A perpendicular is drawn from a point on the line to the plane such that the foot of the perpendicular also lies on the plane . Then, the coordinates of are
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
A point on the given line is P = (1+2t, -1-t, t). The foot Q lies on plane x+y+z=3, and PQ is perpendicular to this plane, so Q = P + s(1,1,1) for some s.
Q = (1+2t+s, -1-t+s, t+s).
From x+y+z=3: (1+2t+s)+(-1-t+s)+(t+s) = 3 → 2t+3s = 3.
From x-y+z=3: (1+2t+s)-(-1-t+s)+(t+s) = 3 → 2+4t+s = 3 → s = 1-4t.
Substituting: 2t+3(1-4t)=3 → 2t+3-12t=3 → -10t=0 → t=0, s=1.
So Q = (1+0+1, -1-0+1, 0+1) = (2, 0, 1).
Q = (1+2t+s, -1-t+s, t+s).
From x+y+z=3: (1+2t+s)+(-1-t+s)+(t+s) = 3 → 2t+3s = 3.
From x-y+z=3: (1+2t+s)-(-1-t+s)+(t+s) = 3 → 2+4t+s = 3 → s = 1-4t.
Substituting: 2t+3(1-4t)=3 → 2t+3-12t=3 → -10t=0 → t=0, s=1.
So Q = (1+0+1, -1-0+1, 0+1) = (2, 0, 1).
[JEE Main 2019 · 10 Apr · Shift II]