JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 24)

Let the point $$(-1, \alpha, \beta)$$ lie on the line of the shortest distance between the lines $$\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$$ and $$\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$$. Then $$(\alpha-\beta)^2$$ is equal to _________.
Answer
25

Explanation

JEE Main 2024 (Online) 5th April Evening Shift Mathematics - 3D Geometry Question 22 English Explanation

$$\overrightarrow{A B} \perp \vec{L}_1 \text { and } \overrightarrow{A B} \perp \vec{L}_2$$

$$\begin{aligned} & \overrightarrow{A B}=(-3 m-2+n+2,4 m+2-2 n+6,2 m+5-1) \\ & =(-3 m+n, 4 m-2 n+8,2 m+4) \\ & \overrightarrow{A B} \perp \vec{L}_1 \\ & \Rightarrow-3(-3 m+n)+4(4 m-2 n+8)+2(2 m+4)=0 \\ & (9 m+16 m+4 m)+(-3 n-8 n)+32+8=0 \\ & \Rightarrow 29 m-11 n+40=0 \quad \ldots(1) \\ & \overrightarrow{A B} \perp \vec{L}_2 \\ & \Rightarrow-1(-3 m+n)+2(4 m-2 n+8)+0(2 m+4)=0 \\ & \Rightarrow 3 m-n+8 m-4 n+16=0 \\ & \Rightarrow 11 m-5 n+16=0 \Rightarrow m=-1, n=1 \\ & \Rightarrow A \equiv(1,-2,3), \quad B \equiv(-3,-4,1) \\ & A B \text { line } \Rightarrow \frac{x-1}{2}=\frac{y+2}{1}=\frac{z-3}{1} \\ & \Rightarrow \alpha=-2, \quad \beta=3 \\ & \Rightarrow(\alpha-\beta)^2=25 \end{aligned}$$

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