Let Q and R be the feet of perpendiculars from the point P(a,a,a) on the lines x=y, z=1 an…

Question

Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a,a,a)$ on the lines $x=y,\ z=1$ and $x=-y,\ z=-1$ respectively. If $\angle QPR$ is a right angle, then $12a^2$ is equal to:

Answer: 12

Accepted Answer

$Q(t,t,1)$: $\vec{PQ}=(t-a,t-a,1-a)$, perpendicular to $(1,1,0)$: $2(t-a)=0\Rightarrow t=a\Rightarrow Q=(a,a,1)$.

$R(s,-s,-1)$: $\vec{PR}=(s-a,-s-a,-1-a)$, perpendicular to $(1,-1,0)$: $2s=0\Rightarrow s=0\Rightarrow R=(0,0,-1)$.

$$\vec{PQ}=(0,0,1-a),\qquad \vec{PR}=(-a,-a,-1-a).$$
$$\vec{PQ}\cdot\vec{PR}=(1-a)(-1-a)=a^2-1=0\implies a^2=1.$$
$12a^2=12.$

Solutions · 1