RD Sharma Solutions Class 12 The Plane Exercise 29.6

RD Sharma Solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.6 is given here. Experts have designed RD Sharma Solutions in a very easy-to-understand manner that helps students solve problems efficiently. Students are advised to practise the RD Sharma Class 12 Solutions regularly to achieve the scores they desire. Students can easily download the PDF consisting of this chapter’s solutions, which are available in the links provided below and can use it for future reference as well.

RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 6

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Access RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 6

EXERCISE 29.6

Q1. i.

Solution:

ii.

Solution:

iii.

Solution:

Q2. i

Solution:

Given planes are:

2x – y + z = 4 and x + y + 2z = 3

We know that angle between two planes,

a₁x + b₁y + c₁z + d₁ = 0 and

a₂x + b₂y + c₂z + d₂ = 0 is given as

So here we have,

a₁ = 2, b₁ = – 1, c₁ = 1

a₂ = 1, b₂ = 1, c₂ = 2

Now let us substitute the values in the above expression, we get

ii.

Solution:

Given planes are:

x + y – 2z = 3 and 2x – 2y + z = 5

We know that angle between two planes,

a₁x + b₁y + c₁z + d₁ = 0

a₂x + b₂y + c₂z + d₂ = 0 is given as

So here we have,

a₁ = 1, b₁ = 1, c₁ = – 2

a₂ = 2, b₂ = – 2, c₂ = 1

Now let us substitute the values in the above expression, we get

iii.

Solution:

Given planes are:

x – y + z = 5 and x + 2y + z = 9

We know that angle between two planes,

a₁x + b₁y + c₁z + d₁ = 0

a₂x + b₂y + c₂z + d₂ = 0 is given as

So here we have,

a₁ = 1, b₁ = – 1, c₁ = 1

a₂ = 1, b₂ = 2, c₂ = 1

Now let us substitute the values in the above expression, we get

iv.

Solution:

Given planes are:

2x – 3y + 4z = 1 and – x + y = 4

We know that angle between two planes,

a₁x + b₁y + c₁z + d₁ = 0

a₂x + b₂y + c₂z + d₂ = 0 is given as

So here we have,

a₁ = 2, b₁ = – 3, c₁ = 4

a₂ = – 1, b₂ = 1, c₂ = 0

Now let us substitute the values in the above expression, we get

v.

Solution:

Given planes are:

2x + y – 2z = 5 and 3x – 6y – 2z = 7

We know that angle between two planes,

a₁x + b₁y + c₁z + d₁ = 0

a₂x + b₂y + c₂z + d₂ = 0 is given as

So here we have,

a₁ = 2, b₁ = 1, c₁ = – 2

a₂ = 3, b₂ = – 6, c₂ = – 2

Now let us substitute the values in the above expression, we get

Q3.i.

Solution:

ii.

Solution:

Given planes,

x – 2y + 4z = 10 and

18x + 17y + 4z = 49

We know that planes

a₁x + b₁y + c₁z + d₁ = 0 and

a₂x + b₂y + c₂z + d₂ = 0 are at right angles if,

a₁a₂ + b₁b₂ + c₁c₂ = 0 …… (1)

Now we have,

a₁ = 1, b₁ = – 2, c₁ = 4 and

a₂ = 18, b₂ = 17, c₂ = 4

Using equation (1), we have,

a₁a₂ + b₁b₂ + c₁c₂ = (1) (18) + (– 2) (17) + (4) (4)

= 18 – 34 + 16 = 0

Hence, the planes are at a right angle to each other.

Q4.i.

Solution:

So, (λ + 4 – 21) = 0

λ = 21 – 4

= 17

Hence, for λ = 17, the given planes are perpendicular.

ii.

Solution:

Given planes are,

2x – 4y + 3z = 5 and

x + 2y + λz = 5

We know that planes

a₁x + b₁y + c₁z + d₁ = 0 and

a₂x + b₂y + c₂z + d₂ = 0 are at right angles if,

a₁a₂ + b₁b₂ + c₁c₂ = 0 …… (1)

We have,

a₁ = 2, b₁ = – 4, c₁ = 3 and

a₂ = 1, b₂ = 2, c₂ = λ

Let us substitute the values in equation (1), we get

a₁a₂ + b₁b₂ + c₁c₂ => (2) (1) + (– 4) (2) + (3) (λ) = 0

=> 2 – 8 + 3λ = 0

6 = 3λ

2 = λ

Hence, for λ = 2, the given planes are perpendicular.

iii.

Solution:

Given planes are,

3x – 6y – 2z = 7 and

2x + y – λz = 5

We know that planes

a₁x + b₁y + c₁z + d₁ = 0 and

a₂x + b₂y + c₂z + d₂ = 0 are at right angles if,

a₁a₂ + b₁b₂ + c₁c₂ = 0 …… (1)

We have,

a₁ = 3, b₁ = – 6, c₁ = – 2 and

a₂ = 2, b₂ = 1, c₂ = – λ

Let us substitute the values in equation (1), we get

a₁a₂ + b₁b₂ + c₁c₂ = (3) (2) + (– 6) (1) + (– 2) (– λ) = 0

=> 6 – 6 + 2λ = 0

0 = – 2λ

0 = λ

Hence, for λ = 0, the given planes are perpendicular to each other.

Q5.

Solution:

We know that the solution of a plane passing through (x₁, y₁, z₁) is given as:

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

The required plane passes through (– 1, – 1, 2),

So the equation of the plane is

a(x + 1) + b(y + 1) + c(z – 2) = 0

⇒ ax + by + cz = 2c – a – b …… (1)

Now, the required plane is also perpendicular to the planes,

3x + 2y – 3z = 1 and

5x – 4y + z = 5

We know that planes

a₁x + b₁y + c₁z + d₁ = 0 and

a₂x + b₂y + c₂z + d₂ = 0 are at right angles if,

a₁a₂ + b₁b₂ + c₁c₂ = 0 …… (2)

Using equation (2), we have,

3a + 2b – 3c = 0 …… (3)

5a – 4b + c = 0 …… (4)

Now let us solve equations (3) and (4), and we get,

So,

a = – 10λ, b = – 18λ, c = – 22λ

Now substitute the values of a, b, c in equation (1), and we get,

(– 10λ)x + (– 18λ)y + (– 22λ)z = 2(– 22)λ – (– 10λ) – (– 18λ)

– 10λx – 18λy – 22λz = – 44λ + 10λ + 18λ

– 10λx – 18λy – 22λz = – 16λ

Divide both sides by (– 2λ), we get

5x + 9y + 11z = 8

Hence, the required equation of the plane is 5x + 9y + 11z = 8.