Here we provide RD Sharma Class 12 Ex 29.6 Solutions Chapter 29 The Plane for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 29.6 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 29 |

Exercise | 29.6 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 29.6 Solutions Chapter 29 The Plane**

**Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed **

### 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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