The perpendicular distance from the point $(-1,1,0)$ to the line joining the points $(0,2,4)$ and $(3,0,1)$ is

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Aicharya · Accepted Answer

Since, equation of line $A B$ is $\frac{x-0}{3}=\frac{y-2}{-2}=\frac{z-4}{-3}$ $\begin{aligned} & \Rightarrow \frac{x}{3}=\frac{y-2}{-2}=\frac{z-4}{-3}=k(\text { Let }) \\ & \Rightarrow(x, y, z)=(3 k,-2 k+2,-3 k+4) \end{aligned}$ Now, direction ratio of $P C$ is $(3 k+1,-2 k+1,-3 k+4)$ So, $3(3 k+1)+(-2)(-2 k+1)-3(-3 k+4)=0$ $\Rightarrow 22 k=11 \Rightarrow k=\frac{1}{2}$ So, $C\left(\frac{3}{2}, 1, \frac{5}{2}\right) \because P C=\sqrt{\frac{25}{4}+0+\frac{25}{4}}=\frac{5}{\sqrt{2}}$.

The perpendicular distance from the point $(-1,1,0)$ to the line joining the points $(0,2,4)$ and $(3,0,1)$ is

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