Q21Written answer
Find the shortest distance between the lines given by 
and 
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Worked solution
Given equations of lines are
NCERT Exemplar · Class 12 · Maths
Chapter 11: Three Dimensional Geometry
Practice each question, then open the answer or worked solution.
Questions
24 questionsQ22Written answer
Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
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Worked solution
The given planes are
P1: 5x + 3y + 6z + 8 = 0
P2: x + 2y + 3z – 4 = 0
P3: 2x + y – z + 5 = 0
Now, the equation of the plane passing through the line of intersection of P1 and P3 is
(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0
(1 + 2λ)x + (2 + λ)y + (3 – λ)z – 4 + 5λ = 0 …. (i)
From the question its understood that plane (i) is perpendicular to P1, then
5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0
5 + 10λ + 6 + 3λ + 18 – 6λ = 0
7λ + 29 = 0
λ = -29/7
Putting the value of ; in equation (i), we get
-15x – 15y + 50z – 28 – 145 = 0
-15x – 15y + 50z – 173 = 0 ⇒ 51x + 15y – 50z + 173 = 0
Thus, the required equation of plane is 51x + 15y – 50z + 173 = 0.
Q23Written answer
The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is
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Worked solution
Given planes are: ax + by = 0 …. (i) and z = 0 …. (ii)
Now, the equation of any plan passing through the line of intersection of plane (i) and (ii) is
(ax + by) + kz = 0 ⇒ ax + by + kz = 0 …. (iii)
Q24Written answer
Find the equation of the plane through the intersection of the planes 
and 
whose perpendicular distance from origin is unity.
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Worked solution
Thus, the required equations of planes are
4x + 2y – 4z – 6 = 0 and -2x – 4y + 4z – 6 = 0.