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

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+2}{-1}=\frac{\mathrm{z}+3}{2}=\lambda \text { (let) } \\ & (2 \lambda+1,-\lambda-2,2 \lambda-3) \\ & \because \overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{n}}=0 \\ & \Rightarrow(2 \lambda-1) 2+(-\lambda+8)(-1)+(2 \lambda-4) 2=0 \\ & \Rightarrow 4 \lambda-2+\lambda-8+4 \lambda-8=0 \\ & \Rightarrow 9 \lambda-18=0 \Rightarrow \lambda=2 \\ & \therefore \mathrm{~A}(5,-4,1) \\ & \therefore \mathrm{AP}=\sqrt{3^2+6^2+0^2}=\sqrt{45}=3 \sqrt{5}\end{aligned}$ *

The perpendicular distance, of the line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$ from the point…

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