# RBSE Solutions for Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.6

## 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 