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# Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.6

• Last Updated : 23 Jul, 2021

### (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).

### (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).

### (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.

### (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

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