The distance of the point (7,10,11) from the line x-4/1=y-4/0=z-2/3 along the line x-9/2=y-13/3=z-17/6 is

Mathematics · 3D Geometry

JEE Main 2025 — 3 April, Evening Shift — Question 35

The distance of the point $(7,10,11)$ from the line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3} \quad$ along the line

$\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is

Equation of line passing through $P(7,10,11)$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is

\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}=\lambda

Let the point on the line is

$Q(2 \lambda+7,3 \lambda+10,6 \lambda+11)$

$Q$ lies on line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$

$3 \lambda+10=4 \Rightarrow \lambda=-2$

$\therefore \quad Q(3,4,-1)$

$P Q=\sqrt{16+36+144}=14$

Answer key and solution verified before publishing.

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