RD Sharma Solutions Class 12 The Plane Exercise 29.8

RD Sharma Solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.8 is provided here for students to grasp the concepts easily and score well in the exams. The subject experts have framed and solved the questions from every section accurately. The RD Sharma Solutions are completely based on the exam-oriented approach to help students score well in their board examination. Students can easily download the PDF consisting of this chapter’s solutions, which are available in the links provided below.

RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 8

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Access RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 8

EXERCISE 29.8

Q1.

Solution:

Given:

Equation of plane is 2x – 3y + z = 0 …… (1)

We know that equation of a plane is parallel to the given plane (1) is given as

2x – 3y + z + k = 0 …… (2)

Given that, plane (2) is passing through point (1, – 1, 2). So it must satisfy plane (2),

2(1) – 3(– 1) + (2) + k = 0

2 + 3 + 2 + k = 0

7 + k = 0

k = – 7

Now, substitute the value of k in equation (2), we get

2x – 3y + z – 7 = 0

Hence, the equation of the required plane is, 2x – 3y + z = 7.

Q2.

Solution:

Q3.

Solution:

We know that the equation of a plane passing through the line of intersection of two planes

a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

(a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0

Given that the equation of the plane is,

2x – 7y + 4z – 3 = 0 and

3x – 5y + 4z + 11 = 0

So the equation of the plane passing through the line of intersection of given two planes is

(2x – 7y + 4z – 3) + k(3x – 5y + 4z + 11) = 0

2x – 7y + 4z – 3 + 3kx – 5ky + 4kz + 11k = 0

x(2 + 3k) + y( – 7 – 5k) + z(4 + 4k) – 3 + 11k = 0 ……(1)

It is given that plane (1) is passing through the point (–2, 1, 3). So it must satisfy the equation (1),

(– 2) (2 + 3k) + (1) (– 7 – 5k) + (3) (4 + 4k) – 3 + 11k = 0

– 2 + 12k = 0

12k = 2

k = 2/12

= 1/6

Now substitute the value of k in equation (1), we get

x(2 + 3k) + y(– 7 – 5k) + z(4 + 4k) – 3 + 11k = 0

Multiply by 6, we get

15x – 47y + 28z – 7 = 0

Hence, the required equation of the plane is 15x – 47y + 28z – 7 = 0.

Q4.

Solution:

Q5.

Solution:

We know that the equation of a plane passing through the line of intersection of two planes

a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

(a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0

So the equation of the plane passing through the line of intersection of given two planes

2x – y = 0 and 3z – y = 0 is

(2x – y) + k(3z – y) = 0

2x – y + 3kz – ky = 0

x(2) + y( – 1 – k) + z(3k) = 0 …… (1)

We know that two planes are perpendicular if

a₁a₂ + b₁b₂ + c₁c₂ = 0 …… (2)

It is given that plane (1) is perpendicular to plane

4x + 5y – 3z = 8 …… (3)

By using equations (1) and (3) in equation (2), we get

(2) (4) + (–1 – k) (5) + (3k) (–3) = 0

8 – 5 – 5k – 9k = 0

3 – 14k = 0

–14k = –3

k = 3/14

Now, substitute the value of k in equation (1), and we get

x(2) + y( – 1 – k) + z(3k) = 0

Multiply by 14, and we get

28x – 17y + 9z = 0

Hence, the required equation of the plane is 28x – 17y + 9z = 0.