Let $\mathrm{Q}$ and $\mathrm{R}$ be the feet of perpendiculars from the point $\mathrm{P}(a, a, a)$ on the lines $x=y, z=1$ and $x=-y, z=-1$ respectively. If $\angle \mathrm{QPR}$ is a right angle, then $12 a^2$ is equal to _________.
Answer (integer)
12
Solution
<p>$$\begin{aligned}
& \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \\
& \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) \\
& \overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \\
& a=r+a-r=0 \\
& 2 a=2 r \rightarrow a=r \\
& \overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} \\
& a-k-a-k=0 \Rightarrow k=0 \\
& A s, P Q \perp P R \\
& (a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0 \\
& a=1 \text { or }-1 \\
& 12 a^2=12
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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