Let Q be the foot of perpendicular drawn from the point P(1,2,3) to the plane x+2 y+z=14.

Question

Let $Q$ be the foot of perpendicular drawn from the point $P(1,2,3)$ to the plane $x+2 y+z=14$. If $R$ is a point on the plane such that $$\angle P R Q=60^{\circ}$$, then the area of  $$\triangle P Q R$$ is equal to :

Accepted Answer

$$
P Q=\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}
$$

$$
Q R=\frac{P Q}{\tan 60^{\circ}}=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}
$$

$$
\text { Area }(\triangle P Q R)=\frac{1}{2} \cdot P Q \cdot Q R=\sqrt{3}
$$

Solutions · 1