Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is

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Aicharya · Accepted Answer

Equation of line through point $(-1,2,1)$ is $\rightarrow$

$\begin{aligned} & \Rightarrow \frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}-(2)=\lambda \\ & \text { So, }\left[\begin{array}{l}x=2 \lambda-1 \\ y=3 \lambda+2 \\ z=4 \lambda+1\end{array}\right.\end{aligned}$
$\operatorname{By}(1) \rightarrow \frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}=\mu($ Let $)$
So, $\left[\begin{array}{l}x=3 \mu-2 \\ y=2 \mu+3 \\ z=\mu+4\end{array}\right.$
For intersection point ' P '
$\begin{aligned}
& \mathrm{x}=2 \lambda-1=3 \mu-2 \\ & \mathrm{y}=3 \lambda+2=2 \mu+3 \\ & \mathrm{z}=4 \lambda+1=\mu+4
\end{aligned}$
So, point $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,5,5)$
\& $\mathrm{Q}(4,-5,1)$
$\begin{aligned}
& \therefore P Q=\sqrt{9+100+16} \\ & =\sqrt{125}=5 \sqrt{5}
\end{aligned}$ ^

Let the line passing through the points $(-1,2,1)$ and parallel to the line…

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