JEE Main 2024 (Online) 31st January Morning ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
3D Geometry
Answer$12$
Solution
Foot on $x=y,z=1$: $Q=(t,t,1)$, $(P-Q)\cdot(1,1,0)=0\Rightarrow t=a$, $Q=(a,a,1)$. Foot on $x=-y,z=-1$: $R=(s,-s,-1)$, $(P-R)\cdot(1,-1,0)=-2s=0$, $R=(0,0,-1)$. $\vec{PQ}=(0,0,1-a)$, $\vec{PR}=(-a,-a,-1-a)$. Right angle: $(1-a)(-1-a)=0\Rightarrow a^2=1$. $12a^2=12$.