Class 12

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 8 Application of Derivatives will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 8 Application of Derivatives

Tangents and normals:
(a) If y = f(x) is the equation of any curve then = The slope of the tangent at P(x₁, y₁).
(b) Slope of the normal at (x₁, y₁)
(c) Equation of tangent at P(x₁, y₁) is y – y₁ =
(d) Equation of normal at P(x₁, y₁) is y – y₁ =
(e) Angle between two curves is the angle between two tangents at the point of contact.

Increasing and decreasing functions:
If y = f(x) is defined in [a, b] then
(i) f'(x) > 0, x ∈ (a, b)
⇒ f is strictly increasing on (a, b).
(ii) f'(x) > 0, x ∈ (a, b)
⇒ f is monotonic increasing on (a, b).
(iii) f'(x) < 0, x ∈ (a, b)
⇒ f is strictly decreasing on (a, b).
(iv) f'(x) < 0, x ∈ (a, b)
⇒ f is monotonic decreasing on (a, b).
(v) f'(x) = 0, x ∈ (a, b)
⇒ f is a constant function on (a, b).

Approximation:
(a) If y = f(x) is a function and δx is a very small change in x then the respective change in y is δy given by
δy = f'(x) δx => dy = f'(x) δx.
∴ The approximate value of y = f(x) at
x = a + δx is f(a + δx)
= f(a) + f(a) δx

Maxima and minima:
(a) First derivative criteria to find max/min of y = f(x)
Algorithm:
Step-1 : Put and solve for x.
Let x = a, b, c ……
Step-2 : If changes sign from (+ve) to (-ve) in then at x = a, ‘f’ has a local maximum. If changes sign from (-ve) to (+ve) in then at x = a, f has a local minimum. If in thenat x = a ‘f’ has neither maxima nor a minima (it may be a point of inflexion).

(b) Second derivative criteria
Algorithm:
Step-1 : Find the roots of f'(x) = 0.
Let they are a, b, c ……
Step-2 : Find f”(x) and put x = a, b, c ……
(i) If f”(a) > 0 then at x = a, f has a local minimum.
(ii) If f”(a) < 0 then at x = a, f has a local maximum.
(iii) If f”(a) = 0 and f”(x) changes sign in (a – δ, a + δ) then x = a is a point of inflexion.
(iv) If f”(a) = 0 and f”(x) does not change sign in then use first derivative criteria to check for maxima/minima.

Mean Value Theorems:
(a) Rolle’s theorem:
If a function f is
(i) continuous on the closed interval [a, b]
(ii) differentiable on the open interval (a, b) and
(iii) f(a) = f(b) then there exists a point c ∈ (a, b) such that f(c) = 0.
Geometrical interpretation:
If f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b) then there exists atleast one point c ∈ (a, b) such that at x = c the tangent is parallel to x-axis.
Algebraic interpretation:
Between two roots ‘a’ and ‘b’ of f(x) there exists atleast one root of f'(x).

(b) Cauchy’s Mean Value theorem:
If ‘f’ and ‘g’ are two functions such that
(i) both are continuous on [a, b]
(ii) both are differentiable on (a, b) and
(iii) g'(x) ≠ 0 for any x ∈ (a, b) then there exists atleast one point c ∈ (a, b) such that
Geometrical interpretation:
The conclusion of Cauchy’s theorem can be written as i.e. the ratio of the mean rate of increase of two functions in an interval equals to the ratio of actual rate of increase at some point of the interval.

(c) Lagrange’s Mean Value theorem:
If a function f is
(i) continuous on the closed interval [a, b]
(ii) differentiable on the open interval(a, b) then there exists atleast one c ∈ (a, b), such that.
Geometrical interpretation:
Between two points A and B of the graph of y = f(x) there exists atleast one point c such that the tangent is parallel to the chord AB.

Indeterminate forms & L’Hospitals rule:
A limit is said to be in indeterminate form if it takes any of the forms.
Note:
If a limit is in indeterminate form then it can be evaluated using the following methods.
(i) Change the function to determinate form (by rationalisation, expansion or any other means) then find the limit.
Or, (ii) Bring to form then use L’Hospitals rule.
L’Hospitals rule:
Let f and g are two functions differentiable on some open interval containing ‘a’ such that g'(x) ≠ 0 for x ≠ a and g(a) = f(a) = 0, then provided the latter limit exists.

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 9 Integration will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 9 Integration

Indefinite integral:
If \(\frac{d}{d x}\)F(x) = f(x) then the indefinite integral of f(x) w.r.t x is
∫f(x)dx = F(x) + C
which represents the entire class of anti-derivatives.

(a) Algebra of integrals:
(i) ∫[f(x) ± g(x)] dx = ∫f(x) dx ± ∫g(x) dx
(ii) ∫af(x) dx = α ∫f(x) dx
(iii) ∫f(x) g(x) dx = f(x) ∫g(x) dx – ∫\(\left[(\frac{d}{d x} f(x)\right) \cdot \int g(x) d x]\) dx (Integration by parts)

(b) Some standard indefinite integrations:
(1) ∫x_n dx = \(\frac{x^{n+1}}{n+1}\) + C, n ≠ (-1)
(2) ∫\(\frac{d x}{x}\) = log_e|x| + C.
(3) ∫sin x dx = -cos x + C
(4) ∫cos x dx = sin x + C
(5) ∫tan x dx = -log |cos x| + C or log |(sec x)| + C
(6) ∫cot x dx = log |(sin x)| + C or -log |(cosec x)| + C
(7) ∫sec x dx = log |sec x + tan x| + C
(8) ∫cosec x dx = log |cosec x – cot x| + C
(9) ∫sec² x dx = tan x +C
(10) ∫cosec² x dx = -cot x + C
(11) ∫sec x tan x dx = sec x + C
(12) ∫cosec x cot x dx = -cosec x + C

[Note: To integrate by parts choose 1st function according to I LATE]
Where I → Inverse trigonometric functions.
L → Logarithmic function
A → Algebraic function
T → Trigonometric function
E → Exponential function

(c) Techniques of integration:

Definite integration:

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 10 Area Under Plane Curves will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 10 Area Under Plane Curves

Area of the portion bounded by x-axis the curve
y = f(x) and two ordinates at x = a and x = b.

= Area of the portion bounded by y-axis,
the curve x = f(y) and two abscissa at y = c and y = d.

Area between two curves y = f(x), y g(x) with g(x) < f(x) in [a, b] and between two ordinates
x = a and x = b is given by \(\int_a^b\){f(x) – g(x)}dx

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 7 Continuity and Differentiability will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 7 Continuity and Differentiability

Definition:
A function f is said to be continuous at a point a D_f if
(i) f(x) has definite value f(a) at x = a,
(ii) lim_x->a f(x) exists,
(iii) lim_x->a f(x) = f(a).
If one or more of the above conditions fail, the function f is said to be discontinuous at x = a. The  above definition of continuity of a function at a point can also be formulated as follows:
A function f is said to be continuous at x = a if
(i) holds and for a given ∈ > 0, there exists a δ > 0 depending on ∈ such that
|x – a| < 8 ⇒ |f(x) – f(a)| < ∈.
A function f is continuous on an interval if it is continuous at every point of the interval.
If the interval is a closed interval [a, b] the function f is continuous on [a, b] if it is continuous on (a, b),
lim_x->a+ f(x) = f(a) and lim_x->b- f(x) = f(b).

Differentiation of a function:
(a) Differential coefficient (or derivative) of a function y = f(x) with respect to x is

Fundamental theorems:

Then to get \(\frac{d y}{d x}\) it is convenient to take log of both sides before differentiation.

Derivative of some functions:

Higher order derivative:

Leibnitz Theorem:
If ‘u’ and ‘v’ are differentiable functions having ‘n’th derivative then
\(\frac{d^n}{d x^n}\)(u.v) = C₀u_nv + C₁u_n-1v₁ + C₂u_n-2v₂ + ….. + C_nuv_n

Partial derivatives and Homogeneous functions:
(a) If z = f(x, y) is any function of two variables then the partial derivative of z w.r.t. x and y are given below.

(b) Homogeneous function:
z = f(x > y) is a homogeneous function of degree ‘n’ if f(tx, ty) = tⁿ f(x, y).

Euler’s Theorem:
If z = f(x, y) is a homogeneous function of degree ‘n’ then \(x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}\) = nf(x, y).

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 6 Probability will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 6 Probability

Important terms:
(a) Random experiment: It is an experiment whose results are unpredictable.
(b) Sample space: It is the set of all possible outcomes of an experiment. We denote the sample space by ‘S’.
(c) Sample point: Each element of a sample space is a sample point.
(d) Event: Any subset of a sample space is an event.
(e) Simple event: It is an event with a single sample point.
(f) Compound event: Compound events are the events containing more than one sample point.
(g) Mutually exclusive events: Two events A and B are mutually exclusive if A ∩ B = φ (i.e. occurrence of one excludes the occurrence of other)
(h) Mutually exhaustive events: The events A₁, A₂, A₃ ……. A_n are mutually exhaustive if A₁ ∪ A₂ ∪ A₃ ∪ A_n = S.
(i) Mutually exclusive and exhaustive events.
The events A₁, A₂, A₃ ……. A_n are mutually exclusive and exhaustive if
(i) A₁ ∩ A₂ = φ for i ≠ j
(ii) A₁ ∪ A₂ ∪ …… A_n = S.
(j) Equally likely events: Two events are equally likely if they have equal chance of occurrence.
(k) Impossible and certain events: φ is the impossible and S is the sure or certain event.
(i) Independent events: The events are said to be independent if the occurrence or non-occurrence of one does not affect the occurrence of non-occurrence of other.

Probability of an event:
Let S be the sample space and A is an event then the probability of A is
\(P(A)=\frac{|A|}{|S|}=\frac{\text { No.of out comes favourable to } A}{\text { Total number of possible outcomes. }}\)
Note:
1. P(φ) = 0
2. P(S) = 1
3. P(A’) = P (not A) = 1 – P(A)
4. P(A) + P(A’) = 1

Odds in favour and odds against an event:
Let in an experiment is the number of cases favourable to A and ‘n’ is the number of cases not in favour of A then

Addition theorem:
If A and B are any two events then
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Note:
If A and B are mutually exclusive then
P(A ∪ B) = P(A) + P(B)
(∴ P(A ∩ B) = P(φ) – 0)

Conditional probability:
Let A and B are any two events and P(B) ≠ 0 then the conditional probability of A when B has already happened
P(A/B) = \( \frac{P(A \cap B)}{P(B)} \)
Note:
1. P(A ∩ B) = P(B) . P(A/B)
2. If A and B are mutually independent events then P(A/B) = P(A).
∴ P(A ∩ B) = P(A) . P(B)
3. If A and B are independent events then (i) A’ and B’ (ii) A’ and B (iii) A and B’ are also independent.
4. P(A₁ ∩ A₂ ∩ A₃ ….. ∩ A_n) = P(A₁) . P(A₂/A₁) . P(A₃/A₂ ∩ A₁) ….. P(A_n/A₁ ∩ A₂ ∩ …. ∩ A_n-1)
5. Let A₁, A₂ ….. A_n are mutually exhaustive and exclusive events and A is any event which occurs with A₁ or A₂ or A₃ … or A_n then
P(A) = P(A₁) . P(A/A₁) + P(A₂) . P(A/ A₂) + …….. + P(A_n) . P(A/A_n).
This is called the total conditional probability theorem.

Baye’s theorem:
If A₁, A₂ …… A_n are mutually exclusive and exhaustive events and A is any event which occurs with A₁ or A₂ or A₃ or …. A_n then
\( P\left(A_i / B\right)=\frac{P\left(A_i\right) \cdot P\left(A / A_i\right)}{\sum_{i=1}^n P\left(A_i\right) P\left(A / A_i\right)} \)

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 4 Matrices will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 4 Matrices

It is a system of mn numbers arranged in a rectangular system of m rows and n columns.
Example : \(\left(\begin{array}{lll}
a_{11} & a_{12} & a_{1 n} \\
a_{21} & a_{22} & a_{2 n} \\
a_{m_1} & a_{m_2} & a_{m_n}
\end{array}\right)\) is a m × n matrix.

Note : We write the above matrix in short as [a_ij]_m×n or [aij]_m×n or ||a_ij||_m×n.

Important types matrices:
(a) Square matrix:
It is a matrix where the number of rows are equal to the number of columns.

(b) Null (zero) matrix:
If all the entries of a matrix are zero, then the matrix is a zero matrix denoted by 0_m×n.

(c) Diagonal matrix:
It is a square matrix in which all the elements except those in main (or leading) diagonal are zero.

(d) Unit matrix:
It is a diagonal matrix where all the elements in the leading (main) diagonal are one.

(e) Scalar matrix:
It is a diagonal matrix with all the elements in the leading (main) diagonal are α (α ≠ 0 or 1).

(f) Singular matix:
A square matrix ‘A’ is singular iff |A| = 0, otherwise it is a non singular matrix.

(g) Symmetric matrix:
A square matrix is symmetric if A = A’ and skew-symmetric if A = -A’.

Matrix algebra:
(a) Addition and subtraction:
If A = [a_ij]_m×n and B = [b_ij]_m×n then A ± B = [a_ij ± b_ij]_m×n

(b) Scalar multiplication:
If A = [a_ij]_m×n then for any scalar ‘k’
kA = [ka_ij]_m×n

(c) Matrix multiplication:

Properties:
1. Matrix addition is commutative as well as associative.
2. 0_m×n is the additive identity.
3. A is the zero additive inverse of A.
4. We can add or subtract matrices if they are of same order.
5. We can multiply two matrices if the number of columns of 1st is equal to the number of rows of 2nd.
6. Matrix multiplication is non commutative but associative.
7. Matrix multiplication is distributive over addition.
8. AB = 0 ≠ A = 0 or B = 0 for two matrices A and B. Also AB = AC ≠ B = C.
9. If A is a square matrix of order n then
I_n =  is the multiplicative identity.

Transpose and adjoint of a matrix:
(a) The transpose of a matrix A = [a_ij]_m×n is A^T or A’ = [a_ji]_n×m.
Properties:
(i) (A’)’ = A
(ii) (A ± B)’ = A’ ± B’
(iii) (AB)’ = B’A’
(iv) (KA)’ = KA’
(v) Any matrix A can be expressed as sum of a symmetric and a skew-symmetric matrix as

(b) Adjoint of a matrix:
If A is a square matrix then Adj A = The transpose of the matrix of co-factors

Properties:
1. (Adj A) A = A (Adj A) = |A|I_n
∴ A^-1 = \(\frac{{Adj} A}{|~A|}\)
2. |Adj A| = A^n-1
3. Adj (AdjA) = |A|^n-2A
4. (AdjA)’ = Adj (A’)
5. (AB)^-1 = B^-1A^-1

Solution of system of linear equations (matrix method):
Let the system of linear equations is
a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃
Let

Note: If the system is homogeneous i.e. d₁ = d₂ = d₃ = 0, then the system has a trivial solution x = 0, y = 0, z = 0 for |A| ≠ 0. In case |A| = 0 then the system has infinitely many solutions.

Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 3 Linear Programming will enable students to study smartly.

CHSE Odisha 12th Class Math Notes Chapter 3 Linear Programming

Definition:
A general linear programming problem LPP is to obtain x₁, x₂, x₃ …… , x_n so as to

Optimize:
Z = c₁x₁ + c₂x₂ + c₃x₃ + ……. + c_nx_n … (A)
subject to
a₁₁x₁ + a₁₂x₂ + ……. + a_1nx_n  ≤ (or ≥) b₁
a₂₁x₁ + a₂₂x₂ + ……. + a_2nx_n  ≤ (or ≥) b₂ … (B)
where x₁, x₂, …… , x_n ≥ … 0 … (C)
and a_ij, b_i, c_j with i = 1, 2, … , m; j = 1, 2, … ,n are real constants.

In the LPP given above, the function Z in (A) is called the objective function. The variables x₁ x₃, ……, x_n are called decision variables. The constants c₁ c₂, ……, c_n are called cost coefficients. The inequalities in (B) are called constraints. The restrictions in (C) are called non-negative restrictions. The solutions which satisfy all the constraints in (B) and the non-negative restrictions in (C) are called feasible solutions.

The LPP involves three basic elements:

1. Decision variables whose values we seek to determine,
2. Objective (goal) that we aim to optimize,
3. Constraints and non-negative restrictions that the variables need to satisfy.

Types of Linear Programming Problems

As we have already discussed we come across different types of problems which we need solve depending on the objective functions and the constraints. Here we discuss a few important types of LPPs. before learning how to formulate them.

(i) Manufacturing Problem
A manufacturer produces different items so as to maximise his profit. He has to determine the number of units of products he must produce while satisfying a number of constraints because each unit of product requires availability of some amount of raw material, certain manpower, certain machine hours etc.

(ii) Diet Problem
Suppose a person is advised to take vitamins/nutrients of two or more types. The vitamins/nutrients are available in different proportions in different types of foods. If the person has to take a minimum amount of the vitamins/nutrients then the problem is to determine appropriate quantity of food of each type so that cost of food is kept at the minimum.

(iii) Allocation Problem
In this type of problem one has to allocate different resources/tasks to different units/persons depending on the nature of the gain or outcome.

(iv) Transportation Problem
These problems involve transporting materials from sources to destinations for sale or distribution of products or collection of raw materials etc. Here the aim is to use various options for transportation such as distance, time etc so as to keep the cost of transportation to a minimum.

Working procedure to solve LPP graphically
Step-1. Taking all inequations of the constraints as equations, draw lines represented by each equation and considering the inequalities of the constraint inequations complete the feasible region.
Step-2. Determine the vertices of the feasible region either by inspection or by and solving the two equations of the intersecting lines.
Step-3. Evaluate the objective function Z = ax + by at each vertex.
Case (i) F.R. is bounded: The vertex which gives the optional value (maximum or minimum) of Z gives the desired optional solution to the LPP.
Case (ii) F.R. is unbounded: When M is the maximum value of Z at a vertex Vmax, determine the open half plane corresponding to the inequation ax + by > M. If this open half plane has no points in common with the F.R. then M is the maximum value of Z and the point Vmax gives the desired solution. Otherwise, Z has no maximum value.

Similarly consider the open half plane ax+by < m when m is the minimum value of Z at the vertex Vmin. If this half-plane has no point common with the F.R. then m is the minimum value of Z and Vmin gives the desired solution. Otherwise Z has no minimum value.

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