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)

Advertisement