RD Sharma Solutions Class 12 The Plane Exercise 29.13

RD Sharma Solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.13 is provided here. Students who aim to score high marks in their board exams can get access to the RD Sharma Solutions for Class 12 and prepare themselves confidently for the final exams. Experts have formulated the solutions in an easily understandable manner for students to grasp the concepts with ease and score good marks in the exams. Students can easily download the PDF consisting of this chapter’s solutions, which are available in the links provided below.

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Access RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 13

EXERCISE 29.13

Q1.

Solution:

Q2.

Solution:

(x + 1) (4 + 3) – (y – 3) (– 6 – 1) + (z + 2) (9 – 2) = 0

7x + 7 + 7y – 21 + 7z + 14 = 0

7x + 7y + 7z = 0

Divide by 7, we get

x + y + z = 0

Hence, the required equation of the plane is x + y + z = 0.

Q3.

Solution:

Substitute a, b, c in equation (2), and we get

ax + b(y – 7) + c(z + 7) = 0

(– 14λ)x + (– 14λ) (y – 7) + (– 14λ) (z + 7) = 0

Dividing by (– 14λ) we get

x + y – 7 + z + 7 = 0

x + y + z = 0

Q4.

Solution:

Given:

We know that equation of the plane passing through (x₁, y₁, z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 …… (1)

Thus, the required plane passes through points (4, 3, 2) and (3, – 2, 0).

So the equation of the required plane is

a(x – 4) + b(y – 3) + c(z – 2) = 0 …… (2)

Plane (2) also passes through (3, – 2, 0),

So,

a(3 – 4) + b(– 2 – 3) + c(0 – 2)

– a – 5b – 2c = 0

a + 5b + 2c = 0 …… (3)

Now plane (2) is also parallel to the line with direction ratios 1, – 4, 5.

So,

a₁a₂ + b₁b₂ + c₁c₂ = 0

(a) (1) + (b) (– 4) + (c) (5) = 0

a – 4b + 5c = 0 …… (4)

Now substitute the values of a, b, c in equation (2), and we get

a(x – 4) + b(y – 3) + c(z – 2) = 0

(11 λ)(x – 4) + (– λ)(y – 3) + (– 3 λ)(z – 2) = 0

11 λx – 44 λ – λy + 3 λ – 3 λz + 6 λ = 0

11 λx – λy – 3 λz – 35 λ = 0

Divide by λ, we get

11x – y – 3z – 35 = 0

Hence, the required equation of the plane is 11x – y – 3z – 35 = 0.

Q5.

Solution:

Another equation of line is

3x – 2y + z + 5 = 0

2x + 3y + 4z – 4 = 0

Let a, b, c be the direction ratio of the line so it will be perpendicular to the normal of 3x – 2y + z + 5 = 0 and 2x + 3y + 4z – 4 = 0

So, using a₁a₂ + b₁b₂ + c₁c₂ = 0

(3) (a) + (– 2) (b) + (1) (c) = 0

3a – 2b + c = 0 …… (2)

Again, (2) (a) + (3) (b) + (4) (c) = 0

2a + 3b + 4c = 0 …… (3)