JEE MAIN - Mathematics (2025 - 3rd April Morning Shift - No. 13)
Line $L_1$ passes through the point $(1,2,3)$ and is parallel to $z$-axis. Line $L_2$ passes through the point $(\lambda, 5,6)$ and is parallel to $y$-axis. Let for $\lambda=\lambda_1, \lambda_2, \lambda_2<\lambda_1$, the shortest distance between the two lines be 3 . Then the square of the distance of the point $\left(\lambda_1, \lambda_2, 7\right)$ from the line $L_1$ is
25
32
40
37
Explanation
_3rd_April_Morning_Shift_en_13_1.png)
$$\begin{aligned} & 3=\frac{|[(\lambda-1) \hat{i}+3 \hat{j}+3 \hat{k}] .(\hat{k} \times \hat{j})|}{|\hat{k} \times \hat{j}|}=|\lambda-1|=3 \\ & \Rightarrow \lambda=4 \text { or }-2, \lambda_2<\lambda_1 \Rightarrow \lambda_2=-2 \\ & \lambda_1=4 \end{aligned}$$
_3rd_April_Morning_Shift_en_13_2.png)
$$\begin{aligned} & (3,-4,4-k) \perp \hat{k} \\ & 4-k=0 \Rightarrow k=4 \\ & \Rightarrow \text { Distance between }(4,-2,7) \&(1,2,7) \\ & \Rightarrow \sqrt{3^2+4^2}=5 \end{aligned}$$
Comments (0)
