Question

Find the $\perp$ Distance of the point $(2,3,4)$ from the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. Also find the coordinates of the foot of the perpendicular.

Answer 1

Consider the given line $L$

and the point $P(2,3,4)$

$\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$

$\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}$

and any point $Q$ on the line

$\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda$

$x=-2\lambda +4,y=6\lambda ,z=-3\lambda +1$

Now, direction ratios of $PQ$

$-2\lambda +4-2,6\lambda -3,-3\lambda +1-4$

Or, $-2\lambda +2,6\lambda -3,-3\lambda -3$

Since $PQ\perp L$

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$-2(-2\lambda +2)+6(6\lambda -3)-3(-3\lambda -3)=0$

$49\lambda -13=0$

$\Rightarrow \lambda =\frac{13}{49}$

Hence the coordinates of $Q$ are

$x=\left(-2\right)\left(\frac{13}{49}\right)+4=\frac{170}{49}$

$y=\left(6\right)\left(\frac{13}{49}\right)=\frac{78}{49}$

$z=\left(-3\right)\left(\frac{13}{49}\right)+4=\frac{10}{49}$

Hence the distance of $PQ$

$=\sqrt{{\left(2-\frac{170}{49}\right)}^2+{\left(3-\frac{78}{49}\right)}^2+{\left(4-\frac{10}{49}\right)}^2}$

$=\sqrt{{\left(\frac{72}{49}\right)}^2+{\left(\frac{115}{49}\right)}^2+{\left(\frac{186}{49}\right)}^2}$

$=4units$