## Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.6

Question 1.

Find the equation of the plane which is perpendicular to x-axis and that passes through the point (2,-1,3).

Solution:

Equation of plane passing through point (2,-1,3)

a(x – 2) + b(y + 1) + c(z – 3) = 0

∵ Plane is perpendicular to T-axis.

∴ b = 0, c = 0

Hence required equation of plane

a(x – 2) + 0(y + 1) + 0(z – 3) = 0

⇒ a(x – 2) = 0

⇒ x – 2 = 0

(∵a ≠ 0)

Question 2.

Find the equation of the plane that passes through X-axis and point (3,2, 4).

Solution:

Equation of plane passing through (3, 2, 4)

a (x – 3) + b(y – 2) + c(z – 4) = 0 …..(1)

∵ Plane passes through X-axis

∴ a = 0,d = 0 ⇒ by + cz = 0 …..(2)

From (1), a = 0 so

b (y – 2) + c(z – 4) = 0 …..(3)

⇒ by – 2b + cz – 4c = 0

⇒ by + cz – 2b – 4c = 0

⇒ – 2b = c

[∵ by + cz = 0 from (2)]

⇒ b = -2c

∴ Equation of plane passing through (3,2,4) and X-axis

b(y- 2) + c(z – 4) = 0

⇒ – 2c(y – 2) + c(z – 4) = 0

⇒ -2y + 4 + z – 4 = 0

⇒ 2y – z = 0

Question 3.

A variable plane passes through the point (p, q, r) and meets the coordinate axis at point A, B and C. Show that the locus of a common point of plane passing through A, B and C and parallel to the coordinate planes, will be

Solution:

Let equation of plane is

∴ Plane passes through point (p,q,r)

Again plane meets the coordinate points.

Coordinates of points are (α,0,0)

Coordinates of point B are (0, β, 0)

and coordinates of point C are (0, 0, γ)

∴ Equation of planes, parallel to coordinate axis and passing through

Point A is x = α …..(3)

Point B is y = β …..(4)

Point C is z = γ …..(5)

∴ Locus of the point of intersection is

\(\frac { p }{ x } \) + \(\frac { q }{ y } \) + \(\frac { r }{ z } \) = 1

Question 4.

Find the vector equation of a plane which is at a distance of 7 unit from the origin and has \(\hat { i }\) as the unit vector normal to it.

Solution:

Given unit vector along normal

\(\hat { n }\) = i

and distance from origin (0, 0, 0) = 7 units

∴ from vector equation of plane

Question 5.

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 6i + 3j – 2k.

Solution:

Unit vector along 6i + 3j – 2k

Question 6.

Reduce the equation 3x – 4y + 12z = 5 or \(\overrightarrow { r } \).(3i – 4j + 12k) = 5 to normal form and hence find the length of perpendicular from the origin to the plane. Also find direction cosines of the normal to the plane.

Solution:

First method : Given

II method : On dividing 3x – 4y + 12z – 5 by its absolute value 5.

Question 7.

Find the vector equation of a plane which is at a distance of 4 units from the origin and direction cosines of the normal to the plane are 2, – 1,2.

Solution:

Given Dc’s of the normal to the plane are 2,-1, 2.

Question 8.

Find normal form of the plane 2x – 3y + 6z + 14 = 0.

Solution:

Equation of given plane is 2x – 3y + 6z + 14 = 0

Dc’s of normal plane are (2, 3, 6).

∴ Dc’s of normal are

Question 9.

Find the equation of plane perpendicular of the origin from the plane is 13 and direction ratios of this perpendicular are 4,-3,12.

Solution:

Given DR’s of normal on plane are 4, -3, 12.

∴ DC’s of normal are

Question 10.

Find a unit normal vector to the plane x + y + z – 3 = 0.

Solution:

Let given x + y + z – 3 = 0