JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 26)
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
12
Explanation
$$\begin{aligned}
& \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \\
& \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) \\
& \overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \\
& a=r+a-r=0 \\
& 2 a=2 r \rightarrow a=r \\
& \overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} \\
& a-k-a-k=0 \Rightarrow k=0 \\
& A s, P Q \perp P R \\
& (a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0 \\
& a=1 \text { or }-1 \\
& 12 a^2=12
\end{aligned}$$
Comments (0)
