Question

A perpendicular is drawn from a point on the line \(\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}\) to the plane \(x+y+z=3\) such that the foot of the perpendicular Q also lies on the plane \(x-y+z=3\). Then the co-ordinates of Q are :

• A. (4, 0, – 1)
• B. (2, 0, 1)
• C. (1, 0, 2)
• D. (– 1, 0, 4)

Hint:

Take a general point on the line, move along the plane normal (1,1,1) to reach Q, and impose that Q lies on both planes.

Answer 1

The Correct Option is B

A point on the line: \(P=(1+2t,\,-1-t,\,t)\). The perpendicular to \(x+y+z=3\) is along its normal \((1,1,1)\), so \(Q=(1+2t+s,\,-1-t+s,\,t+s)\).
Q on \(x+y+z=3\): \((1+2t+s)+(-1-t+s)+(t+s)=2t+3s=3\).
Q on \(x-y+z=3\): \((1+2t+s)-(-1-t+s)+(t+s)=2+4t+s=3\Rightarrow s=1-4t\).
Substituting: \(2t+3-12t=3\Rightarrow t=0\), \(s=1\).
\(Q=(1+0+1,\,-1-0+1,\,0+1)=(2,0,1)\).
Answer: (2) (2, 0, 1)