JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 11)
The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is :
$\sqrt{17}$
$\sqrt{13}$
$\sqrt{15}$
$\sqrt{14}$
Explanation
Let the parallel line is
$$\frac{x-1}{1}=\frac{y-4}{2}=\frac{z-0}{3}$$
so their point of intersection is
$$\begin{aligned} & (\lambda+1,2 \lambda+43 \lambda)=(2 t+2,3 t+6,4 t+3) \\ & \lambda=2 t+1 \end{aligned}$$
$$2 \lambda+4=3 t+6 \Rightarrow t=0$$
so POI is $(2,6,3)$
$$\text { so distance }=\sqrt{(2-1)^2+(6-4)^2+(3-0)^2}=\sqrt{14}$$
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