NCERT Solutions
Class 12 Maths
Three Dimensional Geometry

Ex.Misc.Q.20
Find the vector equation of the line passing through the point (1, 2, − 4) and
perpendicular to the two lines:
Let the required line be parallel to the vector b is given by b = b1i + b2j + b3k
The position vector of the point (1, 2, − 4) is a = i + 2j – 4k
The equation of the line passing through (1, 2, −4) and parallel to vector b is
r = a + λb
=> r = (i + 2j – 4k) + λ(b1i + b2j + b3k) ……….1
The equations of the lines are
……….2
And ……….3
Line 1 and line 2 are perpendicular to each other.
So, 3b1 – 16b2 + 7b3 = 0 ………..4
Also, line 1 and line 3 are perpendicular to each other.
So, 3b1 + 8b2 - 5b3 = 0 ………..5
From equations 4 and 5, we get
=>
=>
So, direction ratios of b are 2, 3, and 6.
So, b = 2i + 3j + 6k
Put value of b in equation 1, we get
=> r = (i + 2j – 4k) + λ(2i + 3j + 6k)
This is the equation of the required line.