Question

Find the perpendicular distance of the point $(1,2,-4)$ from the $\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}$.

Answer 1

From given equation of line we have $A\left(\overline{a}\right)=(3,3,-5)$
$\left(\overline{l}\right)=(2,3,6)$
And $P\left(\overline{l}\right)=$ given point $=(1,2,-4)$
$\overrightarrow{AP}=P-A=(1-3,2-3,-4+5)$
$=(-2,-1,1)$
Also $\overrightarrow{AP}\times \overline{l}=$ $\left|\begin{array}{ccc}i& j& k\\ -2& -1& 1\\ 2& 3& 6\end{array}\right|$
$=i(-6,-3)-j(-12-2)+k(-6+2)$
$=-9i+14j-4k$
$=(-9,14,-4k)$
And $|\overline{l}|=\sqrt{4+9+36}$
$=\sqrt{49}=7$
$\therefore$length of perpendicular from point p
$=\frac{|\overrightarrow{AP}\times \overline{l}|}{|\overline{l}|}$
$=\frac{\left[(-9,14,-4)\right]}{7}$
$=\frac{\sqrt{81+196+16}}{7}=\frac{\sqrt{293}}{7}$
$\therefore$ Required perpendicular distance
$=\frac{\sqrt{293}}{7}$