The perpendicular distance, of the line x-1/2=y+2/-1=z+3/2 from the point P(2,-10,1) , is:

Mathematics · 3D Geometry

JEE Main 2025 — 22 January, Evening Shift — Question 9

The perpendicular distance, of the line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$ from the point $P(2,-10,1)$ , is:

$\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}=\lambda$ (let)

$(2 \lambda+1,-\lambda-2,2 \lambda-3)$

$\because \overrightarrow{\mathrm{PA}} . \overrightarrow{\mathrm{n}}=0$

$\Rightarrow(2 \lambda-1) 2+(-\lambda+8)(-1)+(2 \lambda-4) 2=0$

$\Rightarrow 4 \lambda-2+\lambda-8+4 \lambda-8=0$

$\Rightarrow 9 \lambda-18=0 \Rightarrow \lambda=2$

$\therefore \mathrm{A}(5,-4,1)$

$\therefore\mathrm{AP}=\sqrt{3^{2}+6^{2}+0^{2}}=\sqrt{45}=3 \sqrt{5}$

Answer key and solution verified before publishing.

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