# Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.9

Last Updated : 12 Oct, 2021

### Question 1. Find the distance of the point  from the plane

Solution:

As we know the distance of a point  from a plane  is given by:

Here,  and  is the plane.

Hence,

â‡’

= |-47/13| units

= 47/13 units

Hence, the distance of the point from the plane is 47/13 units.

### Question 2. Show that the points  and  are equidistant from the plane

Solution:

As we know the distance of a point  from a plane  is given by:

Let D1 be the distance of  from the plane

â‡’

= 9/âˆš78 units                  …….(1)

Now, let D2 be the distance between point  and the plane .

â‡’

= 9/âˆš78 units                    ……(2)

From eq(1) and (2), we have

The given points are equidistant from the given plane.

### Question 3. Find the distance of the point (2, 3, âˆ’5) from the plane x + 2y âˆ’ 2z âˆ’ 9 = 0.

Solution:

As we know the distance is given by:

â‡’

= 9/âˆš9

D = 3 units.

### Question 4. Find the equations of the planes parallel to the plane x + 2y âˆ’ 2z + 8 = 0 which are at a distance of 2 units from the point (2, 1, 1).

Solution:

The equation of a plane parallel to the given plane is x + 2y âˆ’ 2z + p = 0.

As we know that the distance between a point and plane is given by:

Given, D = 2 units. Hence,

â‡’

On squaring both sides, we have

â‡’ 36 = (2 + p)2

â‡’ 2 + p = 6           or         2 + p = âˆ’6

â‡’ p = 4                 or           p = âˆ’8

Hence, the equations of the required planes are:

x + 2y âˆ’ 2z + 4 = 0 and x + 2y âˆ’ 2z âˆ’ 8 = 0.

### Question 5. Show that the points (1, 1, 1) and(âˆ’3, 0, 1) are equidistant from the plane 3x + 4y âˆ’ 12z +13 = 0.

Solution:

We know that the distance between a point and plane is given by:

Let D1 be the distance of the point (1,1,1) from the plane.

â‡’

= 8/13 units

Let D2 be the distance of the point.

â‡’

= 8/13 units

Hence, the points are equidistant from the plane.

### Question 6. Find the equation of the planes parallel to the plane x âˆ’ 2y + 2z âˆ’ 3 = 0 which are at a unit distance from the point (2, 1, 1).

Solution:

Equation of a plane parallel to the given plane is x âˆ’ 2y + 2z + p = 0.

As we know that the distance between a point and plane is given by:

Given, D = 1 units. Hence,

â‡’

On squaring both sides, we have

â‡’ 9 = (1 + p)2

â‡’ 1 + p = 3           or         1 + p = âˆ’3

â‡’ p = 2                 or           p = âˆ’4

Hence, the equations of the required planes are:

x âˆ’ 2y + 2z + 2 = 0 and x âˆ’ 2y + 2z âˆ’ 4 = 0.

### Question 7. Find the distance of the point (2, 3, 5) from the xy-plane.

Solution:

As we know the distance of the point from the plane is given by:

D =

D = 5 units

### Question 8. Find the distance of the point (3, 3, 3) from the plane

Solution:

As we know the distance of a point  from a plane  is given by:

D =

D = 9/âˆš78 units.

### Question 9. If the product of distances of the point (1, 1, 1) from the origin and the plane x âˆ’ y + z + p = 0 be 5, find p.

Solution:

The distance of the point (1, 1, 1) from the origin is

Distance of (1, 1, 1) from the plane is

Given:

â‡’ |1 + p| = 5

â‡’ p = 4 or âˆ’6.

### Question 10. Find an equation of the set of all points that are equidistant from the planes 3x âˆ’ 4y + 12 = 6 and 4x + 3z = 7.

Solution:

Now,

Given, D1 = D2

â‡’

Hence, the equations become

37x1 + 20y1 âˆ’ 21z1 âˆ’ 61 = 0 and  67x1 + 20y1 + 99z1 âˆ’ 121 = 0.

### Question 11. Find the distance between the point (7, 2, 4) and the plane determined by the points A(âˆ’2, âˆ’3, 5) and C(5, 3, âˆ’3).

Solution:

The equations of the plane are given as:

âˆ’4a âˆ’ 8b + 8c = 0 and 3a âˆ’ 2b + 0c = 0

Solving the above set using cross multiplication method, we get

â‡’

â‡’

â‡’

â‡’ a = 2p, b = 3p, c = 4p

Thus, the equation of the plane becomes 2x + 3y + 4z âˆ’ 7 = 0.

and, distance = âˆš29 units.

### Question 12. A plane makes intercepts âˆ’6, 3, 4 respectively on the coordinate axis. Find the length of the perpendicular from the origin on it.

Solution:

Since the plane makes intercepts âˆ’6, 3, 4, the equation becomes:

Let p be the distance of perpendicular from the origin to the plane.

â‡’

â‡’

â‡’

â‡’ 1/p2 = 29/144

â‡’ p2 = 144/29

â‡’ p = 12/âˆš29 units.

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