NCERT Solutions
Class 12 Maths
Three Dimensional Geometry

Ex.Misc.Q.19
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes r.(i – j + 2k) = 5 and r.(3i + j + k) = 6
Let the required line be parallel to vector b is given by,
b = b1i + b2j + b3k
The position vector of the point (1, 2, 3) is a = i + 2j + 3k
The equation of line passing through (1, 2, 3) and parallel to b is given by,
r = a + λb
=> r = (i + 2j + 3k) + λ(b1i + b2j + b3k) ………..1
The equations of planes are:
r.(i – j + 2k) = 5 ………2
and r.(3i + j + k) = 6 ………3
The line in equation 1 and plane in equation 2 are parallel.
Therefore, the normal to the plane of equation (2) and the given line are perpendicular.
=> (i – j + 2k). λ(b1i + b2j + b3k) = 0
=> λ(b1 - b2 + 2b3) = 0
=> b1 - b2 + 2b3 = 0 ………..4
Similarly, (3i + j + k). λ(b1i + b2j + b3k) = 0
=> λ(3b1 + b2 + b3) = 0
=> 3b1 + b2 + b3 = 0 ………..5
From equations (4) and (5), we get
=>
Therefore, the direction ratios are −3, 5, and 4.
So, b = b1i + b2j + b3k = -3i + 5j + 4k
Put value of b in equation 1, we get
r = (i + 2j + 3k) + λ(-3i + 5j + 4k)
This is the equation of the required line.