RBSE Solutions for Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.6
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
p /x + q/y + 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 i^ as the unit vector normal to it.
Solution:
Given unit vector along normal
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 
(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
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