The distance of the line x-2/2=y-6/3=z-3/4 from the point (1,4,0) along the line x/1=y-2/2=z+3/3 is :

Mathematics · 3D Geometry

JEE Main 2025 — 23 January, Evening Shift — Question 6

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 :

Let the parallel line is

$\frac{x-1}{1}=\frac{y-4}{2}=\frac{z-0}{3}$

so their point of intersection is

$(\lambda+1,2 \lambda+4,3 \lambda)=(2 t+2,3 t+6,4 t+3)$

$\lambda=2 \mathrm{t}+1$

$2 \lambda+4=3 \mathrm{t}+6 \Rightarrow \mathrm{t}=0$

so POI is $(2,6,3)$

so distance $=\sqrt{(2-1)^{2}+(6-4)^{2}+(3-0)^{2}}=\sqrt{14}$

Answer key and solution verified before publishing.

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