Question

If $P,Q$ and $R$ are foot of perpendiculars drawn from point $A(1,1,1)$ to planes $P_1:x+2y+2z=2$, $P_2:2x-2y+z=-8$ and to line of intersection of $P_1$ and $P_2$ respectively, then area of $\Delta PQR$ is -

• A. $3$
• B. $\frac{3}{2}$
• C. $2$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $\frac{3}{2}$

$P_1\perp P_2$

So $\Delta PQR$ is a right angled triangle.

Perpendicular distance from $A(1,1,1)$ to $P_1=1$

Perpendicular distance from $A(1,1,1)$ to $P_2=3$

So area of $\Delta =\frac{1}{2}\times 1\times 3=\frac{3}{2}$