Class 12 RD Sharma Solutions - Chapter 29 The Plane - Exercise 29.6

Question 1. Find the angle between the given planes:

(i) and 

Solution:

Given 

and 

As we know that the angle between the planes  is given by,

Here, 

So, 

= 

= 

= 

Therefore, .

(ii)  and 

Solution:

Given 

 and 

As we know that the angle between the planes  is given by,

Here, 

So, 

= 

= 

= -4/21

Therefore, θ = cos^-1 (-4/21).

(iii) and 

Solution:

Given 

and 

As we know that the angle between the planes,  is given by,

Here, 

So, 

= 

= 

= -16/21

Therefore, θ = cos^-1 (-16/21).

Question 2. Find the angle between the planes:

(i) 2x − y + z = 4 and x + y + 2z = 3

Solution:

Given, 2x − y + z = 4 and x + y + 2z = 3

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

So, the angle between 2x - y + z = 4 and x + y + 2z = 3 is given by,

= 

= 

= 3/6

= 1/2

Therefore, θ = cos^-1 (1/2) = π/3.

(ii) x + y − 2z = 3 and 2x − 2y + z = 5

Solution:

Given, x + y − 2z = 3 and 2x − 2y + z = 5

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

So, the angle between x + y - 2z = 3 and 2x - 2y + z = 5 is given by,

= 

= 

= -2/3√6

Therefore, θ = cos^-1 (-2/3√6).

(iii) x − y + z = 5 and x + 2y + z = 9

Solution:

Given, x − y + z = 5 and x + 2y + z = 9

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

So, the angle between x - y + z = 5 and x + 2y + z = 9 is given by,

= 

= 

= 0

Therefore, θ = cos^-1 (0) = π/2.

(iv) 2x − 3y + 4z = 1 and − x + y = 4

Solution:

Given, 2x − 3y + 4z = 1 and − x + y = 4

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

So, the angle between 2x - 3y + 4z = 1 and -x + y + 0z = 4 is given by,

= 

= 

= 

Therefore, .

(v) 2x + y − 2z = 5 and 3x − 6y − 2z = 7

Solution:

Given, 2x + y − 2z = 5 and 3x − 6y − 2z = 7

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

So, the angle between 2x + y - 2z = 5 and 3x - 6y - 2z = 7 is given by,

= 

= 

= 4/21

Therefore, θ = cos^-1 (4/21).

Question 3. Show that the following planes are at right angles.

(i) and 

Solution:

Given, and 

As we know that the planes  are perpendicular to each other only if .

Here, 

Now, we have

= -2 + 1 + 1 

= 0

So, the given planes are perpendicular.

Hence proved.

(ii) x − 2y + 4z = 10 and 18x + 17y + 4z = 49

Solution:

Given, x − 2y + 4z = 10 and 18x + 17y + 4z = 49

As we know that the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 are perpendicular to each other only if,

=> a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

The given planes are x - 2y + 4z = 10 and 18x + 17y + 4z = 49.

Here, a₁ = 1, b₁ = - 2, c₁ = 4, a₂ = 18, b₂ = 17 and c₂ = 4

Now, we have

a₁ a₂ + b₁ b₂ + c₁ c₂ = (1) (18) + (- 2) (17) + (4) (4)

= 18 - 34 + 16 

= 0

So, the given planes are perpendicular.

Hence proved.

Question 4. Determine the value of λ for which the following planes are perpendicular to each other.

(i) and 

Solution:

Given, and 

As we know that the planes,  are perpendicular to each other only if,  .

Here, 

The given planes are perpendicular. So, we have

λ + 4 - 21 = 0

λ - 17 = 0

λ = 17

Therefore, the value of λ is 17.

(ii) 2x − 4y + 3z = 5 and x + 2y + λz = 5

Solution:

Given, 2x − 4y + 3z = 5 and x + 2y + λz = 5

As we know that the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 are perpendicular to each other only if,

=> a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

The given planes are 2x - 4y + 3z = 5 and  x + 2y + λz = 5.

Here, a₁ = 2, b₁ = - 4, c₁ = 3, a₂ = 1, b₂ = 2 and c₂ = λ.

It is given that the given planes are perpendicular. So, we get

a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

(2) (1) + (-4) (2) + (3) (λ) = 0

2 - 8 + 3λ = 0

3λ = 6

λ = 2

Therefore, the value of λ is 2.

(iii) 3x − 6y − 2z = 7 and 2x + y − λz = 5

Solution:

Given, 3x − 6y − 2z = 7 and 2x + y − λz = 5

As we know that the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 are perpendicular to each other only if,

=> a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

The given planes are 3x - 6y - 2z = 7 and 2x + y - λz = 5.

Here, a₁ = 3, b₁ = - 6, c₁ = - 2, a₂ = 2, b₂ = 1 and c₂ = -λ

Here, the given planes are perpendicular.

a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

(3) (2) + (-6) (1) + (-2) (λ) = 0

6 - 6 + 2λ = 0

2λ = 0

λ = 0

Question 5. Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.

Solution:

The equation of any plane passing through  (-1, -1, 2) is,

a (x + 1) + b (y + 1) + c (z - 2) = 0  . . . . (1)

It is given that the above equation is perpendicular to each of the planes 3x + 2y - 3z = 1 and 5x - 4y + z = 5. 

3a + 2b - 3c = 0   . . . . (2)

5a - 4b + c = 0    . . . . (3)

On solving eq (1), (2) and (3), we get,

-10 (x + 1) - 18 (y + 1) - 22 (z - 2) = 0

-5 (x + 1) - 9 (y + 1) - 11 (z - 2) = 0

5x + 5 + 9y + 9 + 11z - 22 = 0

5x + 9y + 11z - 8 = 0

Hence, the equation of the plane is 5x + 9y + 11z - 8 = 0

Question 6. Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.

Solution:

The equation of any plane passing through (1, -3, -2) is,

a (x - 1) + b (y + 3) + c (z + 2) = 0 . . . . (1)

It is given that above equation is perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.

a + 2b + 2c = 0    . . . . (2)

3a + 3b + 2c = 0  . . . . (3)

On solving eq (1), (2) and (3), we get,

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

-2x + 2 + 4y + 12 - 3z - 6 = 0

2x - 4y + 3z - 8 = 0

Hence, the equation of the plane is 2x - 4y + 3z - 8 = 0

Question 7. Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y − z = 1 and 3x − 4y + z = 5.

Solution:

The equation of any plane passing through the origin (0, 0, 0) is,

a (x - 0) + b (y - 0) + c (z - 0) = 0  . . . . (1)

ax + by + cz = 0 

It is given that above equation is perpendicular to the planes x + 2y - z = 1 and 3x - 4y + z = 5 .

a + 2b - c = 0   . . . . (2)

3a - 4b + c = 0  . . . . (3)

On solving eq(1), (2) and (3), we get,

- 2x - 4y - 10z = 0

x + 2y + 5z = 0

Hence, the equation of the plane is x + 2y + 5z = 0

Question 8. Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.

Solution:

The equation of any plane passing through (1, -1, 2) is,

a (x - 1) + b (y + 1) + c (z - 2) = 0  . . . . (1)

It is given that above equation is passing through (2, -2, 2). So,

a (2 - 1) + b (-2 + 1) + c (2 - 2) = 0 

a - b = 0  . . . . (2)

It is given that above equation is perpendicular to the plane 6x - 2y + 2z = 9. So,

6a - 2b + 2c = 0

3a - b + c = 0  . . . . (3)

On solving eq(1), (2) and (3), we get,

-1 (x - 1) - 1 (y + 1) + 2 (z - 2) = 0

- x + 1 - y - 1 + 2z - 4 = 0

x + y - 2z + 4 = 0

Hence, the equation of the plane is x + y - 2z + 4 = 0

Question 9. Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.

Solution:

The equation of any plane passing through (2, 2, 1) is,

a (x - 2) + b (y - 2) + c (z - 1) = 0  . . . . (1)

It is given that the above equation is passing through (9, 3, 6). So,

a (9 - 2) + b (3 - 2) + c (6 - 1) = 0

7a + b + 5c = 0  . . . . (2)

It is given that the above equation is perpendicular to the plane 2x + 6y + 6z = 1. So,

2a + 6b + 6c = 0

a + 3b + 3c = 0 . . . . (3)

On solving eq(1), (2) and (3), we get

-12 (x - 2) - 16 (y - 2) + 20 (z - 1) = 0

3 (x - 2) + 4 (y - 2) - 5 (z - 1) = 0

3x + 4y - 5z = 9

Hence, the equation of the plane is 3x + 4y - 5z = 9

Question 10. Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5.

Solution:

The equation of any plane passing through (-1, 1, 1) is,

a (x + 1) + b (y - 1) + c (z - 1) = 0  . . . . (1)

It is given that the above equation passed through (1, -1, 1). So,

a (1 + 1) + b (-1 - 1) + c (1 - 1) = 0 

2a - 2b = 0  . . . . (2)

It is given that the above equation is perpendicular to the plane x + 2y + 2z = 5. So,

a + 2b + 2c = 0  . . . . (3)

On solving eq(1), (2) and (3), we get

-4 (x + 1) - 4 (y - 1) + 6 (z - 1) = 0

2 (x + 1) + 2 (y - 1) - 3 (z - 1) = 0

2x + 2y - 3z + 3 = 0

Hence, the equation of the plane is 2x + 2y - 3z + 3 = 0

Question 11. Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane.

Solution:

The equation of the plane parallel to the plane ZOX  is,

y = b  . . . . (1)

According to the question, it is given that this plane passes through (0, 3, 0). So,

=> b = 3

On substituting this value in eq(1), we get

y = 3, 

Hence, the equation of the plane is

y = 3 

Question 12. Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.

Solution:

The equation of any plane passing through (1, -1, 2) is,

a (x - 1) + b (y + 1) + c (z - 2) = 0  . . . . (1)

It is given that the above equation is perpendicular to the plane 2x + 3y - 2z = 5. So, 

2a + 3b - 2c = 0  . . . . (2)

It is given that the above equation is perpendicular to the plane x + 2y - 3z = 8. So,

a + 2b - 3c = 0   . . . . (3)

So, on solving eq (1), (2) and (3), we get

-5 (x - 1) + 4 (y + 1) + 1 (z - 2) = 0

5x - 4y - z = 7

Hence, the equation of the plane is 5x - 4y - z = 7

Question 13. Find the equation of the plane passing through (a, b, c) and parallel to the plane  = 2.

Solution:

Given equation of the plane is \vec{r} \cdot \left( \hat{i}  + \hat{j}  + \hat{k}  \right)  = 2

So, on substituting  in the given equation of the plane, we get

=> x + y + z - 2 = 0   . . . . (1)

The equation of a plane which is parallel to plane (eq 1) is of the form,

x + y + z = k   . . . . (2)

It is given that above equation of plane is passing through the point (a, b, c). So,

a + b + c = k

On substituting this value of  k in eq(2), we get

x + y + z = a + b + c

Hence, the equation of the plane is

x + y + z = a + b + c 

Question 14. Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Solution:

The equation of any plane passing through point (-1, 3, 2) is,

a (x + 1) + b (y - 3) + c (z - 2) = 0  . . . . (1)

It is given that the above equation is perpendicular to the plane x + 2y + 3z = 5. So, 

a + 2b + 3c = 0   . . . . (2)

It is given that the above equation is perpendicular to the plane 3x + 3y + z = 0. So,

3a + 3b + c = 0   . . . . (3)

On solving eq (1), (2) and (3), we get

-7 (x + 1) + 8 (y - 3) - 3 (z - 2) = 0

7x - 8y + 3z + 25 = 0

Hence, the equation of the plane is

7x - 8y + 3z + 25 = 0

Question 15. Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10. 

Solution:

The equation of any plane passing through (2, 1, -1) is,

a (x - 2) + b (y - 1) + c (z + 1) = 0  . . . . (1)

It is given that the above equation passes through (-1, 3, 4). So,

a (-1 - 2) + b (3 - 1) + c (4 + 1) = 0

-3a + 2b + 5c   . . . . (2)

It is given that the above equation is perpendicular to the plane x - 2y + 4z = 10. So,

a - 2b + 4c = 0  . . . . (3)

On solving eq (1), (2) and (3), we get

18 (x - 2) + 17 (y - 1) + 4 (z + 1) = 0

18x + 17y + 4z - 49 = 0

Hence, the equation of the plane is

18x + 17y + 4z - 49 = 0