Question

Find the length of perpendicular from the point (4, -5, 3) to the line $\frac{x-5}{3}=\frac{y+2}{-4}=\frac{z-6}{5}$

Answer 1

Consider the problem

Let,

$\frac{x-5}{3}=\frac{y+2}{-4}=\frac{z-6}{5}=r$

$x=3r+5,y=-4r-2,z=5r+6$

So, general point on the given line is

$(3r+5,-4r-2,5r+6)$

Let the foot of perpendicular be

$(3r+5,-4r-2,5r+6)$

Then, direction ratios of the perpendicular

$(3r+5-4,-4r-2+5,5r+6-3)$

which is

$3r+1,-4r+3,5r+3$

Since it is perpendicular to the given line

then,

$3(3r+1)+(-4)(-4r+3)+5(5r+3)=0$

therefore,

$9r+3+16r-12+25r+15=0$

then,

$r=\frac{3}{25}$

Thus foot of perpendicular is

$=(\frac{134}{25},\frac{-62}{25},\frac{165}{25})$

then,

Length of perpendicular

$=\sqrt{(\frac{134}{25}-4{)}^2+(\frac{-62}{25}+5{)}^2+(\frac{165}{25}-3{)}^2}$

$=\sqrt{(\frac{34}{25}{)}^2+(\frac{63}{25}{)}^2+(\frac{90}{25}{)}^2}$

$=\sqrt{\frac{1156+3969+8100}{625}}$

$=\frac{23}{5}$ units.