# 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 a

_{1}x + b_{1}y + c_{1}z + d_{1}= 0 and a_{2}x + b_{2}y + c_{2}z + d_{2}= 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 a

_{1}x + b_{1}y + c_{1}z + d_{1}= 0 and a_{2}x + b_{2}y + c_{2}z + d_{2}= 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 a

_{1}x + b_{1}y + c_{1}z + d_{1}= 0 and a_{2}x + b_{2}y + c_{2}z + d_{2}= 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

_{1}x + b_{1}y + c_{1}z + d_{1}= 0 and a_{2}x + b_{2}y + c_{2}z + d_{2}= 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

_{1}x + b_{1}y + c_{1}z + d_{1}= 0 and a_{2}x + b_{2}y + c_{2}z + d_{2}= 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 a

_{1}x + b_{1}y + c_{1}z + d_{1}= 0 and a_{2}x + b_{2}y + c_{2}z + d_{2}= 0 are perpendicular to each other only if,=> a

_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}= 0The given planes are x – 2y + 4z = 10 and 18x + 17y + 4z = 49.

Here, a

_{1}= 1, b_{1}= – 2, c_{1}= 4, a_{2}= 18, b_{2}= 17 and c_{2}= 4Now, we have

a

_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}= (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 a

_{1}x + b_{1}y + c_{1}z + d_{1}= 0 and a_{2}x + b_{2}y + c_{2}z + d_{2}= 0 are perpendicular to each other only if,=> a

_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}= 0The given planes are 2x – 4y + 3z = 5 and x + 2y + λz = 5.

Here, a

_{1}= 2, b_{1}= – 4, c_{1}= 3, a_{2}= 1, b_{2}= 2 and c_{2}= λ.It is given that the given planes are perpendicular. So, we get

a

_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}= 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 a

_{1}x + b_{1}y + c_{1}z + d_{1}= 0 and a_{2}x + b_{2}y + c_{2}z + d_{2}= 0 are perpendicular to each other only if,=> a

_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}= 0The given planes are 3x – 6y – 2z = 7 and 2x + y – λz = 5.

Here, a

_{1}= 3, b_{1}= – 6, c_{1}= – 2, a_{2}= 2, b_{2}= 1 and c_{2}= -λHere, the given planes are perpendicular.

a

_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}= 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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