Question

Find the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{x+4}{3}$. Also find the length of the perpendicular.

Answer 1

Given line is

$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=r$ (say) ..........(i)

And point $P(0,2,3)$

Co-ordinates of any point on (i) may be taken as, $(5r-3,2r+1,3r-4)$

Let, $Q=(5r-3,2r+1,3r-4)$

Direction ratio's of $PQ$ are $(5r-3,2r-1,3r-7)$

Direction ratio's of $AB$ are $(0,2,3)$

Since, $PQ\perp AB$

$\therefore 0(5r-3)+2(2r-1)+3(3r-7)=0$

$⟹4r-2+9r-21=0$

$⟹13r=23$

$⟹r=\frac{23}{13}$

$\therefore Q=(5r-3,2r+1,3r-4)$

$⟹Q=(\frac{115}{13}-3,\frac{46}{13}+1,\frac{69}{13}-4)$

$⟹Q=(\frac{76}{13},\frac{59}{13},\frac{17}{13})$

Now, Length of perpendicular $PQ=\sqrt{{(0-\frac{76}{13})}^2+{(2-\frac{59}{13})}^2+{(3-\frac{17}{13})}^2}$

$=\sqrt{\frac{5776}{169}+\frac{1089}{169}+\frac{484}{169}}$

$=\sqrt{\frac{7349}{169}}$

$=\frac{\sqrt{7349}}{13}$ Units