# Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.8

### Question 1. Find the equation of the plane which is parallel to 2x â€“ 3y + z = 0 and passes through the point (1, â€“1, 2).

Solution:

We know that the equation of a plane parallel to 2x â€“ 3y + z = 0 is given by:

2x â€“ 3y + z + Î» = 0

Since the plane passes through the point (1, â€“1, 2), we have:

2(1) â€“ 3(â€“1) + 2 + Î» = 0

â‡’ Î» = â€“7

On substituting the value of  Î» in the equation, we have:

2x â€“ 3y + z + (-7) = 0

2x â€“ 3y + z â€“ 7= 0 is the required equation.

### Question 2. Find the equation of the plane through (3, 4, â€“1) which is parallel to the plane

Solution:

The given plane passes through the vector . Thus,

(3)(2) + (4)(-3) + (-1)(5) + Î» = 0

â‡’  Î» = 11

On substituting the value of  Î» in the equation, we have:

is the required equation.

### Question 3. Find the equation of the plane passing through the line of intersection of the planes 2x â€“ 7y + 4z â€“ 3 = 0 and 3x â€“ 5y + 4z + 11 = 0 and the point (â€“2, 1, 3).

Solution:

The equation of the plane passing through the line of intersection of the given planes is:
(2x â€“ 7y + 4z â€“ 3) + Î»(3x â€“ 5y + 4z + 11) = 0

â‡’ x(2 + 3Î») + y(â€“7 â€“ 5Î») + z(4 + 4Î») â€“ 3 + 11Î» = 0

Also, since the plane passes through the point (â€“2, 1, 3), we have:

(â€“2)(2 + 3Î») + (1)(â€“7 â€“ 5Î») + (3)(4 + 4Î») â€“ 3 + 11Î» = 0

â‡’ Î» = 1/6

On substituting the value of  Î» in the equation, we have:

x(2 + 3(1/6)) + y(â€“7 â€“ 5(1/6)) + z(4 + 4(1/6)) â€“ 3 + 11(1/6) = 0

15x â€“ 47y + 28z = 7 is the required equation.

### Question 4. Find the equation of the plane passing through the point and passing through the line of intersection of the planes and

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

Also, since the plane passes through point , we have:

â‡’  Î» = 6

On substituting the value of  Î» in the equation, we have:

is the required equation.

### Question 5. Find the equation of the plane passing through the intersection of 2x â€“ y = 0 and 3z â€“ y = 0 and perpendicular to 4x + 5y â€“ 3z = 8.

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

2x â€“ y + Î»(3z â€“ y) = 0

â‡’ 2x + y(â€“1 â€“ Î») + z(3Î») = 0

Since the planes are perpendicular, we have:

2(4) + (â€“5)(â€“1 â€“ Î») + (â€“3)(3Î») = 0

â‡’ Î» = 3/14

On substituting the value of  Î» in the equation, we have:

2x + y(â€“1 â€“ 3/14) + z(3(3/14)) = 0

28x â€“ 17y + 9z = 0 is the required equation.

### Question 6. Find the equation of the plane which contains the line of intersection of planes x + 2y + 3z â€“ 4 = 0 and 2x + y â€“ z + 5 = 0 and is perpendicular to the plane 5x + 3y â€“ 6z + 8 = 0.

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

x + 2y + 3z â€“ 4 + Î»(2x + y â€“ z + 5) = 0

â‡’ x(1 + 2Î») + y(2 + Î») + z(3 â€“ Î») â€“ 4 + 5Î» = 0

Since the planes are perpendicular, we have:

5(1 + 2Î») + 3(2 + Î») + (â€“6)(3 â€“ Î») = 0

â‡’  Î» = 7/19

On substituting the value of  Î» in the equation, we have:

x(1 + 2(7/19)) + y(2 + 7/19) + z(3 â€“ 7/19) â€“ 4 + 5(7/19) = 0

33x + 45y + 50z â€“ 41 = 0 is the required equation.

### Question 7. Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z + 4 = 0 and  x â€“ y + z + 3 = 0 and passing through the origin.

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

x + 2y + 3z + 4 + Î»(x â€“ y + z + 3) = 0

â‡’ x(1 + Î») + y(2 â€“ Î») + z(3 + Î») + 4 + 3Î» = 0

Also, since the plane passes through the origin, we have:

0(1 + Î») + 0(2 â€“ Î») + 0(3 + Î») + 4 + 3Î» = 0

â‡’ Î» = -4/3

On substituting the value of  Î» in the equation, we have:

x(1 + (-4/3)) + y(2 â€“ (-4/3)) + z(3 + (-4/3)) + 4 + 3(-4/3) = 0

x â€“ 10y â€“ 5z = 0 is the required equation.

### Question 8. Find the vector equation in scalar product form of the plane containing the line of intersection of the planes x â€“ 3y + 2z â€“ 5 = 0 and 2x â€“ y + 3z â€“ 1 = 0 and passing through (1, â€“2, 3).

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

x â€“ 3y + 2z â€“ 5 + Î»(2x â€“ y + 3z â€“ 1) = 0

â‡’ x(1 + 2Î») + y(â€“3 â€“ Î») + z(2 + 3Î») â€“ 5 â€“ Î» = 0

Also, since the plane passes through the origin, we have:

1(1 + 2Î») + (â€“2)(â€“3 â€“ Î») + 3(2 + 3Î») â€“ 5 â€“ Î» = 0

â‡’  Î» = -2/3

On substituting the value of  Î» in the equation, we have:

x(1 + 2(-2/3)) + y(â€“3 â€“ (-2/3)) + z(2 + 3(-2/3)) â€“ 5 â€“ (-2/3) = 0

is the required equation.

### Question 9. Find the equation of the plane which contains the line of intersection of planes x + 2y + 3z â€“ 4 = 0 and 2x + y â€“ z + 5 = 0 and is perpendicular to the plane 5x + 3y + 6z + 8 = 0.

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

x + 2y + 3z â€“ 4 + Î»(2x + y â€“ z + 5) = 0

â‡’ x(1 + 2Î») + y(2 + Î») + z(3 â€“ Î») â€“ 4 + 5Î» = 0

We know that two planes are perpendicular when

â‡’ 5(1 + 2Î») + 3(2 + Î») + 6(3 â€“ Î») = 0

â‡’ Î» = -29/7

On substituting the value of  Î» in the equation, we have:

x(1 + 2(-29/7)) + y(2 + (-29/7)) + z(3 â€“ (-29/7)) â€“ 4 + 5(-29/7) = 0

51x + 15y â€“ 50z + 173 = 0 is the required equation.

### Question 10. Find the equation of the plane passing through the line of intersection of the planes and and which is at a unit distance from the origin.

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

x(1 + 3Î») + y(3 + Î») â€“ 4zÎ» + 6 = 0

Distance from plane to the origin = 1

â‡’

â‡’ Î» = Â±1

Hence, 4x + 2y â€“ 4z + 6 = 0 and â€“2x + 2y + 4z + 6 = 0 are the required equations.

### Question 11. Find the equation of the plane passing through the line of intersection of the planes 2x + 3y â€“ z + 1 = 0 and x + y â€“ 2z + 3 = 0 and perpendicular to the plane 3x â€“ 2y â€“ z â€“ 4 = 0.

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

2x + 3y â€“ z + 1  + Î»(x + y â€“ 2z + 3) = 0

â‡’ x(2 + Î») + y(3 + Î») + z(â€“1 â€“ 2Î») + 1 + 3Î» = 0

We know that two planes are perpendicular when

â‡’ 3(2 + Î») + (â€“1)(3 + Î») + (â€“2)(â€“1 â€“ 2Î») = 0

â‡’  Î» = -5/6

On substituting the value of  Î» in the equation, we have:

x(2 + (-5/6)) + y(3 + (-5/6)) + z(â€“1 â€“ 2(-5/6)) + 1 + 3(-5/6) = 0

7x + 13y + 4z â€“ 9 = 0 is the required equation.

### Question 12. Find the equation of the plane that contains the line of intersection of the planes and and which is perpendicular to the plane

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

â‡’

We know that two planes are perpendicular if

â‡’

â‡’ 5(1 + 2Î») + 3(2 + Î») + (â€“6)(3 â€“ Î») = 0

â‡’  Î» = 7/19

On substituting the value of  Î» in the equation, we have:

33x + 45y + 50z â€“ 41 = 0 is the required equation.

### Question 13. Find the vector equation of the plane passing through the intersection of planes and and the point (1, 1, 1).

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

â‡’ x(1 + 2Î») + y(1 + 3Î») +z(1 + 4Î») = 6  â€“ 5Î»

Also, since the plane passes through the point(1, 1, 1), we have:

1(1 + 2Î») + 1(1 + 3Î») +1(1 + 4Î») = 6  â€“ 5Î»

â‡’ Î» = 3/14

On substituting the value of  Î» in the equation, we have:

x(1 + 2(3/14)) + y(1 + 3(3/14)) +z(1 + 4(3/14)) = 6  â€“ 5(3/14)

is the required equation.

### Question 14. Find the equation of the plane passing through the intersection of the planes and and the point (2, 1, 3).

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

â‡’

Also, since the plane passes through the point (2, 1, 3) we have:

9Î» = â€“7

â‡’ Î» = -7/9

Substituting the value of  Î» in the equation, we have:

is the required equation.

### Question 15. Find the equation of the plane passing through the intersection of the planes 3x â€“ y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

3x â€“ y + 2z â€“ 4  + Î»(x + y + z  â€“ 2) = 0

Also, since the plane passes through the point (2, 2, 1), we have:

Î» = -2/3

On substituting the value of  Î» in the equation, we have:

3x â€“ y + 2z â€“ 4  + (-2/3)(x + y + z  â€“ 2) = 0

7x â€“ 5y + 4z = 0 is the required equation.

### Question 16. Find the vector equation of the plane through the line of intersection of the planes x + 2y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x â€“ y + z = 0.

Solution:

The equation of the plane passing through the line of intersection of the given planes is:

x + 2y + z â€“ 1  + Î»(2x + 3y + 4z â€“ 5) = 0

â‡’ x(1 + 2Î») + y(1 + 3Î») +z(1 + 4Î») = 1 + 5Î»

We know that two planes are perpendicular when

â‡’ 1(1 + 2Î») + (â€“1)(1 + 3Î») + 1(1 + 4Î») = 1 + 5Î»

â‡’ Î» = -1/3

On substituting the value of  Î» in the equation, we have:

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

x â€“ z + 2 = 0 is the required equation.

### Question 17. Find the equation of the plane passing through (a, b, c) and parallel to the plane

Solution:

Equation of the family of planes parallel to the given plane =

Since the plane passes through (a, b, c), we have:

a + b + c = d

Substituting the above equation in the equation of family of planes we have:

Hence, x + y + z = a + b + c is the required equation of the plane.

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