Geometry Ex 13(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 13 Three Dimensional Geometry Exercise 13(a)

Question 1.
Fill in the blanks in each of the following questions by choosing the appropriate answer from the given ones.
(a) The number of lines making equal angles with coordinate axes is _______. [1, 2, 4, 8]
Solution:
8

(b) The length of the projection of the line segment joining (1, 3, -1) and (3, 2, 4) on z-axis is _______. [1, 3, 4, 5]
Solution:
5

(c) If a line is perpendicular to z-axis and makes an angle measuring 60° with x-axis, then the angle it makes with y-axis measures _______. [30°, 60°, 90°, 120°]
Solution:
30°

(d) If the distance between the points (-1, -1, z) and (1, -1, 1) is 2 then z = _______. [1, √2, 2, 0]
Solution:
1

Question 2.
Which of the following statements are true (T) or false (F):
(a) The line through (1, -1, 2) and (-2, -1, 2) is always perpendicular to z-axis.
Solution:
True

(b) The line passing through (0, 0, 0) and (1, 2, 3) has direction cosines (-1, -2, -3)
Solution:
False

(c) if l, m, n be three real numbers proportional to the direction cosines of a line L, then l² + m² + n² = 1.
Solution:
False

(d) If α, β, γ be any three arbitrary angles then cos α, cos β, cos γ can always be considered as the direction cosines of a line.
Solution:
False

(e) If two lines are perpendicular to a third line, then the direction ratios of the two lines are proportional.
Solution:
False

Question 3.
(a) Show that the points (3, -2, 4), (1, 1, 1) and (-1, 4, -1) are collinear.
Solution:
Let A = (3, -2, 4), B = (1, 1, 1) and C = (-1, 4, -2)
D. rs of AB are < -2, 3, -3 >
D. rs of BC are < -2, 3, -3 >
As D.rs of AB are same as d.rs of BC it follows that A, B, C lie on the same straight line.
So the points are collinear. (Proved)

(b) Show that points (0, 1, 2), (2, 5, 8), (5, 6, 6) and (3, 2, 0) form a parallelogram.
Solution:
Let A = (0, 1, 2), B = (2, 5, 8), C = (5, 6, 6), D = (3, 2, 0)

∴ BC = AD
Thus the opposite sides of the quadrilateral ABCD are equal, hence it is a parallelogram.

Question 4.
(a) Find the co-ordinates of the foot of the perpendicular from the point (1, 1, 1) on the line joining (1, 4, 6) and (5, 4, 4).
Solution:

Let P = (1, 1, 1), A = (1, 4, 6) and B = (5, 4, 4)
Let Q be the foot of the perpendicular drawn from P on AB. Suppose that Q divides AB into the ratio m:n.

⇒ 4 (4m) – 2 (3m + 5n) = 0
16m – 6m – 10n = 0
⇒ 10m 10n = 0
⇒ m = n
∴ Q = (3, 4, 5)

(b) Find the co-ordinates of the point where the perpendicular from the origin meets the line joining the points (-9, 4, 5) and (11, 0, -1).
Solution:
Let A = (-9, 4, 5) and B = (11, 0, -1)

Let OM be the perpendicular drawn onto the line segment AB. Suppose that M divides AB into the ratio m:n.
Then co-ordinates of

D.rs. of AB are < 20, -4, -6 >
As OM is perpendicular on to AB, then
20 (11m – 9n) – 4 (4n) – 6 (-m + 5n) = 0
⇒ 220m – 180n – 16n + 6m – 30n = 0
⇒ 226m – 226n = 0
⇒ m = n
∴ The ratio is \(\frac{m}{n}=\frac{1}{1}\)
∴ Co-ordinates of M are (1, 2, 2).

(c) Prove that the points P(3, 2, -4), Q (5, 4, -6) and R (9, 8, -10) are collinear.
Solution:
Given that P = (3, 2, -4), Q = (5, 4, -6), R = (9, 8, -10)
D. rs. of PQ are < 2, 2, -2 >
D. rs. of QR are < 4, 4, -4 >
i.e., < 2, 2, -2 >
Thus D.rs. of PQ and QR are same. So P, Q, R lie on the same straight line.
Hence P, Q, R are collinear. (Proved)

(d) If P (1, y, z) lies on the line through (3, 2, -1) and (-4, 6, 3) find y & z.
Solution:
Let A = (3, 2, -1), B = (-4, 6, 3)
Suppose that P (1, y, z) divides AB into the ratio m:n.

Question 5.
(a) If A, B, C, D are the points (6, 3, 2), (3, 5, 7), (2, 3, -1) and (3, 5, -3) respectively, then find the projection of \(\overline{A B} \text { on } \overleftrightarrow{CD}\).
Solution:
Given that
A = (6, 3, 2), B = (3, 5, 7), C = (2, 3, -1), D = (3, 5, -3)

(b) The projections of a line segment \(\overline{OP}\), through origin O, on the coordinate axes are 6, 2, 3. Find the length of the line segment \(\overline{OP}\) and its direction cosines.
Solution:

(c) The projections of a line segment of x, y and z-axis respectively are 12, 4, 3. Find the length and the direction cosines of the line segment.
Solution:
Let the length of the line segment be ‘r’ and suppose that it makes angles α, β, γ with the axes.
Then r cos α = 12, r cos β = 4 and r cos γ = 3

Question 6.
(a) If A, B, C are the points (1, 4, 2), (-2, 1, 2) and (2, -3, 4) respectively then find the angles of the triangle ABC.
Solution:

Given that A = (1, 4, 2), B = (-2, 1, 2) and C = (2, -3, 4)
D.r.s. of AB are < -3, -3, 0 >
D.r.s. of BC are < 4, -4, 2 >
D.r.s. of AC are < 1, -7, 2 >

(b) Find the acute angle between the lines passing through (-3, -1, 0), (2, -3, 1) and (1, 2, 3), (-1, 4, -2) respectively.
Solution:
Let A = (-3, -1, 0), B = (2, -3, 1), C = (1, 2, 3), D = (-1, 4, -2)
D.rs. of AB are < 5, -2, 1 >
D.rs. of CD are < -2, 2, -5 >

(c) Prove that measure of the angle between two main diagonals of a cube is cos^-1\(\frac{1}{3}\).
Solution:

Let OABCDEFG be a cube of side ‘a’. Choose the vertex ‘O’ as origin, the line OA as x-axis, OE as y-axis and OC as z-axis.
Then
A = (a, 0, 0), B = (a, 0, a), C = (0, 0, a), D = (0, a, a), E = (0, a, 0), F = (a, a, 0) and G = (a, a, a)
The main diagonals are OG and BE
D.rs. of OG are < a, a, a >.

(d) Prove that measure of the angle between the diagonal of a face and the diagonal of a cube, drawn from a vertex is cos^-1\(\sqrt{\frac{2}{3}}\).
Solution:
See the figure of (c)
Consider the face OABC.
Let us find the angle between AC and AD.
D.rs. of AC are < -a, 0, a >

(e) Find the angle which a diagonal of a cube makes with one of its edges.
Solution:
See the figure of (c).
Let us find the angle between the edge AB and diagonal AD.

(f) Find the angle between the lines whose dcs. l, m, n are connected by the relation, 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0
Solution:
Given that 3l + m + 5n = 0 … (1)
6mn – 2nl + 5lm = 0 … (2)
From (1) we get
n = –\(\frac{3 l+m}{5}\)
Putting it in (2) we get
(6m – 2l) . (-\(\frac{3 l+m}{5}\)) + 5lm = 0
⇒ -18lm + 6l² – 6m² + 2lm + 25lm = 0
⇒ 6l² – 6m² + 9lm = 0
⇒ 2l² – 2m² + 3lm = 0
⇒ 2l² + 4lm – lm – 2m² = 0
⇒ 2l (l + 2m) – m (1 + 2m)
⇒ (l + 2m) (2l – m) = 0
⇒ l + 2m = 0 or, 2l – m = 0

Case (i) Let l + 2m = 0

Case (ii) Let 2l – m = 0

Question 7.
Show that measures of the angles between the four diagonals, of a rectangular paralle-lopiped whose edges are a, b, c are cos^-1\(\left(\frac{a^2 \pm b^2 \pm c^2}{a^2+b^2+c^2}\right)\).
Solution:

Let OABCDEFG be a rectangular parallelopiped whose edges are a, b, c.
Choose O as origin, OA as x-axis, OE as y-axis and OC as z-axis.
Then A = (a, 0, 0), B = (a, 0, c), C = (0, 0, c), D = (0, b, c), E = (0, b, 0), F = (a, b, 0), G = (a, b, c)
The four diagonals are OG, EB, AD and CF.
D. rs. of OG are < a, b, c >
D. rs. of EB are < a, -b, c >
D. rs. of AD are < -a, b, c >
D. rs. of CF are < a, b, -c >
Angle between OG and EB.

Similarly find the angles between the other diagonals. Thus the angles between the four diagonals are cos^-1\(\left(\frac{a^2 \pm b^2 \pm c^2}{a^2+b^2+c^2}\right)\).

Question 8.
If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines show that the d.cs. of the line perpendicular to both of them are m₁n₂ – n₁m₂, n₁l₂ – l₁n₂, l₁m₂ – m₁l₂
Solution:
Let < l₁, m₁, n₁ > and < l₂, m₂, n₂ > be the d.cs. of two mutually perpendicular straight lines.
Then l₁l₂ + m₁m₂ + n₁n₂ = 0 … (1)
Let < l, m, n > be the d.cs. of the straight line perpendicular to the above lines.
Then ll₁ + mm₁ + nn₁ = 0 … (2)
ll₂ + mm₂ + nn₂ = 0 … (3)
Solving (2) and (3) for l, m, n we get

Odisha State Board CHSE Odisha Class 12 Approaches to English Book 2 Solutions Grammar Word Order Textbook Exercise Questions and Answers.

CHSE Odisha 12th Class Alternative English Grammar Word Order

1. Fill in the blanks with the suitable verb forms relating to the subjects.
(a) One number of the books ____________ been lost (here / has).
(b) Neither boy ____________ Ramesh (is/are)
(c) Whether Sita or Gita wins ____________ does not matter.
(d) Neither Sushma nor Shasmita ____________ well, (do / plays)
(e) Your knowledge in many languages ____________ poor, (is/are)
(f) Dr. Das with his sons and daughters ____________ gone to Puri, (has/have)
(g) A gang of thieves ____________ arrested, (was/were)
(h) My family ____________ early sisters, (is/are)
(i) Rains ____________ welcome (is/are)
(j) The cattle ____________ grazing in the field (is/are)
(k) The public ____________ not accept your view, (do/does)
(l) The committee ____________ divided in their opinion, (was/were)
(m) Your spectacles ____________ nice on me. (look / looks)
(n) The rich ____________ always unhappy, (is/are)
(o) The boy as well as the girls ____________ guilty, (is/are)
(p) Ten minutes ____________ not enough for me. (is/are)
(q) Show and steady ____________ the race (win/wins)
(r) Bread and butter ____________ my favorite subject, (is/are)
(s) The United nations ____________ not functioning well, (is/are)
(t) Mathematics ____________ my favorite subject, (is/are)
(u) Measles ____________ a dreadful disease, (is/are)
(v) Rice and dal ________________ eaten in Odisha. (is/are)
(w) I ____________ your brother, should be respected, (is / am / are)
(x) The father of these children ____________ an innocent fellow, (is/are)
(y) A pair of spectacles ____________ lying on the table, (is/are)
(z) Ten kilometers ____________ not a long way to walk, (is/are)

Answer:
(a) One number of the books has been lost (here / has).
(b) Neither boy is Ramesh (is/are)
(c) Whether Sita or Gita wins does not matter.
(d) Neither Sushma nor Shasmita plays well, (do / plays)
(e) Your knowledge in many languages is poor, (is/are)
(f) Dr. Das with his sons and daughters has gone to Puri, (has/have)
(g) A gang of thieves was arrested, (was/were)
(h) My family are early sisters, (is/are)
(f) Rains are welcome (is/are)
(j) The cattle are grazing in the field (is/are)
(k) The public do not accept your view, (do/does)
(l) The committee were divided in their opinion.
(m) Your spectacles looks nice on me.
(n) The rich are always unhappy.
(o) The boy as well as the girls is guilty.
(p) Ten minutes is not enough for me.
(q) Show and steady wins the race
(r) Bread and butter is my favourite subject.
(s) The United nations is not functioning well.
(t) Mathematics is my favourite subject.
(u) Measles is a dreadful disease.
(v) Rice and dal is eaten in Odisha.
(w) I am your brother, and should be respected.
(x) The father of these children is an innocent fellow.
(y) A pair of spectacles is lying on the table.
(z) Ten kilometers is not a long way to walk.

2. Supply the correct forms of the verbs given in the brackets in relation to their respective subjects.
(a) Economics ____________ difficult subject, (is/are)
(b) “Lines written in March” ____________ a good poem, (is/are)
(c) The audience ____________ enjoying the speech, (was/were)
(d) Three deers ____________ grazing in the park (is/are)
(e) Neither John nor Mary ____________ responsible for the breakage, (is/are)
(f) None of you ____________ correct in your answer, (is/are)
(g) Everyone in the class has ____________ identify cards. (his / their)
(h) Either Mary or her sister ____________ in the house, (was/were)
(i) The number of students ____________ large, (have/has)
(j) A large number of students ____________ applied for admission. (have/has)
(k) The poet and teacher ____________ deed, (is/are)
(l) Two fives ____________ ten. (is/are)
(m) Twice twelve ____________ twenty-four, (is/are)
(n) Two and two ____________ four, (make / makes)
(o) More then one person ____________ signed, (have/has)
(p) No person of the name ____________ here, (lives / five)
(q) In the past, goods ____________ cheap. (was/were)
(r) The man together with his wife ____________ been killed, (have/has)
(s) Rs 50,000 ____________ a big sum (is/are)
(t) The unemployed ____________ living in a precious situation, (is/are)
(u) You who ____________ my brother should not say so. (is/are)
(v) The military ____________ called out to tackle the situation, (was/were)
(w) There ____________ many trees on both sides of the road, (is /are)
(x) Each of the boys ____________ a pen. (have/has)
(y) A committee ____________ appointed to enquire into the matter, (was/were)
(z) He, you, and they ____________ ready, (is/are)

Answers:
(a) Economics is a difficult subject.
(b) “Lines written in March” is a good poem.
(c) The audience were enjoying the speech.
(d) Three deer are growing in the park.
(e) Neither John nor Mary is responsible for the breakage.
(f) None of you are correct in your answer.
(g) Everyone in the class has his identification cards.
(h) Either Mary or her sister was in the house.
(1) The number of students is large.
(j) A large number of students have applied for admission.
(k) The poet and teacher is the deed.
(l) Two fives are ten.
(m) Twice twelve is twenty-four.
(n) Two land two makes four.
(o) More than one person has signed.
(p) No person of the name lives here.
(q) In the past, goods were cheap.
(r) The man together with his wife has been killed.
(s) Rs 50,000 is a big sum.
(t) The unemployed are living in a precious situation.
(u) You who is my brother should not say so.
(v) The military were called out to tackle the situation.
(w) There are many trees on both sides of the road.
(x) Each of the bo vs has a pen.
(y) A committee was appointed to enquire into the matter.
(z) He, you, and they is ready.

3. Concord of nouns, pronouns, possessive, adjectives:
(a) They had a dog. Once ____________ was ill. (he/she / it)
(b) Give the baby ____________ toy. (his / her / its)
(c) The cow lives ____________ calf, (it’s / her)
(d) The car is ____________ (your / yours). The car is (my, mine)
(e) This book is new, ____________ books are old. (these / those)
(f) Everyone of the boys is keen on ____________ success, (his / their)
(g) The cabinet has taken ____________ decision, (it’s / their)
(h) The flock of sheep has done ____________ part well, (her / their)
(i) Neither of the girls have done ____________ part well, (her / their)
(j) I am not one of the ____________ who betray friends, (these / those)
(k) I ____________ your friend, should not be misbehaved, (am / are)
(l) Neither is capable of making ____________ mind, (his / their)

Answers:
(a) They had a dog. Once it was ill.
(b) Give the baby its toy.
(c) The cow lives with her calf.
(d) The car is yours. The car is mine.
(e) This book is new, those books are old.
(f) Every one of the boys is keen on his success.
(g) The cabinet has taken its decision.
(h) The flock of sheep has done its part well.
(i) Neither of the girls has done her part well.
(j) I am not one of those who betray friends.
(k) I am your friend, should not be misbehaved.
(l) Neither is capable of making his mind.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Additional Exercise Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Additional Exercise

Question 1.
Prove that the sum of the vectors directed from the vertices to the midpoints of opposite sides of a triangle is zero.
Solution:
Let D, E, F are midpoints of side BC, CA and AB of ΔABC

Question 2.
Prove by vector method that the diagonals of a quadrilateral bisect each other iff it is a parallelogram.
Solution:

Question 3.
If G is the centroid of a triangle ABC, prove that \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\).
Solution:

Question 4.
If M is the midpoint of the side B͞C of a triangle ABC, prove that \(\overrightarrow{AB}+\overrightarrow{AC}=2 \overrightarrow{AM}\).
Solution:

Question 5.
If \(\vec{a} \text { and } \vec{b}\) are the vectors represented by the adjacent sides of a regular hexagon, taken in order, what are the vectors represented by the other sides taken in order?
Solution:

Question 6.
If the points with position vectors 10î +3ĵ, 12î – 5ĵ and aî +11ĵ are collinear, find the value of a.
Solution:
Let the points are A, B and C.
As A, B and C are co-llinear.
We have \(\overrightarrow{AB}=\overrightarrow{AC}\)
\(\overrightarrow{AB}=\lambda \overrightarrow{AC}\)
⇒ (12i – 5j) – (10i + 3j)
⇒ λ [(ai + 11j) – (10i + 3j)
⇒ 2i – 8j = (a – 10)λ i + 8λ j
Comparing the components we have
8λ = -8
⇒ λ = -1
(a – 10) (-1) = 2
⇒ -a + 10 = 2
⇒ a = 8

Question 7.
Prove that the four points with position vectors \(2 \vec{a}+3 \vec{b}-\vec{c}, \vec{a}-2 \vec{b}+3 \vec{c}, 3 \vec{a}+4 \vec{b}-2 \vec{c}\) and \(\vec{a}-6 \vec{b}+6 \vec{c}\) are coplanar.
Solution:
Let the points are A, B, C

Question 8.
For any vector \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) prove that \(\vec{r}=(\vec{r} \cdot \hat{i}) \hat{i}+(\vec{r} \cdot \hat{j}) \hat{j}+(\vec{r} \cdot \hat{k}) \hat{k}\).
Solution:

Question 9.
If two vectors \(\vec{a} \text { and } \vec{b}\) are such \(|\vec{a}|={3},|\vec{b}|=2 \text { and } \vec{a} \cdot \vec{b}={6}\text { find }|\vec{a}+\vec{b}| \text { and }|\vec{a}-\vec{b}|\).
Solution:

Question 10.
If \(\vec{a}\) makes equal angles with î, ĵ and k̂ and has magnitude 3, prove that the angle between \(\vec{a}\) and each of î ,ĵ and k̂ is \(\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\).
Solution:

Question 11.
If \(\vec{a}, \vec{b}, \vec{c}\) are such that \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}\) then show that \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\vec{c} \text { or } \vec{a}\) is perpendicular to \(\vec{b}-\vec{c}\)
Solution:

Question 12.
If \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0},|\vec{a}|={3},|\vec{b}|={5}\text { and }|\vec{c}|=7\), find the angle between \(\vec{a} \text { and } \vec{b}\).
Solution:

Question 13.
If \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), find the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\).
Solution:

Question 14.
Find the angles which the vector \(\vec{a}=\hat{i}-\hat{j}+\sqrt{2} \hat{k}\) makes with the coordinate axes.
Solution:

Question 15.
Find the angle between \(\vec{a} \text { and } \vec{b}\) if \(|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}\).
Solution:

Odisha State Board CHSE Odisha Class 12 Approaches to English Book 2 Solutions Short Stories Chapter 3 The Watchman Textbook Exercise Questions and Answers.

CHSE Odisha 12th Class Alternative English Solutions Short Stories Chapter 3 The Watchman

Section- I

Questions For Discussion:

Question 1.
How does the story writer present the setting of the story? Does it arouse the thrill and excitement of a detective story?
Answer:
The angler was angling in the tank water. He continued into the dead of night. There was nobody there. All of a sudden, he saw about a hundred yards away. A shadowy figure moving down the narrow stone steps that led to the water’s edge. He thought it might be a ghost. This arouses the thrill and excitement of a detective story.

Question 2.
What does the watchman think about the figure at a distance?
Answer:
The watchman saw a figure at a distance while he was angling in the tank. He thought it to be a ghost. Then he dismissed the idea and from the top step of the tank he observed that it was a woman’s form who stopped over the last step and placed something on it.

Question 3.
“Unmistakable signs – always to be followed by the police and gives some details…” What does this statement of the watchman indicate?
Answer:
Women coming to the tank and committing suicide in its water had caused the police to rush to the scene for actions. This had brought a lot of it reputation to the tank. Keeping this in mind, the watchman had uttered the statement.

Question 4.
What kind of tone do you mark in the words of the watchman in his first encounter with the girl?
Answer:
The watchman told the girl to come? out of the water otherwise she would be caught cold. He hurriedly lighted his lamp and discovered the letter on the step. He formed a notion that she had gone there to commit suicide. He said to himself as to why everyone was going to the same rank. His words mark a notion of suicide of the stranger girl.

Question 5.
Why did the girl want to commit suicide? Was the reason she advanced good enough to warrant such a step?
Answer:
The girl wanted to commit suicide because she was not selected for scholarship for studying medicine. She was going to be given in marriage very soon. This made her commit suicide.

Question 6.
What impression do you get about the girl from her narration? Do her words indicating her self-reliance and strong desire to study and her decision to commit suicide go together?
Answer:
The girl had lost her mother and her father married again. However, the step-mother was co-opearative and careful. She took the girl’s care as a mother. She had a strong desire to study. But she was not chosen for the scholarship to study medicine and she was to be given in marriage very soon. This incident had made her decide to commit suicide. Her words indicating her self-reliance and strong desire to study and her decision to commit suicide go together.

Vocabulary:
A. Derive adverbs from the following:

splash	temperament
except	topic
obstinate	tranquil
sad	silent
possible	stubborn
helpless	furious
interesting	poor
idea	formidable
logical	obey
surprise	race
wonder	pity
afford	pliable
worry	symmetry
comprehend	satire
death	sorrow
fantastic

Answer:
Words – Adverbs
splash -splashingly
except- exceptionally
obstinate- obstinately
sad- sadly
possible- possibly
helpless- helplessly
interesting- interestingly
idea- ideally
logical- logically
surprise- surprisingly
wonder- wonderfully
afford- affordably
worry- worryingly
comprehend- comprehensively
death- deadly
fantastic- fantastically
temperament- temperamentally
topic- tropically
tranquil- tranquility
silent- silently
stubborn- stubbornly
furious- furiously
poor- poorly
formidable- formidably
obey- obediently
race- racingly
pity- pitiably
pliable- pliably
symmetry- symmetrically
satire- satirically
sorrow- sorrowfully

B. Make sentences with the following:

feint	bury
horizon	trouble
survey	retort
obstinate	burden
trespass	livelihood
dismiss	worry
investigate	afford
reputation	exclaim
gaze	announce
bright	breakdown
hurriedly	comprehend
murmur	hysterically

Answer:
feint- He looks very faint today.
horizon – The sunset looks beautiful in the western horizon.
survey- The government has started survey of population.
obstinate – This boy is very obstinate.
trespass – Trespassing is a serious offense.
dismiss – He dismissed his proposal of going to Puri.
investigate – The police investigated the case.
reputation – Hari has an ill reputation.
gaze- The girl gazed boldly at my face.
bright – He had a bright chance to pass the exam.
hurriedly- He ran to me hurriedly.
murmur- I enjoy murmurs of the river.
bury- The dead body was buried in the graveyard.
trouble- I can face any trouble.
retort- She strongly retorted to refute her husband’s argument.
burden- Donkey is an animal of burden.
livelihood- She worked hard to earn her livelihood.
worry- He is very worried about his result.
afford- His father could not afford his study expenses.
exclaim- He exclaimed that it was a very terrible sight.
announce – The results was announced in time.
breakdown – The bus came to a breakdown.
comprehend- You should comprehend the topic thoroughly.
hysterically – The woman rushed hysterically to the site to sea the son’s dead body.

C. Write the antonyms of the following:

alive	interesting
alert	poor
inlet	obey
appear	sad
possible

Answer:
Words – Antonyms
alive – dead
alert – dormant
inlet – outlet
disappear – appear
possible – happy
interesting – disinteresting
poor – rich
obey – discovery

D. Supply synonyms to the following:

abandon	barbarous
abound	bare
absurd	base
adept	beginning
adequate	behaviour
admission	belie
adversary	beseech
adversity	betray
affable	bias
affection	bizarre
affront	blame
aggressive	bold
awful	fearful

Answer:
Words – Synonyms
abandon – forsake
abound – increase
absurd – ridiculous
adept – skilled
adequate – sufficient/ enough
admission – entry
adversary – opponent /enemy
adversity – misfortune
affable – gracious
affection – love /inclination
affront – insult
aggressive – attacking
awful – fearful
barbarous – uncivilised
bare – nude
base – mean /worthless
beginning – start
behaviour – conduct/manner
belie – contradict
beseech – pray
betray – mislead
bias – inclination
bizarre – strange
blame – disapprove
bold – valiant /daring

Section – II

Questions For Discussion:

Question 1.
Do you notice a shift of focus in this part of the story?
Answer:
This part of the story marks a shift of focus. The even of the first section tells about the girl’s sorrows and pangs for her not being selected for scholarship to study medicine. Her inclination to committing suicide has been delineated there. But the second section shifts from that even to the village life of the watchman

Question 2.
Give an account of the part life of the watchman.
Answer:
In the distant past, in his little village home an epidermic of cholera laid out of his father, mother and brothers on the same day and he was the sole surviver. He was turned out of his ancestral home through the trickery of his father’s kinsmen and he wandered as an orphan suffering indescribable hunger and privatition.

Question 3.
Everyone has his own miseries. If people tried to kill themselves for each one of them. I don’t know how often they would have to drown”. How does this statement of the watchman affect the girl?”
Answer:
The statement head a great deal of effect on the girl who was in the tank to commit suicide. The watchman made understand the matter saying the everyman has his own sorrow and problems. He said that the girl was very young and did not know what sorrow was. His wife bore his eight children out of whom only one lived. She had only one daughter living with him. This affected the girl and she looked back at him in bewilderment.

Question 4.
What you think was the outcome of the conversation between the watchman and the girt?
Answer:
Nothing came out of the conversation between the watchman and the girl. As the
Taluk going stuck again he advised her to go home, but she declined saying that she had no home to go to. The watchman warned her of not being obstinate. He picked up his lantern and staff and get up. He put her letter where he had found it. He said if she was going to be obstinate she would leave her alone. He went up steps and left.

Question 5.
What did the watchman believe when he saw the letter on the steps next morning? How did he feel thereafter?
Answer:
When the watchman came back to duty next morning he hurried down the stone steps. The letter lay where he had dropped it on the previous night. He believed that the girl had committed suicide. He tore it up and flung it on the water. He blamed himself for leaving her and going away on the suicide in the tank. He could never look at the blue expanse of water gain with an easy mind.

Question 6.
If you believe the watchman’s words he recognised the married young woman as the girl he thought was dead. Why then should she look away from the watchman?
Answer:
Years later, one evening as the stood on the bond and took a final survey before going home, he saw a car draw up on the road below. A man a woman and there children emerged from the car and climbed the bund. The watchman recognised the married young woman as the girl he thought was dead. She looked away from the watchman least she should have been exposed before her husband and children through their conversation.

Question 7.
Whom do you like more – the girl or the watchman? Justify your answer.
Answer:
In fact, the watchman is obviously the person whom anybody can like. He inspires and persuads the girl for going back home and the girl does not leave. He leaves for home and comes back there the next morning. He takes for granted that the girl had committed suicide. He takes for granted that the girl had committed suicide. He blames himself for at least one suicide. Years later, when he sees children. He recognizes her whom he wants to speak to, but she avoids talking to him deliberately. The watchman shows her respect without reciprocity from the women. This speaks volumes of the goodness of the watchman as contrasted with the nature of the girl.

Composition:

Question 1.
Attempt an appreciation of the character of the watchman.
Answer:
The short story “The Watchman” is undoubtedly the most typical masterpiece of R.K. Narayan, an eminent and outstanding. Indian storyteller of the twentieth century. He is really a superb master in depicting the realistic and vivid characters skillfully chosen from the then prevalent Indian society. His characters, mostly belonging to the lower middle class and poorer sections, exhibit an uncanny sense offear, in their actions and behaviour, the fear and uncertainties beaming out of foreign domination and a servile attitude of the people who suffer, sometimes they are the victim of social evil and corrupt order.

Such a typical character is the watchman in this discussing story. However, the story writer R.K. Narayan gives a graphic sketch of the character and personality of the watchman in his story names after this protagonist. The story embodies two characters the watchman and the girl the characters of whom need a close study. The watchman demonstrates a greater understanding of man’s predicament and endurance to encounter boldly the onslaughts of misfortune. Moreover, the watchman has a deep concentration on angling in the tank. All the people who had come for evening had returned to their homes. Not a soul anywhere except that obstinate angle at the northern end who sat with his feet in water, sadly gazing on his rod. He would sit there till midnight with the hope of catching fish. Of course, he has a sense of conscience. He advised the girl and dissuaded her from taking any attempt for committing suicide. She declines to go home as she has not any. He gets irritated and leaves for his own home.

Coming back from home the next morning he discovers the same letter lying on the steps of the tank bund, he presumes that the girl is dead. His conscience bites him and he holds himself responsible for the death of the girl. He thinks if he had not left her alone, she wouldn’t have committed suicide. He speculates that he isat least responsible for one suicide in the tank. Again, we mark that he has a sense of respect years later, one evening as he stood on the bund and took a final survey before going home, he sees a car draw up on the road below. A man, a woman and three children emerged from the car and climb the bund.

When they approach the watchman feels a start at his heart. The figure and face of the woman seems familiar to him. The woman is altered by years, ornaments and dresses, he thinks that he has recognised the face of the woman. He feels excited at the discovery. He has numerous question to ask. He brings together his palms and salutes her respectfully. As a matter of fact, the watchman happens to be pivotal character of the story. He is hold, courageous, helpful, generous and benevolent. The way Narayn depics his protagonist is superb and fantastic. It is inspiring, elevating and heart-enduring.

Question 2.
Sketch the character of the girl in contrast with the character of the girl.
Answer:
In fact the short story “The Watchman” is R.K. Narayan’s super band fantastic creation. He is in fact, an outstanding master in the work of sketching vivid and lively and realistic characters. His characters, mostly belonging to the lower middle class and poorer section, exhibit an uncanny sense of fear and in their actions and behaviour sometimes, they are a victim of social evil and corrupt. Such of a character is the girl in “The Watchman” of R.K. Narayan. However, between the two important characters in the story ‘The Watchman,” the girl is equally essential to the formation of its structure. The girl is ambitious, believe in self-reliance and detests being a liability or living on charity “I won’t like an anybody’s charity” she says.

To the watchman is suggestion of accepting marriage, she hysterically reveals that her cowardly thought of suicide is not in harmony with the bold statement she makes before the watchman. Like most of Narayan’s characters, she being poor suffers from a fear of uncertainty. She looks a strong willpower to arrange money shown in order to pursue her studies. Moreover, the watchman saw the girl in the tank bund while he was angling in the tank water concentratedly late at night, she had kept a letter on the step and was in the water. The watchman lit his lantern and discovered the girl in the tank which had an ill-reputation for committing suicide. The young girl was in tears. He asked her to go home. Sputtering through her sob, she said she had not home to go to.

She lost her mother some years back and her father married again. Her stepmother was also kind and cooperative to her, when the watchman said that her problems would be solved if she got married, she said that she did not want to marry as she wanted to study and become a doctor and earn her livelihood. Till that she had been hoping to get a scholarship which would have helped her. But another to get a scholarship which would have helped her. But another not she was given the scholarship. Her mother was going to settle her in marriage which she did not like. As a matter of fact, from the above sketch it is clear that in spite of her obstinacy, the girl is a lovable and lively character. Such type of situation and happenings are general and common in our society and so Narayan’s exhibition of such is superb and outstanding. On the whole both of the characters are watchman and the girl are inspiring, elevating and thought enduring.

Question 3.
Comment on Narayan’s art of storytelling. With reference to the short story “The Watchman”.
Answer:
In fact, R.K. Narayan is universally acknowledged as an outstanding and superb master in the realm of storytelling. Not only in India, but also in the entire cosmos internationally, he has put a burning stamp to have name and fame which is rarely marked and depicted on the part of his other contemporaries. He is an adept fabricator of stories. He is undoubtedly one of the greatest Indian in English, having to his credit a number of novels, volumes of short stories’- travelogues, retold legends and an autobiography.

“Swamy and Friends”,‘The Bachelor of ‘The English Teacher”, “Mr. Sampath”,‘The Guide” etc. are some of his important Astrologer’s Day and Other Stories”, “LawlyRoad and Other Stories, “Malgudi Days”, “Under the Banyan Tree and Other Stories”, include his best known stories. “My Dateless Diary” is a travelogue and “My Days” is the writer’s autobiography. Having an intimate knowledge of Indian life and society based with orthodoxy and superstition, he becomes Indian’s greatest storyteller, an astute observer of Indian psyche growing through the pre and post independence times. His characters mostly belonging to the lower middle class and poorer section, exhibit an uncanny sense offear in their actions and behaviour, the fear and uncertainty stemming mainly out of foreign dominant and a servile attitude of the people who suffer.

Sometimes they are the victims of social evils and a corrupt order of which his writings are humorous and satiric exposition. Irony at times lapses into cynicism in this writings. Narayan will live forever his beautiful imaginary town ofMalgudi faithfully representing Indian ethos, milie, and sensibility. However, the story extracted from “Malgudi Days” enacts an incident of a late evening on a tank bund which ultimately brings into focus a contrast between two characters a girl and a watchman. The girl is ambitious, believes in self-reliance and detests being a liability or living on charity. “I won’t live on anybody’s charity, she says to the watchman’s suggestion of accepting marriage the historically reacts. “No, no… I don’t want to marry. But a deeper look into her character reveals that he cowardly thought of suicide is not in harmony with the bold statements she makes before the watchman.

Like most of Narayan’s characters she being poor suffers from a fear of uncertainty. She lacks a strong willpower to arrange money somehow in order to pursue her studies. In contrast, the watchman demonstrates a greater underlying of man’s predicament and endurance to encounter boldly the slaughts of misfortune. The story ends with a note of ambiguity and uncertainty as regards the identity of the woman. As a matter of fact, the story with its elements of irony and undertone of cynicism is a typical masterpiece of Narayan’s fictional writing. Judged from all sides this story becomes a perfect embodiment of Narayan’s method of storytelling keeping all his qualities in feet.

Vocabulary:
A. Say what parts of speech to these following words belong to:

faint	wet
western	tears
watchman	sudden
final	pity
obstinate	obey
satisfy	her
trespass	she
cattle	sputter
hundred	old
prayer	money
investigate	spend
gruesome	dinner
reputation	angrily
hurriedly	good
same	enough
yourself	dark
under	livelihood
light	far

Answer:

Words – Parts of Speech
feint – adjective
western – adjective
watchman – noun
final – adjective
obstinate – adjective
satisfy – verb
trespass – verb
cattle – noun
hundred – adjective
prayer – noun
investigate – verb
gruesome – adjective
reputation – noun
hurriedly – adverb
same – adverb
yourself – pronoun
under – preposition
light – noun
wet – noun
tears – adjective
sudden – adjective
pity – noun
obey – noun
her – pronoun
she – pronoun
sputter – verb
old – adjective
money – noun
spend – verb
dinner – noun
angrily – adverb
good – adjective
enough – adjective
dark – noun/ adjective
livelihood – noun
far – adjective

B. Make sentences using the following from the text:

Western	get up
obstinate	trouble
trespass	pause
investigate	pickup
sobbing	flung
wavelets	responsible
far off	draw up
epidemic	familiar
survivor	numerous
ancestral	resolve
trickery	episode
privation	excited
miseries	alter
bewilderment	recognise

Answer:

Western – We should not follow western culture.
obstinate – The child is very obstinate.
trespass – He trespassed my backyard to pluck flowers.
investigate – The police investigated the matter.
sobbing – I don’t care for her sobbing.
wavelets – The boy threw a stone into the pond and so wavelets appeared in the water.
far off – He has come from a far off place.
epidemic – Cholera is an epidermic
survivor – The boy is the sole survivor of his family after the super cyclone.
ancestral – Nobody wants to part with his ancestral home.
trickery – The boy fell into the trickery of his friend.
privation – The girl has nobody to fall back on. She is in a state of privation.
miseries His life is full of miseries.
bewilderment – When the son misbehaved his father his father was bewilderment
get up – We should get up from bed early, in the morning.
trouble – He is in a great trouble
pause – You should not take much pasues in your speech.
pickup – He picked up a stone and threw it at the dog.
flung – He flung himself back on the sofa.
responsible – I am not responsible for your problem.
drew up – He drew up his car near the house.
familiar – He is a Familiar face here.
numerous – He faced a numerous problem to do his work.
resolve – He resolved to do better in studies.
episode – This episode is very attractive
excited – He got excited to see his enemy.
alter – She altered her mind to marry.
recognise – I can’t recognise my friend as he becomes.

C. The three forms of irregular verbs:

Present – Past – Past participle
abide – abode – abode
arise – arose – arisen
awake – awoke – awaken
be – was – been
bear – bore – borne /bom
beat – beat – beaten
become – became – become
befail – befell – befallen
beget – begot – begotten
begin – began – begun
behold – beheld- beheld
bend – bent – bent
bereave – bereft/bereaved – bereft/bereaved
beseech – besought – besought
bet – betted – betted
bid – bade – bidden
bind – bound – bound
bite – bit- bitten
bleed – bled – bled
blow – blew – blown
break – broke – broken
bring – brought – brought
bread – bred – bred
broadcast – broadcast – broadcast
build – built – built
bum – burnt – burnt
burst – burst – burst
buy – bought – bought
can – could – could
cast – cast – cast
catch – caught – caught
chide – chid – chid
choose – chose -chosen
cleave – cleft /cleaved – cleft /cleaved
cling – clung – clung
clothe – clothed – clothed
come – came – come
cost – cost – cost
creep – crept – crept
crow – crowed – crowed
cut – cut – cut
dare – dared / durst – dared /durst
deal – dealt – dealt
dig – dug – dug
do – did – done
draw – drew – drawn
dream – dreamt – dreamt
drink – drank – drunk
drive – drove – driven
dwel – dwelt – dwelt
eat – ate – eaten
fall – fell – fallen
feed – fed – fed
feel – felt – felt
fight – fought – fought
find – found – found
flee – fled – fled
fling – flung – flung
fly – flew – flown
forbear – forbore – forborn
forbid – forbade – forbidden
forget – forgot – forgotten
forgive – forgave – forgiven
forsake – forshook – forsaken
freeze – froze – frozen
get – got – got
gild – gilt/glided – gilt/glided
gird – girt – girt
give – gave – given
go – went – gone
grind – ground – ground
grow – grew – grown
hang – hung – hung
here – had – had
hear – heard – heard
hew – hewed – hewn
hide – hid – hidden
hit – hit – hit
hold – held – held
hurt – hurt – hurt
keep – kept – kept
kneel – knelt – knelt
know – knew – known
lay – laid – laid
lead – led – led
lean – learnt – learnt
leak – leapt – leapt
learn – learnt – learnt
leave – left – left
lend – lent – lent
let – let – let
lie – lay – lain
light – lit – lit
lose – lost – lost

Section- I

Pre-reading Activity:
Have you ever given thought to why people commit suicide? Give it a thought now and list possible reasons here: What is your view of the people who commit suicide?
Focussing Questions: In this unit you will read a story dealing with a case of attempting suicide. As you read story the focusing questions you will bear in mind are:
(i) Why didn’t the girl want to live?
(ii) It is a justifiable reason for committing suicide?

The Writer:
R.K. Narayan who passed away on 13th May 2001 at the age of 94 is undoubtedly one of the greatest Indian writers in English having to credit a number of novels, volumes of short stories, travelogues, eclogues, retold legends and an autobiography. “Swamy and Friends”‘The Bachelor of Arts”, ‘The English Teacher”. Mr. Sampath, ‘The Guide”, “The Vendor of Sweets”, “Waiting for Mahatma”, “The Man-eater of Malgudi”, “A Tiger of Malgudi: are some of the important novels.

“An Astrologist Day and Other Stories”, “Lawly Road and Other Stories”, include his best known stories “MyDateless Dinner” is a travelogue and‘My Days” is the writer’s autobiography. Having an intimate knowledge of Indian life and society be set with orthodoxy and superstition, he becomes India’s great story teller, an astute observer of Indian psyche growing through the pre as well as post-independence times. His characters, mostly belonging to lower middle class and poorer sections, exhibit an uncanny sense of fear in their actions and behaviour the fear and uncertainty teeming out foreign domination and servile attitude of the people who suffer sometimes they are the victims of social evils and a corrupt order of which his writing are a humours and satiric exposition. Irony at time lapses into cynicism in his writings. Narayan will five for ever for his beautiful imaginary town ofMalgudi faithfully representing Indian ethos, milieu and sensibility.

The Story:
The story extracted from “Malgudi Days” enacts an incident of a late evening on attend bound which ultimately brings into focus a contrast between characters- a girl and a watchman. The girl is ambitious, believes in self-reliance and detests being a liability or living on charity. 1 don’t live on anybody’s charity” she says. To the watchman’s suggestion of accepting marriage she hysterically reacts. “No no … I don’t want to marry. I want to study. “But a deeper look into her character reveals that her cowardly thought of suicides is not in harmony with the bold statements she makes before the watchman like most of Narayan’s characters, she being poor suffers from a fear of uncertainty. She lacks a strong willpower to arrange watchman demonstrates a great understanding of man’s predicament and enduracne to encounter boldly the unslaughts of misfortune. The story ends with a note of ambiguity and uncertainty as regards the identity of the women. With its element of irony understanding of cynicism the story is typical Narayan’s fictional writing.

Gist:
Paragraphs (1-2)
There was still a faint splash of red on the western horizon. The watchman stood on the tank bund and took a final survey. All the people who had come for evening walks had returned to their homes not a soul anywhere – except that obstinate angler, at the northern and who set with his feet in water, sadly gazing on his rod. It was no use bothering about him, he would sit there still midnight, hoping for a catch.
The Taluk office struck nine. The watchman was satisfied that no trespassing cattle had sneaked in though the wire fencing. As he turned to go he saw about a hundred yards away a shadow figure moving down the narrow stone steps that led to the water’s edge. He thought for a second that it might be a ghost. He dismissed the idea and went up to investigate if it was anyone who came to bath at this hour. He observed that it was a woman’s form. She stopped over the 1st step and placed something on it – possibly a letter. She then stopped into knee deep water and stood there; hands pressed together in prayer.

Gist:
Paragraphs (3-6)
He shouted for coming. He raced down the steps and picked up the letter. He hurriedly lit the lamp and the light fell upon the other’s face. It was a young girl’s wet with tears. He told her to be seated. He sat down on the last step between her and the water and placed the lantern on the step, took out a piece of tobacco and put it in his month. She began to sob. He asked why she did not go home. She said that she had no home in the world. The watchman enquired how she grew up without a home. She lost her mother when she was only five. She grew under the i care of her stepmother. He said that he was sixty – five and asked whether her step mother troubled her. She replied in negative she was well looked after. He told her to leave as it was late but she retorted that she had not home to go to. But she went to saying why she should go to become a burden to her again. She never to live on anybody’s charity.

Gist:
Paragraphs (7-9)
He suggested her waiting till her mother was able to find her a husband. She glared at him in the dark. But she did not with to go that either, she wanted to study and became a doctor and earn her livelihood. She never wanted to marry. She often catch mother talking for into the right to her eldest son, worrying about her future and about her marriage. She knew that they could not afford to keep her in college very long because it was very expensive. It cost twenty rupees. The watchman exclaimed “twenty rupees”. Because it was his monthly salary, she thought she would get a scholarship. But it was announced that evening that another not she was entitled to be given scholarship. She broke down as her name was not there. The watchman looked at her in surprise. He understood very little of all the situation. She was unhappy as someone was coming to see her the next day. She said again the she did not wish to many, she wished to study.

Analytical Outlines:

• There was still a faint splash of red on the western horizon.
• The watchman stood on the tank bund.
• He took a final survey.
• The people were in evening walk.
• They all had returned to their homes.
• The obstinate angler was at the northern end.
• He sat with his feet in water.
• He was sadly gazing at his rod.
• It was no use bothering about him.
• He would sit there still midnight.
• He hoped for a catch.
• The Taluk office struck nine.
• The wet man was satisfied.
• No trespassing cattle had sneaked in thought the wire fencing.
• He returned to go.
• He saw a shadowy figure.
• It was moving down the narrow stone steps.
• It led to the water’s edge.
• It was about a hundred yards away.
• He thought for a second.
• It might be a ghost.
• He dismissed the idea.
• He went up to investigate.
• It was anyone who came to bath at this hour.
• He observed that it was a woman’s form.
• She stopped over the last step.
• He placed something on it.
• It was possibly a letter.
• She then stopped into knee deep water.
• She stood there.
• Her hands pressed together in prayer.
• He shouted for coming.
• He raced down the steps.
• He picked up the letter.
• He hurriedly lit the lamp.
• The light fell upon the other’s face.
• It was a young girl’s wet with tears.
• He told her to be seated.
• He sat down on the last step.
• It was between her and the water.
• He placed the lantern on the step.
• He took out a piece of tobacco.
• He put it on his mouth.
• She began to sob.
• He asked why she did not go home.
• She said that she had no home in the world.
• He enquired how she grew up without a home.
• She lost her mother.
• She was at the age of five then.
• She grew under the care of her step mother.
• He said that he was sixty-five.
• He asked whether her stepmother troubled her.
• She replied negatively.
• She was well looked after.
• He told her to leave.
• Because, it was late.
• But she retorted.
• She had no home to go to.
• But she said why she should go to.
• She did not want to be a burden to her again.
• She never liked to live on anybody’s charity
• He suggested her to wait.
• Till her mother was able to find her a husband.
• She glanced at him in the dark.
• But she did not wish to do that either.
• She wanted to study.
• She wanted to become a doctor.
• So that she would earn her livelihood.
• She never wanted to marry.
• She often catch her mother.
• She talked to her eldest son in night.
• She worried about her future.
• She also worried about her marriage.
• She knew that they could not afford to keep her in college very long.
• Because it was very expensive.
• It costs twenty rupees.
• The watchman exclaimed, ‘Twenty rupees”!
• Because, it was his month’s salary.
• She thought she would get a scholarship.
• But it was announced that evening.
• Another one was entitled to get it.
• So she was deprived of it.
• She broke down as her name was not there.
• The watchman looked at her in surprise.
• He understood very little of all the situation.
• She was unhappy.
• Because, someone was coming to see her next day.
• She said again that she did not want to marry.
• She wished to study.

Meaning Of Difficult Words:

splash – small area of bright colour
trespassing – entering unauthorized
gruesome – horrible, ghastly
sputter – to speak indistinctly
hysterical – violently emotional
survey investigation
obstinate – stubborn, opinionated
bothering – brooding over, thinking
trespassing – entering without permission
sneaked in – secretly entered
gruesome – causing fear
gaze – look, discern
hurriedly – in a hurry, quickly
glared at – looked strongly and boldly at
exclaimed – surprised, wondered, astonished
comprehended- understood felly.

Section – II

Gist:
Paragraphs (11 – 12)
The silent night was stabbed by her sobbing and some night birds rustled the water and wavelets bear upon the shore. Seeing her suffer, he found his own sorrows in life came to his mind; how in those far-off times, in his little village, home an epidemic of cholera laid out his father and mother and brothers on the someday and he was the sole surviver. How he was turned out of his ancestral home through the trickery of his father kinsmen and he wandered as an orphan suffering hunger and privation. He said that everyone has his own miseries. If people tried to kill themselves for each one of them, he didn’t know how often they would have to drown. He remembered further incidents and his voice shook with sorrow. He remained silent and sob broke out of him as he said. He prayed to all the gods in the world for a son. His wife bore him eight children. Only one daughter lived, and none of the others saw the eleventh year. The girl looked at him with bewilderment.

Gist:
Paragraphs (13 – 14)
The Taluk gong struck again and he said that she had better get up and go back home. But she replied that she had no home. He felt irritated. He said that she should not be so obstinate and leave the place as soon as possible. The girl said that the watchman did not know her trouble. He picked up his lantern and staff and got up. He put her letter down where he found it. He warned her that she was becoming very obstinate. And he was leaving her there alone. He should not blame. He paused for a moment, looked at her and went up the steps, not a word passed between them again.

Gist:
Paragraph (15)
The moment he came back to duty next morning, he hurried down the stone steps. He picked up the letter and gazed on it, helplessly, wishing that it could tall him about the fete of the girl after he had left her. He tore it up and flung it on water. As he watched the bits float off on ripples, he blamed himself for leaving her and going away on the previous night. He said that he was responsible for at least one suicide in that rank. He could never look at the blue expense of water again with an easy. Even many months later he could not be certain that he remains if a body would not come up all of a sudden.

Gist:
Paragraph (16)
Years, later, one evening as he stood on the bund and took a final survey before going home, he saw a car drew up on the road below. Aman, woman and three children emerged from the care and climbed the bund. When they approached, the watchman felt a start at his heart, the figure and face woman seemed familiar to him. Though the woman was altered by years and ornaments and dress, be brought that he had then recognized the face he had once seen by the lantern light. He felt excited at his discovery. He had numerous questions to ask. He brought together his palms and saluted her respectfully. He expected she would stop and speak to him. But she merely threw at himan indifferent glance and passed on. He stood staring after her for a moment. Baffled, he said to himself that perhaps she was someone else and turned to go home resolving dismiss the whole episode from his mind.

Analytical Outlines:

• The silent night was stabbed by her sobbing.
• Some night birds rustled the water.
• Wavelets beat upon the shore.
• He saw her suffering.
• He found his own sorrows in life.
• It automatically came to his mind.
• It was the happening of that for off off times.
• His home was in a little village.
• An epidermic of cholera laid out his father.
• It also laid out his mother and brothers.
• All it happened on his same day.
• He was the sole survior.
• He was turned out of his ancestral home.
• It happened due to the trickery of his father’s kinsmen.
• He wandered as on orphan.
• He had to undergo suffering hunger and privation.
• He said that everyone has his own miseries.
• But people did not kill themselves for this.
• Rather they struggle for existence.
• He remembered further incidents.
• His voice shook with sorrow.
• He remained silent.
• A sob broke out of him.
• He prayed to all the gods in the world.
• He prayed for a son.
• His wife bore him eight children.
• Only lived one daughter.
• None of the others saw the eleventh year.
• The girl looked at him with bewilderment.
• The talking struck again.
• He said that she had better got up.
• She should go back home.
• But she replied that she had no home.
• He felt irritated.
• He said that she should not be so obstinate.
• He left the place immediately.
• The girls said that the watchman did not know her trouble.
• He picked up his later and got up.
• He put her letter down where he found it.
• He warned her that she was becoming very obstinate.
• He left her there alone.
• He should not be blamed.
• He paused for a moment.
• He looked at her.
• He went up the steps.
• Not a word passed between them again.
• Next morning, he came back to his duty.
• He hurried down the stone steps.
• He picked up the letter and gazed on it.
• He helplessly wished something.
• It could tell him about her fete.
• He tore it up.
• He flung it on water.
• He watched the bits float offon ripples.
• He blamed himself for leaving her.
• He went away in that previous night.
• He said that he was responsible for at least one suicide.
• He could never look at the blue expanse of water again with an easy mind.
• Even many months later, he could not be certain.
• The remains of a body would not come up all of a sudden.
• Years later one evening, he stood on the bund.
• He took a final survey before going home.
• He saw a car draw up on the road below.
• Aman, woman and three children emerged from the car.
• They climbed the bund.
• They approached towards the watchman.
• The face of the woman seemed familiar to him.
• The woman was altered by years ornaments and dress.
• Still then he recognised her face.
• He had once seem that face by the lantern light.
• He felt excited at this discovery.
• He had a lot of questions to ask.
• He saluted her respectfully.
• He expected she would stop and speak to him.
• But she only threw at him an indifferent glance.
• Then she passed on.
• He stood staring after her for a moment.
• He baffled and said that perhaps she was someone else.
• She turned to go home.
• So, she resolved to dismiss whole episode from his mind.

Meaning Of Difficult Words:

rustle – to produce a soft whispering sound (as of dry leaves) by stirring.
wavelets – small waves, ripples.
privation – state of being deprived of something, especially of what is necessary for comfort.
staff – a stick carried in the hand as a support.
sobbing – weeping, lamenting with tears in eyes.
wandered – travelled, roamed, moved about
expanse – a vast stretch of
emerged from – came out of, emanated
numerous – innumerable, a number of
muttered – made indistinct sound
episode – chapter

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Ex 12(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Exercise 12(c)

Question 1.
Each question given below has four possible answers out of which only one is correct. Choose the correct one.
(i) (î + k̂) × (î + ĵ + k̂) = ______.
(a) î – k̂
(b) k̂ – î
(c) k̂ – 2î – ĵ
(d) 2
Solution:
(î + k̂) × (î + ĵ + k̂) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= î (0 – 1) – ĵ (1 – 1) + k̂ (1 – 0)
= -î + k̂ = k̂ – î

(ii) A vector perpendicular to the vectors î + ĵ and î + k̂ is ______.
(a) î – ĵ – k̂
(b) ĵ – k̂ + î
(c) k̂ – ĵ – î
(d) ĵ + k̂ + î
Solution:
A vector perpendicular to the vectors î + ĵ and î + k̂ is
(î + ĵ) × (î + k̂) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 0 \\
1 & 0 & 1
\end{array}\right|\)
= î (1 – 0) – ĵ (1 – 0) + k̂ (0 – 1)
= î – ĵ – k̂

(iii) The area of the triangle with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1) is ______.
(a) \(\frac{1}{2}\)
(b) 1
(c) \(\frac{\sqrt{3}}{2}\)
(d) 2
Solution:

(iv) If â and b̂ are unit vectors such that â × b̂ is a unit vector, then the angle between â and b̂ is ______.
(a) of any measure
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{2}\)
(d) π
Solution:
|a × b| = ab sin θ = sin θ
⇒ sin θ = 1
⇒ θ = \(\frac{\pi}{2}\)

(v) If \(\vec{a}, \vec{b} \text { and } \vec{c}\) are non-zero vectors, then \(\vec{a} \times \vec{b}=\vec{a} \times \vec{c}\) ______.
(a) \(\vec{b}=\vec{c}\)
(b) \(\vec{a} \|(\vec{b}-\vec{c})\)
(c) \(\vec{b} \| \vec{c}\)
(d) \(\vec{b} \perp \vec{c}\)
Solution:

Question 2.
Let \(\vec{a}\) = 2î + ĵ, \(\vec{b}\) = -î + 3ĵ + k̂ and \(\vec{c}\) = î + 2ĵ + 5k̂ be three vectors. Find
(i) \(\vec{c} \times \vec{a}\)
Solution:

(ii) \(\vec{a} \times(-\vec{b})\)
Solution:

(iii) \((\vec{a}-2 \vec{b}) \times \vec{c}\)
Solution:

(iv) \((\vec{a}-\vec{c}) \times \vec{c}\)
Solution:

(v) \((\vec{a}-\vec{b}) \times(\vec{c}-\vec{a})\)
Solution:

Question 3.
Find the unit vectors perpendicular to the vectors
(i) î, k̂
Solution:

(ii) î + ĵ, î – k̂
Solution:

(iii) 2î + 3k̂, î – 2ĵ
Solution:

(iv) 2î – 3ĵ + k̂, -î + 2ĵ – k̂
Solution:

Question 4.
Determine the area of parallelogram whose adjacent sides are the vectors
(i) 2î, ĵ
Solution:

(ii) î + ĵ, -î + 2ĵ
Solution:

(iii) 2î + ĵ + 3k̂, î – ĵ
Solution:

(iv) (1, – 3, 1), (1, 1, 1).
Solution:

Question 5.
Calculate the area of the traingle ABC (by vector method) where
(i) A (1, 2, 4), B (3, 1, -2), C (4, 3, 1)
Solution:

(ii) A (1, 1, 2), B (2, 2, 3), C (3, -1, -1).
Solution:

Question 6.
Determine the sine of the angle between the vectors
(i) 5î – 3ĵ, 3î – 2k̂
Solution:

(ii) î – 3ĵ + k̂, î + ĵ + k̂
Solution:

Question 7.
Show that \((\vec{a} \times \vec{b})^2\) = a²b² – \((\vec{a}, \vec{b})^2\)
Solution:

Question 8.
If \(\vec{a} \times \vec{b}=\vec{b} \times \vec{c} \neq \overrightarrow{0}\), prove that \(\vec{a}+\vec{c}=m \vec{b}\), where m is a scalar.
Solution:

Question 9.
If \(\vec{a}\) = 2î + ĵ – k̂, \(\vec{b}\) = -î + 2ĵ – 4k̂, \(\vec{c}\) = î + ĵ + k̂, find \((\vec{a} \times \vec{b}) \cdot(\vec{a} \times \vec{c})\).
Solution:

Question 10.
If \(\vec{a}\) = 3î + ĵ – 2k̂, \(\vec{b}\) = 2î – 3ĵ + 4k̂ then verify that \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).
Solution:

Question 11.
Find the area of the parallelogram whose diagonals are vectors 3î + ĵ – 2k̂ and î – 3ĵ + 4k̂.
Solution:

Question 12.
Show that \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\). Interpret this result geometrically.
Solution:

= Vector area of the parallelogram ABCD.
Hence twice the vector area of a parallelogram ABCD is equal to the vector area of the parallelogram whose adjacent sides are the diagonals of the parallelogram ABCD.

Odisha State Board CHSE Odisha Class 12 Approaches to English Book 2 Solutions Short Stories Chapter 2 The Tree Textbook Exercise Questions and Answers.

CHSE Odisha 12th Class Alternative English Solutions Short Stories Chapter  2 The Tree

Section – I

Questions For Discussion

Question 1.
How does the writer describe the river in the spate? Did it cause havoc in the village: What was the time of the flood?
Ans:
When the river was in full spate, it bulged up sounding like a thousand hooded hissing cobra. It did not play havoc in the village. The flood came a little past midnight.

Question 2.
What did the villagers do to meet the crisis?
Answer:
As people suddenly realized that the situation was much more grave than they had imagined they raised their lanterns. They ran to take shelter under the banian tree.

Question 3.
The leaves chattered incessantly in their familiar language of hope and courage. How does this statement portray the banian tree?
Answer:
The leaves of the banian tree made sounds as they usually do. Inspiteofthe approaching danger, they behaved in a similar way without paying any attention to the flood and its associated dangers. They were a symbol of courage and boldness which attribute the same qualities to the banian tree.

Question 4.
What were the story behind the mound and the tree?
Answer:
The mound contained the ruins of a certain king’s palace. The king dared to cut down a few branches of the tree to make room for his palace. Perhaps he planned to destroy the tree, but before he could do so a terrific storm had broken out. The palace collapsed and formed a mound. The tree, it was said had taken off down to the Himalayas or other such meaningful places at the command of a certain great soul who lived under it.

Question 5.
Write briefly about the “banian goddess”. What does this portrait reveal about the village? Do you make an element of humor and satire in the description?
Answer:
The “banian goddess” had to regular priest attached to her. Whoever so desired could approach her and sprinkle vermilion on her. In course of generations, the vermilion crust had come to account for the greater part of the goddess’s body. Devotees usually did not prostrate to her, they bowed when they passed by complex and formidable matters were referred to the deities of distant temples whereas small issues were put forth before her. Children in particular found her quite helpful in regard to crises arising out of undone home works or the ill-humored primary school pundits. It works with an element of humor and mild satire in the description.

Question 6.
How does the author describe the different trunks of the banian tree? Are the descriptions given in a serious or light-hearted vein? Do they reveal the orthodoxy and superstition of the villagers?
Answer:
At the foot of one of the trunks rested the tiny “banian goddess”. The revered and sacred bull of the village relaxed beside another trunk, eyes shut and jaws moving. An old woman from the neighboring village on her way back from the bi-weekly market set learning against another trunk. In a hollow at the foot of another trunk raised a family of snakes that had earned the reputation of being conscientious and harmless. The description reveals the orthodoxy and superstition of the villagers.

Question 7.
How was the tree godly? What did it symbolize?
Answer:
The tree was taken to be immortal by all without anybody having to be told about it. Immortality being an attribute of the gods, it was godly. Nobody would flaunt a decision that had been arrived at in a meeting under the tree. There was the seal of some power invisible and inaudible.

Vocabulary
Make nouns from the following:

terrible
collective sensible
advise
patient
remote
syllabic
strong courageous
protect
generate
respond expect
helpful dark
aware
full
cloudy
starry
descend
scandalize
expand crazy
Answer:
Words – Noun Forms
terrible – terror
collective – collection
aware – awareness
full – fullness
cloudy – cloud
starry – star
descend – descent
scandalize – scandal
expand – expansion
crazy – craze
respond – response
helpless – helplessness
dark – darkness
sensible – sensibility
advise – advice
patient -patience
remote – remoteness
syllabic – syllable
strong – strength
courageous – courage
protect – protection
generate – generation
expect – expectation

Section – II

Questions For Discussion

Question 1.
What is the central incident in the section?
Answer:
The banian tree being swept away by the flood and its fall and uprooting consist in the central incident in this section.

Question 2.
What attitude of the villagers is reflected in the words of the retired head pundit? How does he interpret the fleeing of the birds and snakes from the tree?
Answer:
The retired head-pundit said, “far from the good sign, snakes, and birds fleeing this great shelter”. It indicates the head pundits’ superstitious attitude. Fleeing of the birds and snakes from the tree had taken place owing to the human sins to the excess. They were going to face the consequences for their ill deeds.

Question 3.
How do the villagers react to the words of the college-going young men? What is distinctive about the attitude of the young men
Answer:
When the college-going young men said that the banian tree was felling which would take away a chunk of earth. But the villagers reacted against it telling that they were studying in college, and would they be able to save the tree with their English, Algebra, and all that abracadabra.

Question 4.
What does Srikant Das, the Vaishnav, observe about the impending fall of the tree? Does it reflect the orthodoxy and superciliousness of the villagers?
Answer:
Srikant Das, the Vaishnav observed that not only those boys but everybody in the village has his share of sin. And if the tree was going to collapse it was because it could not bear the burden of its sins. It reflects the orthodoxy and superciliousness of the villager.

Question 5.
Do you find a contrast between the two generations in this section? Explain
Answer:
Two generations such as the old and young are contrasted in this section. The old cling to the ancient orthodox values and the young college students provide a reaction of the student against old and hackneyed beliefs.

Grammar

Fill in the blanks with the right verbs, (for practice test)
l. You ________ to read this book. It is really very fine.
2. You _______ came to work on time.
3. __________ I came in Sir?
4. He is driving recklessly, he_________ face an accident.
5._________ I speak to Mr. Panigrahi.
6. You________ honour your superiors
7______ I see your camera
8. My dear son you ________ do what your teacher says
9. Come soon, there ________ be a crowd here.
10 I __________ read this book well.
11. You_______ take my umbrealla.
12. All the doors_______ be closed because storm appears.
13.______ you say so?
14. You should buya camera now, press_________ go up.
15. As he was strong he____________ swim for 3 hours.
16. I ________ never seen you again
17. You _______ be here by six.
18. It _________ rain, you’d better take an umbrella.
19. He ___________ be on the next bus
20. He__________ to go to river bank.
21. You___________ comb your hair so stylistically.
22. He said that it _________ rain
23. You_______ tell her about it. He knows it.
24. When I was a boy I_________ understand all.
25_________ you read this letter?
26. You _________ be here in time because your presence is highly essential
27. You_________ reach here in time.
28. Students________ not enter into the room with a book
30_______ I borrow your umbrella?
31. When I first went to London. I ______ easily understand them
32. I_______ prepare a lesson plan for you
33. I suggest that you _____ go for a picnic.
34. We ___________ play here.
35. It is natural that you__________ behave like this.
36. A leader ________ be a man of good character
37. He__________ reach here after a few minutes.
38. You ___________ stop drinking, otherwise, death is sure
39.________ came for this insult?
40. I_________ lend you fifty thousand.
41. You _______ use my bicycle
42_________ we go for a picnic?
43. You_________ run so fast because you have much time to spare.
44. I _________ drive this car
45. I wish you _________ wait five minutes for me.
46. She_________ be waiting me there.
47. You __________ be punished.
48________ you help me lifting this stone?
49. You _________ be punished.
50. When he was a child, he _________ get set up early in the morning.
51. We__________ to help the helpless.
52. Work had lest you __________ fail.
53. You_________ read this play.
54.___________ I work for you?
55. He_________ eat all the sweets.
56._________ I discuss with him
57. __________ that I was handsome, I would win her heart
58.__________ I came in?
59. It is too late you ___________ be hurried.
60. An accident __________ happen
61. I _________ help you.
62. You stop _______ here because there is danger ahead.
63. _______ you pass that tea?
64. I_________work for him
65. ________ you show me the way to the Ashoka Hotel?
66._______ you shut that door?
67._________ I carry your bag?
68. I wish you ______ success in life
69. How __________ I distribute the sweets?
70. You __________ read regularly.
71._________ his soul get peace.
72.__________ you like to stay with us?
73. ________ 1 have a glass of water?
74.___________ you have a journey with me?

Section – III

Questions For Discussion

Question 1.
What is the primary concern of the villagers in this section? In which context did Manoj Das bring in the names of some important villagers? What tone do you mark in the description?
Answer:
The primary concern of the villagers in this section is to protect them as the representatives of peoples of that sort. Manoj Das brings in the names of important villagers to let the readers know how much individuals take the lead in superstitious attitudes. There is a tone of mind satire in the description.

Question 2.
What did the old Brahmin do when the tree was about to be swept away by the flood?
Answer:
The old Brahmin rushed to the remnants of the tree. He sat down on the muddy ground a spot which had been considered dangerously unsafe even by the snakes and mustering all his strength pulled up the small stone that has stuck to the spot. Holding the unrooted crowd that watched him breathlessly.

Question 3.
How did the villagers behave when the tree was about to be swept away by the flood?
Answer:
When the tree was gone, the villagers were excited to utter, “Haribol, Haribol”. They kept up the poignant chant with all their hearts all looking stupefied and some aping.

Question 4.
How does the story end? Is it a satisfactory ending?
Answer:
The story ends with the frantic attitude of the villagers towards the idea of superstition in connection with the appearance of the banian Goddess and the unreasoning belief of the people that the ending is not satisfactory ending. The story is brought to an end abruptly with the child’s words.

Questions For Composition

Question 1.
Give an account of the life and attitude of the villagers as portrayed in the story.
Answer:
The Story Tree” is written by Manoj Das an eminent, popular, and outstanding storyteller of Odisha. Actually, Manoj Das is a bilingual writer of international repute. Here, Das very interestingly depicts the life and attitude of the people of an Odishan village, their life governed by rituals, orthodoxy, and superstitions. In feet, this is Das’s most typical piece of writing which depicts the true and realistic picture of society. However, right from the time, the reason was on the verge of monsoons, the villager’s elders had begun to look grave.

The sinister cloud formation on the mountains several miles away and a widening of uncanny awareness around the moon had informed them that there were terrible days ahead. The villagers felt scandalized every time their familiar tame river expanded and looked alien and began hissing. It gave the sort of shock which are experienced when the domestic animal suddenly went craggy behaving widely and not responding to any amount of endearment. Moreover, the villagers had a strong belief. There was a mound that contains the rains of a certain king’s palace. It was neither possible nor necessary to recall the name of the king who had built it whether he had been of the solar or the lunar dynasty.

What was frequently recalled was that he had dared to cut down a few branches of the tree to make room for his palace. Perhaps he had planned to cut more, perhaps even to totally destroy the tree, but before he could do so a terrific storm had broken out. The palace collapsed. The king and his family took shelter under the tree and were saved. The king elapsed the tree and wept and the story was mitigated. Further back in time, it was said, the tree had taken off and flown to the Himalayas or other such meaningful places, at the behest of a certain great sound who lived under it. On the other hand, the villagers were also superstitious.

They had believed that at the foot of one of the trunks of the banian tree rested the tiny “banian goddess” who had a regular priest attained to her. Whoever so desired could approach her and sprinkle vermilion on her. In the course of generations, the vermilion crust had come to account for the greater part of the goddess’s body. Devotees did not ordinarily prostrate themselves to her, but everybody while passing before her bowed enough for her to take contingency of his or her devotion. In complex or formidable matters, the villagers prayed for the intervention of famous deities of distant temples.

But small issues were referred to her from time to time. Children, in particular, found her quite helpful in regard to crises arising from undone homework or the ill humor of the pundits of the primary school. Again, a Brahmin rushed to the felling tree and sat down on the muddy ground which had been considered dangerously unsafe even by the snakes, and getting all his might pulled up the stone that had struck the spot for God knows how many ages. Holding the uprooted goddess close to his bosom as though to protect her from invisible enemies, he returned to the crowd that watched him breathlessly.

Someone spread a towel on the grass. The Brahmin put down the Goddess and patted her. Bishnu Jena had seated himself before the banian Goddess, several people rushed to their homes and brought out cymbals and drums, and conch shells. These had to be played close to his ears as loudly as possible. He began by shivering. Bishnu Jena was thought to have been possessed. He said, “I will be born as a thousand trees here, there and everywhere” Such was the life and attitude of the villagers as portrayed in the story. As a matter of feet, Manoj Das’s portrayal of village life and attitude is superb and fantastic. It is realistic, elaborate, vivid, and outstanding. On the whole, the storyteller’s description is inspiring and thought-provoking.

Question 2
Discuss how the tree is an integral part of the life of the villagers.
Answer:
The short story Tree” is undoubted, the most typical masterpiece of Manoj Das, a prominent, famous, and outstanding angle – Indian writer of this present era. This story woven around an age-old banian tree of a village typically portrays the behavioral pattern and psychic responses of its inhabitants at a critical moment when the tree is uprooted and carried away by the river in spate. ‘ However, trees are a part of rustic life. This story also embodies the same things associated with a banian tree which was an age-old tree and had seen many generations.

The story, however, revolves around the age-old banian tree of a village. The story writer very interestingly depicts the life and attitudes of the people of an Odishan village, their life governed by rituals, present and future is intimately connected with the villagers and other life pattern. They cannot bear the nerve-racking incident of the tree being swept away by the flood. They even think of the rebirth of the tree in thousands the discussion among the different categories of villagers is really amusing and humorous. The elements of ivory and satire are discernible even on the occasion of a terrible loss to the village. However, there was a mound that contained the rains of a certain King’s palace.

It was neither possible nor necessary to recall the name of the King who had built it or whether he had been of the solar or the lunar dynasty. What was frequently recalled was that he had dared to cut down as few branches of the tree as to make room for his palace. Perhaps he had planned to cut nappe, perhaps even to totally destroy the tree, but before he could do so a terrific storm had broken out. The palace collapsed. The king and his family took shelter under the tree and were saved. The King clasped the tree and wept. The storm was averted. Moreover, many years back, it was said, the tree had taken off and flown to the Himalayas or other such meaningful places, at the behest of certain great souls who lived under it.

But that was an era of truth, and in the absence of some concrete evidence like the mound to support this legend, elders of the present generation spoke relatively less about it, and branches spreading over an acre resting on these trunks had become an institution long ago. Of course, at the foot of one of the trunks rested the tiny banian goddess. She had no regular priest attached to her. Whoever desired could approach her and sprinkle vermilion on her In the course of generations, the vermilion crust had come to account for the greater part of the goddess’s body. Children, in particular, found her quite helpful in regard to crises arising from undone homework or the ill humor of the pundits of the primary school.

The area before another neighboring the reversed sacred bull is used to relax eyes shut and jaws moving. An old woman coming from a village on the horizon sat leaning against another trunk with a sack half filled with greens and drumsticks. In the hollow at the foot, of another trunk resided a family of snakes which have earned the reputation of being conscientious and harmless above, rested a legion of birds. Again, the tree was taken to be immortal by all without anybody have been to be told about it. Immortality being an attribute of the gods, it was goodly.

Nobody would easily flout a decision that had been arrived at in a meeting under the tree, for even when the decision was unpalatable to the party, it knew that behind it, there was the seal of some power, invisible and inaudible thoughts. As a matter of feet, the banian tree plays a most integral and vital part in the story. It serves as the inevitable part of the life of the villagers. The tree provided a lot of emotional and philosophical sustenances that the villagers had derived throughout their life. The way Manoj Das has presented it in the story is superb and fantastic on the whole, the story is inspiring, elevating, thought-provoking, and heart-touching.

Question 3.
Describe the feelings and reactions of the villagers when the age-old tree is swept away by the flood.
Answer:
In fact, ‘The Tree” is Manoj Das’s best typical masterpiece. Like his other stories, here Das substantiates a vivid and elaborate realistic picture of village society where the banian tree cures the lives of the villagers to a great extent. However, the storytellers’ presentation of a banian tree with so much concern for the villagers is superb and fantastic. It serves as a vitally integral part of their day-to-day life. However, the short – story by Manoj Das portrays the feelings and reactions of the villagers when the age-old banian tree of a village, typically portrays the behavioral pattern and psychic responses of its inheritance at a critical moment when the tree is uprooted and carried away by the river in spate.

The story writer very interestingly depicts the life and attitudes of the people of an Odishan village, their life governed by rituals, orthodox, and superstitions. The old tree symbolizing the past, present, and future is intimately connected with the villagers and their life patterns. They cannot bear the nerve-racking incident of the tree being swept away by the flood. Theyeventhinkofthe rebirth of the tree in thousands. The discussion among the different categories of villagers is really amusing and humorous. The elements of irony and satire are discernible even on the occasion of a terrible loss to the village.

Moreover, as the sky in the east grew brighter it was observed that the ground between the tree and the river had already tilted towards the river. The young men tried to appear engrossed in discussing something highly sophisticated. “Not only these boys, but we all have our shares of sin,” he said. He asked one to confess his sin addressing the spirit of the tree, silently in hearts: But it sounded like a cry of lamentation. The crowd swelled rapidly. Almost all the villagers, women and children included were gathered there. Of course, the M.L.A. arrived there walking at a running space.

People asked him,’“Do you see the situation, M.L.A. Baboo? We are doomed !”The M.L.A. ordered to bring as much hope as they could but it was a force. The tree slowly went into the river. Everybody was pained and sorrowful struck to see the pathetic sight. Actually, the villagers had grown accustomed to the tree for years together. It has been for them sympathizers in their active life.

The sweeping away of the banian tree brought the village people a lot of emotional shocks. Hence, the villages had an instant love towards the banian tree and so its loss became extremely intolerable on their part. As a matter of fact, the way Manoj Das has presented the emotions and reactions of the villagers to the loss of the banian tree is most fascinating and elevating. In short, the description is, very much thought-provoking and heart-provoking.

Question 4.
Bring out the elements of humor and satire in the story.
Answer:
In feet, the short story, ‘The Tree” is true, the most typical masterpiece of Manoj Das, an eminent popular and outstanding Odishan storyteller of this current era. He has so far written a large volume of stories. His stories are realistic, ironic, humorous, and satirical in nature. This discussion story serves as a burning example expressing these qualities. However, the story “The Tree” by Manoj Das is a masterpiece that is replaced with humor and satire. There is a vein of satire and humor in the story throughout.

Actually, satire is a literary device designed for the other hand something which automatically arises laughter on the part of the reader. Both of them are used by the writer to depict the social follies and foibles and their rectification thereby. These are used to make a piece of writing interesting, enjoyable, and colorful. In this preview, this work is a satire on social and political issues. Mr. Das has pointed out the social evil by means of mild satire. Moreover, the trunk of the tree was the abode, of an Albanian goddess’. It has a satire on human attitude. Snakes and birds felt the spot was a sign of even women. This symbolized the approach of imminent danger. This harps on human superstition and people were shouting.

“Haribol” to save it from falling was another example of humor and satire. Sridhar Mishra, an eminent homeopath was able to save many from certain death. This was equally humorous. The approach of the M.L.A. is also a humorous and satirical expression of the political avenue. He ordered to fetch as much as possible but to no avail. Locating a stone on the spot and attributing it to godly qualities is another sample specimen of satire and humor. Satire and humor touch the zenith when people come with drums and cymbals to install the goddess amidst “Haribol”: Old Bishnu Jena had sealed himself before the banian goddess. He shook before the goddess and he was thought to have been possessed.

It is a mere expression of satire and humor. In this way, the entire; story is reminiscent of humor and satire. As a matter of fact, the humor and satire used by Mr. Das are superb and marvelous. The story writer has tried his level best to show the social follies and foibles through the use of mixed humor and satire throughout. On the whole, the story is most inspiring, enjoyable, elevating, and heart-touching.

Vocabulary
Derive nouns from the followings:

furious, diurnal
allocate, womanly
mental, distant
see, serve
live, handful
narrate, remain
systematic, sacred
young, conscientious
bright, excite
rotate, authoritative
diagnose, appear
fantastic, abandon
materialistic,declare
dutiful, extinguish
circular, grave
synonymous, extensive
antonymous, explain
popular, friendly
relax,
Answer:
Words – Noun Forms
furious – file
allocate – allocation
mental – mind
see – sight
live – life
narrate – narration
systematic – system
young – youth
bright – brightness
rotate – rotation
diagnose – diagnosis
fantastic – fantasy
materialistic – materialism
dutiful – duty
circular – circle
synonymous – synonym
antonymous – antonym
popular – popularity
relax – relaxation
diurnal – day
womanly -woman
distant – distance
serve – service
handful – hand
remains – remainder
sacred – sacredness
conscientious – conscience
excite – excitement
authoritative – authority
appear – appearance
abandon – abandonment
declare – declaration
extinguish – extinction
grave – gravity
extensive – extension
explain – explanation
friendly – friend

B. Indicate against each word (given below) the parts of speech they belong to:

stop, imply
wind, pity
touch, generation
now, tauntingly
suddenly, meek
splashed, mild
school, brightly
crept, collapse
hollow, sin
wriggled, silently
thousand, crowd
long, horizon
carry, helpless
huge, rapidly
develop, gathered
commented, throne
pleased, irony
Answer:
Words – Parts ofSpeech
stop – noun
wind – noun
touch – noun, verb
now – adverb
suddenly – adverb
splashed – verb
school – noun
crept – verb
hollow – noun
wriggled – verb
thousand – adjective
tong – adjective
carry – verb
huge – adjective
develop – verb
commented – verb
pleased – verb
imply – verb
pity – noun
generation – noun
tauntingly – adverb
meek – adjective
mild – adjective
brightly – adverb
collapse – verb
sin – noun
silently – adverb
crowd – noun
horizon – noun
helpless – adjective
rapidly – adverb
gathered – verb
throne – adverb
irony- noun

Section – I

Pre-Readingactivity:
Have you seen a banana tree or a pipal tree with its huge trunk and leafy branches spreading all around? Such a tree in a village is considered holy and sacred. The old and ageless tree standing at the end of the village since time immemorial has been a mute spectator of changes in culture and civilization. It has always been a part of a violent storm or a devastating flood? Here is a story about an old banian tree up-rooted by a heavy flood in a village in Odisha. As you read the first part of the story by Manoj Das, recollect the ways in which the banian tree is significant in the life of the villagers.

The Writer:
Manoj Das (Born – 1934), a devoted disciple ofSreeAurobindo, is a bilingual writer of international repute, writing in both English and Odia. Born in a small village in Odisha, he has a rich understanding of the life of his people, their rites and rituals, their orthodoxy and superstitions, “Cyclones” (a novel), “The submerged valley, and other stories, “A Bride inside a Casket and other Tales”, “Man who Lifted the Mountain and other Fantasies” are some of his important contributions to Indian English literature.

His Katha O Kahani won him the most prestigious SahityaAkademiAward. The world of his sort of story is not merely a world of shark reality but blended into a world of dreams and fantasy, which he creates at the psychic level. The elements of irony, humor, and satire add to the beauty and charm of his writings. The novelty and innovation both in theme and form exhibited in his fictional world have brought him immense fame. He now teaches English at the Sree Aurobindo International Centre of Education in Pondichery. He has been recently honored with the UtkalRatnaSamman for his outstanding contribution to literature.

The Story:
This story“The Tree” woven around the age-old banian tree of a village typically portrays the behavioral pattern and psychic responses of its inhabitants at a critical moment, when the tree is uprooted and carried away by the river in spate. Das very interestingly depicts the life and attitude of the people of Odishan village, their life governed by rituals, orthodoxy, and superstitions. The old tree symbolizing the past, present, and fixture is intimately connected with the villagers and their life patterns. They cannot bear the nerve-racking incident of the tree being swept away by the flood. They even think of the rebirth of the tree in thousands. The discussion among the different categories of villagers is really amusing and humorous. The elements of irony and satire are discerning, even on the occasion of a terrible loss to the village.

GIST:
Paragraphs1-4
The village elders have begun to look at graves right from the time the season was on the brink of monsoons. The formation of clouds on the mountains end wick ring of natural aura around the moon had informed them that there were terrible days ahead. The flood approached at late midnight. People called out to each other and were reassured of collective awareness and gathered on the river bank with lanterns or torches of dirty twigs. The flames began dancing in the air. The moon was clouded and the stars looked pale. The river came up hissing like a thousand hooded cobra. Floodwater never entered this village although hardly a season passed without the river playing havoc with the villagers a couple of miles downstream.

Paragraphs5-11
The villages felt scandalized every time their familiar tame river expanded and looked alien and began hissing. They suddenly realized that the situation was much more grave than they had imagined. They raised their lanterns. Boats generally came from the forest at the foot of the mountains where they went to collect timber.
The wind grew stronger and cooler and was soon accompanied by a thin shower. All ran to take shelter under the banian tree. The wicks of the lanterns were turned low. The elders would point to a mound covered with grass and shrubs, not far from the tree white living the ancient most proof of this fact. The mound had decayed through centuries but it was still “as high as two men”.

Paragraph 12
The mound contained the ruins of a certain king’s palace. It was neither possible nor necessary to recall the name of the king who had built it or whether he had been of the solar or lunar dynasty. What was frequently recalled was that he had dared to cut down a few branches of the tree to make room for his place. Perhaps he had planned to cut more even to totally destroy the tree. But before they could do so a terrific storm had broken old. The palace collapsed. The king and his family took shelter under the tree and were saved. The king clasped the tree and wept. The storm subsided.

GIST:
Paragraphs 13-15
It was again said that the tree had taken off and flown to the Himalayas or other such meaningful places at the order of a great soul who lived under it. The trunk that had once been clasped by the king had decayed and disappeared since time immemorial other sending down numerous shoots which have formed new trunks. The tree was resting on those trunks. At the foot of one of the trunks rested the tiny “banian goddess”. No regular priest had gotten attached to her. Whoever so desired could approach her and sprinkles vermilion on her. Devotees ordinarily did not prostrate themselves to her but everybody passing by usually bowed, complex and formidable problems were put before the deities or distend temples but small issues were referred to her from time to time. Children in particular found her quite helpful in regard to crises arising from undone homework or the ill humor of the pundits of the primary school

Paragraphs16- 19
The area before another trunk was the usual site for the village meetings. Relaxing beside a neighboring trunk, eyes shut and Jews moving in a leisurely rhythm, could be found the much revered sacred bull of the village. In the afternoons of the bi-weekly marked days, on old, women from a village on the horizon set leaving against another trunk with a sack half-filled with greens or drumsticks.

At sunsets, they would rise and offer a handful of whatever still remained in her sack to the scared bull. In a hollow at the front of another trunk resided a family of naked which had earned the reputation of being conscientious and harmless and in the branches above rested a legion of birds. The tree was taken to be immortal. Immortality is an attribute of gods. It was gods. Nobody would easily flout a decision that had been arrived at in a meeting under the tree even when the decision was unpleasant to a party.

Analytical Outlines

• The village elders had begun to look grave.
• The season was on the brink of monsoon.
• There was the formation of clouds in the mountains.
• There was a wide ring of natural aura around the moon.
• All those had informed them that there was a terrible day’s moon.
• The flood approached late midnight.
• People called out to each other.
• They reassured of collective awareness.
• They gathered on the river bank.
• They gathered with lanterns or arches of dry twigs.
• The flames began dancing in the air.
• The moon was clouded.
• It looked pale.
• The river came up hissing like a thousand hooded cobra.
• Flood waters never entered this village.
• The season passed without the river playing havoc.
• The villagers felt scandalized every time.
• Their familiar tame river expanded.
• It looked alien.
•  It began hissing.
•  They had imagined the situation.
•  It was rather much more grave.
• They raised their lanterns.
•  Boats generally came from the forest.
• They had gone to collect timber from it.
• The wind grew strongest and cooler.
• Then it was accompanied by a thin shower.
• All ran to take shelter under the banian tree.
• The wicks of the lanterns were turned low.
• The elders would point at a mound.
• It was covered with grass and shrubs.
• It was not far from the tree.
• The mound had decayed through centuries.
• But it was still “as high as two men”.
• The mound contained the ruins of a certain king’s palace.
• It was not possible to recall certain kings’ names.
• That king had built it.
• He might belong to the solar or lunar dynasty.
• It is not remembered by him.
• The king had dared to cut down a few branches.
• He made room for his palace by that.
• Perhaps he had planned to cut more.
• Even he had planned to totally destroy the tree.
• But a terrific storm had broken out.
• The palace collapsed.
• The king and his family took shelter under it.
• They were saved.
• The king elapsed the tree and wept.
• The storm subsided.
• It was again said that the tree had taken off.
• It had flown to the Himalayas. Or it had flown to other meaningful places.
• It is said that a great would have lived under it.
•  The trunk clasped by the king had decayed.
•  It had even disappeared.
•  Numerous shoots had been produced.
•  They had formed new trunks.
•  The tree was resting on these trunks.
•  At the foot of one of the trunks rested the tiny “banian goddess
•  No regular priest had gotten attached to her.
•  Whoever so desired could approach her.
• He could sprinkle vermillion on her.
• Devotees ordinarily did not prostrate themselves to her.
• But everybody passing by usually bowed.
• They put complex problems before her.
• They also put formidable problems.
• But small issues were referred to her from time to time.
• Children in particular found her quite helpful.
• They approached her for their crises arising from under homework.
• She is helpful for the ill humor of the pundits of the primary school.
• The area before another trunk was the usual site for village meetings.
• The neighboring trunk could be found much reversed sacred bull of the village.
• In the afternoon of the bi-weekly market days, an old woman sat leaning against another trunk.
• She was with a sack half-filled with greens or drunk sticks.
• She would rise in summer.
• She would offer a handful of things from the sack to the sacred bull
• A family of snakes resided in the hollow of another trunk.
• It had earned the reputation of being conscientious and harmless.
• A legion of birds rested above this branch.
• The tree was taken to be immortal.
• Immortally being an attitude of gods.
• It was godly.
• Nobody would easily flout a decision.
• It had been arrived at in a meeting under the tree.
• The decision was unpleasant to a party.

Meaning Of Difficult Words

sinister – that which makes you feel that something evil is going to happen.
uncanny – unearthly, supernatural
draped – covered
hiss – to make a sibilant sound, to condemn by hissing.
hood – expanded head of a snake.
havoc – a devastation.
scandalize – make a malicious gossip
expand – to spread out, to develop, to amplify
timber – carpenter’s wood, a beam
accompany – to keep company with
mound – bank of the earth or stone, heap, hillock
solar – influenced by the sun
lunar – influenced by the moon
terrific – fearful
collapse – to fall away, to break down
clasp – embrace
subside – to abate, to sink down

Section – II

Gist:
Paragraphs: 21 – 25
The rain stopped though no wind. A crashing sound stunned the people. Suddenly the earth seemed to rock. A few who were nearest the river were splashed. Had they been standing a few feet farther they would have been gone forever. Nirakar Das, the retired head-pundit of the primary school called everybody to leave the place and go away to a safer place. All obeyed his outright. A few snakes crept out of the hollow under the tree and wriggled away toward the mound. It appeared like the exodus of a thousand snakes. It was now about dawn. Nirakar Das advanced near the tree and looked up for a long time. He declared that his eyes were gone. He called one of his ex-pupils. Ravindra, the founder proprietor of the village’s sole, grocery and asked him to look up and see if there were any birds on the tree

Paragraphs: 26-38
Ravindra and others gazed up into the branches for a while and reported their finding that nothing of that sort was there. He asked the people of his age group and the reply was the same. He said it was not a good sign because snakes and birds have fled the natural shelter. Ravindra and others detected an extensive crack in the shape of a sickle with both ends pointing toward the river. The semi-circle embraced the tree. If the tree falls, it will carry the whole huge chunk along with it into the river for its unnumerable roots have made much of earth like a single cake.

As the sky in the east grew brighter it was observed that the ground between the tree and the river had already tilted towards the river. The young men tried to appear engrossed in discussing something highly sophisticated among themselves. Srikanta Das raised his voice and whispered that not only the boys but all adults have their share of it. And if the tree is going to collapse, it is because cannot one of us confess his sin, addressing the spirit of the tree silently in “our hearts”. Let us pray to be pardoned by saying “Haribol”.

GIST:
Paragraphs: 39 – 43
All shouted “Haribol” but it sounded like a cry of lamentation. When they stopped, the silence seemed bitingly sharp. With the gradual brightening of the sky, the seriousness of the situation became more and more apparent. A few kites that were circling above the whirling waters at times swooped down on the crowd as though to show the contempt of those who could dwell at such height and see all that was happening from horizon to horizon for the wretched men below regarding their situation with other helplessness.

The crowd swelled rapidly. Almost the villagers women and children included were then gathered there. In different words, all asked the same question. “What is to be done?” Apart from the tree was clearly learning about the river. Once the college boys had been humbled there was no hesitation to openly discuss the impending fall of the tree. Something no doubt had to be done. Only one knew what that was.

Analytical Outlines

• The rain stopped though not the wind.
•  A crashing sound stunned the people.
• Suddenly the earth seemed to rock.
• A few near the river were splashed.
• Had they been standing a few feet farther?
• They would have been gone forever.
• Nirakar Das was the retired Head-pundit of the primary school.
• He called everybody to leave the place.
• He called them to go away to a safer place.
• All obeyed his outright.
• A few snakes crept out of the hollow under the tree.
• They wriggled away towards the mound.
• It appeared like the exodus of thousand snakes.
• It was about dawn then.
• Nirakar Das advanced near the tree.
• He looked up for a long time.
• He declared that his eyes were gone.
• He called one ofhis expupiL
• It was Ravindra.
• He was the founder and proprietor of the village’s sole grocery.
• He asked him to see if there were any birds on the tree.
• Ravindra and others gazed up into the branches.
• He reported their findings.
• Nothing of that sort was there.
• He asked the people of his age group.
• The reply was the snake.
• He said it was not a good sign.
• Because snakes and birds have fled the natural shelter.
• Ravindra and others detected an extensive crack.
• It was the shape of a sickle.
• It’s both ends pointed towards the river.
• The semi-circle embraced the tree.
• If the tree falls.
• It will carry the whole huge chunk into the river.
• Its innumerable roots have made much of the earth like a single cake.
• The sky in the east grew brighter.
• The ground of the tree is titled towards the river.
• The young men appeared there.
• They discussed among themselves the situation.
• Srikanta Das raised his voice.
• He whispered that both the boys and adults love their share of sin.
• The tree is going to collapse.
• It is because it can’t bear the burden of its sins.
• He said them to confess their sins.
• He addressed the spirit of the tree silently.
• He inspired them to pray to be pardoned.
• He inspired them to say “Haribol”.
• Now, shouted, “Haribol”.
• But it sounded like lamentation.
• They stopped then.
• The silence seemed bitingly sharp.
• The seriousness of the situation became more and more apparent.
• A few flying kites swooped down on the crowd.
• It showed the situation of the wretched men with utter helplessness.
• The crowd swelled rapidly.
• All the villagers, women, and children gathered there.
• All asked the same question in different words.
• It was, “What is to be done? Apart from the tree was clearly leaning towards the river.
• In college, boys had been humbled.
• There was no hesitation about the fall of the tree.
• Something, no doubt had to be done.
• Nobody knew what to do

Meaning Of Difficult Words

exodus – a going out on a mass scale.
sniffly – sniff to draw in air sharply and audibly through the nose.
whimper – to carry the nose.
swoop – to become down with a sweeping rush
edge – border, at the end of something
cashing sound – a great sound like that of a crash
stunned – bewildered, in a state of trance.
countless – uncountable, numerous, innumerable
crack – fissure, chasm
notorious – mischievous, wicked, naughty
tauntingly – sarcastically, caustically, satirically
implored – requested earnestly, entreated
titled – bent, bending towards the earth
apparent – clear, smooth, easy
contempt – hatred, decision, dislike
swelled – become thicker and thicker, grew in number
impending – imminent, very nearer

Section – III

Gist:
Paragraphs (44- 48)
Sridhar Mishra was a well-known homeopath. He had saved so many from certain death. When the people looked expectantly at him, his lips quivered as they always deed when he was about to diagnose a disease. The villagers were accustomed to reading in that quiver the promise of remedy. But as now the quivering did not stop even when the people had looked at him for a long time, they focussed their attention on Raghu Dalbehera, the only villager to possess a gun. When Raghu realized that the crowd had already been staring at him for five minutes, he raised his gun at an audaciously swooping kite, took aim, and continued to take aim. “Don’t Raghu, point”. Nirakar Das warned and Raghu brought down his gun in relief.

Gist:
Paragraphs (48-62)
Just then someone brought the news that the honorable member of the legislative assembly has been observed only by one on a nearby road perhaps heading for the next village. Freed from the obligation to think or do anything now that the M.L.A. had been located and summoned, all stood peacefully looking towards the bend of the road where he was expected to appear. People were sad that their sheer ill luck had spoiled them.

They had been doomed. The leader retorted that was why they had such pessimism. He also added that people downstream are greatly in trouble. These people are better off than the people downstream. The three college boys were elbowed their way forward, throwing glances back at the crowd as if defying it to stop them from confronting the leader.

They were of course two or three years below the voting age, but they were determined to regain face after their earlier humiliation. The M.L.A. paled, but ignored the boys and asked the elders “What would you like me to do ?” Someone said to him to stop the tree from falling. The M.L.A. said to fetch as many ropes as they could. He guided up his loins and got ready. But suddenly a part of the tree resting on several trunks slid into the river.

Gist:
Paragraphs (63-71)
The crowd stood thunderstruck. The silence was broken by an anxious voice “What will happen to the banian and goddess ?” No sooner had this been said than the ill-tempered old Brahmin was seen rushing to the remnants of the tree. He sat down on the muddy ground which was dangerously unsafe even by the snakes and mustering all his strength pulled up the small stone that has struck the spot. Holding the uprooted goddess close to his bosom as though to protect her from invisible enemies, he returned to the crowd that watched him breathlessly.

While thronging around the Brahmin the people said excitedly to give place to the goddess someone spread a towel on the grass. The Brahmin put down the goddess and patted her. Several people rushed to their homes and brought out symbols and conch shells had to be played close to his ears at loudly as possible. He began by shivering. Then he would fall down in a swoon and rise up with the face beaming supernaturally, eyes wild with inexplicable experiences.

Gist:
Paragraphs (72 – 74)
Bishnu was in France for at least two decades. Those who used to play the instruments close to his ears had purses, they were doing their best. Bishnu opened his mouth. The sound stopped. The voice from him ran: I will be born as a thousand trees here, there, and everywhere. The instruments played louder as the younger ones took over from the tired old hands. Along with Bishnu danced Nirakar Das, Srikanta Das, Vaishnav, and several others.

Analytical Outlines

• Sridhar Mishra was a well-known homeopathic.
• He had saved so many from certain death.
• People looked expectantly at him.
• He was about to diagnose a disease.
• His lips quivered at that time.
• The villagers were accustomed to reading in that quiver.
• That quiver was the promise of remedy.
• But now the quivering did not stop.
• They had looked at him for a long time.
• They focused their attention on Raghu Dalbehera.
• He was the only villager to possess a gun.
• Raghu realized that the crowd had already been staring at him.
• He raised his gun at an audaciously swooping kite.
• “Don’t Raghu, don’t Nirakar Das warned.
• Raghu brought down his gun in relief.
• Just then someone brought the news.
• It was that M.L.A. would be coming to visit a nearby road of villages.
• They all stood peacefully looking towards the bent of the road.
• The M.L.A. was expected to appear there.
• People said that their sheer ill luck had spoiled them.
• They had been doomed.
• They retorted that why they had such a pessimism
• He also added that people downstream are greatly in trouble.
• These people are better off than the people downstream
• The three college boys were elbowed their way forward.
• They threw glances back at the crowd.
• They defined it to stop them from confronting the leader.
• Of course, they were two or three years below the voting age.
• But they were determined to regain face.
• It was after their earlier humiliation.
• The M.L.A. paled but ignored the boys.
• He asked the elders, “What would you like me to do?
• Someone said him to stop the tree from falling.
• The M.L.A. said to fetch as much rope as they could.
• He girded up his loins.
• He got ready.
• But suddenly a part of the tree slid into the river.
• It rested on several trunks.
• The crowd stood thunderstruck.
• The silence was broken by an anxious voice.
• It was “What will happen to the banian goddess”?
• No sooner had this been said than the ill-tempered old Brahmin was soon rushing to the remnants of the tree.
• He sat down on the muddy ground.
• It was dangerously unsafe even by the snakes.
• He mustered all his strength.
• He pulled up the small stone.
• It has struck the spot.
• He held the uprooted goddess close to his bosom.
• He did this to protect her from invisible enemies.
• He returned to the crowd.
• They watched him breathlessly.
• They thronged around the Brahmin.
• The people said excitedly to give place to the Goddess.
• Someone spread a towel on the grass.
• The Brahmin put down the Goddess.
• He patted her.
• Old Bishnu Jena had seated himself before the banian Goddess.
• Several people rushed to their homes.
• They brought out cymbals
• They brought out drums.
• They also brought out conch shells.
• Drums, cymbals, and conch shells had to be played close to her ears as loudly as possible.
• He began by shivering.
• Then he would fall down in a swoon.
• He would rise up with a face bearing supernatural eyes.
• He bears wild eyes with inexplicable experience.
• Bishnu was France after at least two decades.
• The instrument players had grown old.
• They were doing their best with their sagging skin flapping like empty purses.
• Bishnu opened his mouth.
• Their voice from him ran “I will be born as a thousand trees here, there, and everywhere.
• The instruments played louder.
• The voice from him ran. “I will be born as a thousand trees here, there, and everywhere.
• The younger ones took over from the tired old hands.
• Nirakar Das, Srikanta Das etc, danced with Bishnu.

Meaning Of Difficult Words

clarion call – to blind around, to make first by a belt or girdle.
remnants – the battle cry of an ancient trumpet.
thronging – surviving and remaining person after destruction.
poignant – crowding or pressing.
gird – deeply moving.
sagging – bent down with age, wrinkled
pals – mates, chums.