Question

If $Q$ be the foot of perpendicular from $P(2,4,3)$ on the line joining the points $A(1,2,4)$ and $B(3,4,5)$, find the coordinates of $Q$.

Answer 1

Equation of line AB is given by

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$\frac{x-1}{3-1}=\frac{y-2}{4-2}=\frac{z-4}{5-4}$

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

So dr's of line AB is $(2,2,1)$

since Q is on the line AB so let

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

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

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

$y=2\lambda +2$

of Q$(2\lambda +1,2\lambda +2,\lambda +4)$ & P$(2,4,3)$ is

$(2\lambda +1-2)$ $2\lambda +2-4$ $(\lambda +4-3)$

$(2\lambda -1)$, $(2\lambda -2)$, $(\lambda +1)$

We know for perpendicular lines $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$2(2\lambda -1)+2(2\lambda +2)+\lambda +4=0$ so $x=2\lambda +1=2\times \frac{-2}{3}+1=\frac{-4}{3}+1$

$x=\frac{-1}{3}$

$y=\frac{-4}{3}+2$

$y=\frac{2}{3}$

$z=\frac{-2}{3}+4$

$z=\frac{10}{3}$

$4\lambda -2+4\lambda +4+\lambda +4=0$

$9\lambda +6=0$

$\lambda =\frac{-6}{9}=\frac{-2}{3}$

$Q(\frac{-1}{3},\frac{2}{3},\frac{10}{3})$.