Question

If $Q(0,-1,-3)$ is the image of the point $P$ in the plane $3x-y+4z=2$ and $R$ is the point $(3,-1,-2),$ then the area (in sq. units) of $\Delta PQR$ is :

• A. $\frac{\sqrt{91}}{4}$
• B. $2\sqrt{13}$
• C. $\frac{\sqrt{65}}{2}$
• D. $\frac{\sqrt{91}}{2}$

Answer 1

The correct option is D $\frac{\sqrt{91}}{2}$

Let $M$ be a point on the plane such that $PQ$ passes through $M$.
The point $R$ lies on the plane : $3x-y+4z-2=0$
$PM=\left|\frac{1-12-2}{\sqrt{9+1+16}}\right|=\sqrt{\frac{13}{2}}$
$PR=\sqrt{9+1}=\sqrt{10}$
Now , using pythagorus theorem in $\Delta PMR$, we have
$RM=\sqrt{10-\frac{13}{2}}=\sqrt{\frac{7}{2}}$
$\Rightarrow \Delta PQR=2\times \frac{1}{2}\times \sqrt{\frac{13}{2}}\times \sqrt{\frac{7}{2}}=\frac{\sqrt{91}}{2}$