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 :

Question

Aicharya · Accepted Answer

Line passing through $(1,4,0)$ and parallel to $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is $L: \frac{x-1}{1}=\frac{y-4}{2}=\frac{z}{3}$
Any point on $L:(\lambda+1,2 \lambda+4,3 \lambda)$
Any point on $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ is $(2 \mu+2,3 \mu+ 6,4 \mu+3)$
$\left.\begin{array}{l}
\lambda+1=2 \mu+2 \\ 2 \lambda+4=3 \mu+6 \\ 3 \lambda=4 \mu+3
\end{array}\right\} \lambda=1 \mu=0$
Point: $(2,6,3)$
$\begin{aligned}
& \text { Distance }=\sqrt{(2-1)^2+(6-4)^2+(3-0)^2} \\ & =\sqrt{1+4+9}=\sqrt{14}
\end{aligned}$ *

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…

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