Question

Find the shortest distance between the line $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and the line of intersection of the planes $2x+5y-z+47=0$ and $2x+y+z+7=0$.

Answer 1

Let the DR of the line $<a,b,c>$

Since the line lies on the plane $2x+5y-z=-14$ and $2x+y+z+7=0$

$\therefore 2a+5b-c=0-\left(1\right)$

$2a+5b+c=0-\left(2\right)$

on solving

$\frac{a}{5b+1}=\frac{-b}{2+2}=\frac{c}{2-10}$

$\Rightarrow a=3,b=-2,c=-4$

putting $x=0$ in the planes

$5y-z=-47-\left(3\right)$

$y+z=-7-\left(4\right)$

on adding : $6y=-54$

$y=-9$

putting $y$ in (4)

$z=-7+9$

$=2$

Co-ordinate of a point in the line = $(0,-9,2)$

$\therefore$ Equation of the line :

$\frac{x-0}{3}=\frac{y+9}{-2}=\frac{z-2}{-4}$

Shortest distance :

$=\frac{\left|\begin{array}{ccc}0-3& 9-8& 2-3\\ 3& -1& 1\\ 3& -2& -4\end{array}\right|}{\sqrt{(4+2{)}^2+(3+12{)}^2+(-6+3{)}^2}}$

$=\frac{\left|\begin{array}{ccc}-2& 1& -1\\ 3& -1& 1\\ 3& -2& -4\end{array}\right|}{\sqrt{6^2+{15}^2+3^2}}$

$=-2(4+2)-1(-12-3)-1(-6+3)$

$=\frac{-12+15+3}{\sqrt{6^2+{15}^2+3^2}}$

$=\frac{6}{\sqrt{270}}units=\sqrt{\frac{2}{15}}units$