UPSC NDA Math Model Paper 7

Let A (a, b, c) is foot of perpendicular of point P (0, -2, -1)
Given plane is x - 3y + 3z = 3.
So, PA will be normal to the given plane, so direction ratios of PA will be proportional to (1, -3, 3).
∵ PA passes from (0, -2, 1) and have direction ratios (1, -3, 3).
∴ Equation of line PA:

x−0

1

=

y+2

−3

=

z−1

3

=r
So, point A (a, b, c) in the form of r is (r, -3r - 2, 3r - 1)
Since A (a, b, c), lies on the given plane, so,
Plane x - 3y + 3z = 3 ⇒ 1. (r) – 3. (- 3r - 2) + 3(3r - 1) – 3 = 0
∴ r = 0
So, foot of perpendicular A (r, -3r - 2, 3r - 1) will be (0, -2, -1).