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,

⇒

= |-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

⇒

= 9/√78 units                  …….(1)

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

⇒

= 9/√78 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, −5) from the plane x + 2y − 2z − 9 = 0.

Solution:

As we know the distance is given by:

⇒

= 9/√9

D = 3 units.

Question 4. Find the equations of the planes parallel to the plane x + 2y − 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 − 2z + p = 0.

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

Given, D = 2 units. Hence,

⇒

On squaring both sides, we have

⇒ 36 = (2 + p)2

⇒ 2 + p = 6           or         2 + p = −6

⇒ p = 4                 or           p = −8

Hence, the equations of the required planes are:

x + 2y − 2z + 4 = 0 and x + 2y − 2z − 8 = 0.

Question 5. Show that the points (1, 1, 1) and(−3, 0, 1) are equidistant from the plane 3x + 4y − 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.

⇒

= 8/13 units

Let D2 be the distance of the point.

⇒

= 8/13 units

Hence, the points are equidistant from the plane.

Question 6. Find the equation of the planes parallel to the plane x − 2y + 2z − 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 − 2y + 2z + p = 0.

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

Given, D = 1 units. Hence,

⇒

On squaring both sides, we have

⇒ 9 = (1 + p)2

⇒ 1 + p = 3           or         1 + p = −3

⇒ p = 2                 or           p = −4

Hence, the equations of the required planes are:

x − 2y + 2z + 2 = 0 and x − 2y + 2z − 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/√78 units.

Question 9. If the product of distances of the point (1, 1, 1) from the origin and the plane x − 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:

⇒ |1 + p| = 5

⇒ p = 4 or −6.

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

Solution:

Now,

Given, D1 = D2

⇒

Hence, the equations become

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

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

Solution:

The equations of the plane are given as:

−4a − 8b + 8c = 0 and 3a − 2b + 0c = 0

Solving the above set using cross multiplication method, we get

⇒

⇒

⇒

⇒ a = 2p, b = 3p, c = 4p

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

and, distance = √29 units.

Question 12. A plane makes intercepts −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 −6, 3, 4, the equation becomes:

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

⇒

⇒

⇒

⇒ 1/p2 = 29/144

⇒ p2 = 144/29

⇒ p = 12/√29 units.

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