Three Dimensional Geometry Question 21
Question 21 - 2024 (31 Jan Shift 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 $\angle Q P R$ is a right angle, then $12 a^{2}$ is equal to
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Answer (12)
Solution
$\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1)$
$\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1)$
$\overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k}$
$a=r+a-r=0$
$2 a=2 r \rightarrow a=r$
$\overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k}$
$a-k-a-k=0 \Rightarrow k=0$
As, $P Q \perp P R$
$(a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0$
$a=1$ or -1
$12 a^{2}=12$
