Question

Find the length and the foot of perpendicular from the point $(1,\frac{3}{2},2)$ to the plane 2x-2y+4z+5=0.

Answer 1

Equation of the given plane is 2x-2y+4z+5=0. ...(i)
$\overrightarrow{n}=2\hat{i}-2\hat{j}+4\hat{k}$
So, the equation of line through $(1,\frac{3}{2},2)$ and parallel to $\overrightarrow{n}$ is given by
$\frac{x-1}{2}=\frac{y-3/2}{-2}=\frac{z-2}{4}=\lambda \Rightarrow x=2\lambda +1,y=-2\lambda +\frac{3}{2}andz=4\lambda +2$
If this point lies on the given plane, then
$2(2\lambda +1)-2(-2+\frac{3}{2})+4(4\lambda +2)+5=0$ [using Eq.(i)]
$\Rightarrow 4\lambda +2+4\lambda -3+16\lambda +8+5=0\Rightarrow 24\lambda =-12\Rightarrow \lambda =\frac{-1}{2}$
$\therefore$ Required foot of perpendicular $=[2\times \left(\frac{-1}{2}\right)+1,-2\times \left(\frac{-1}{2}\right)+\frac{3}{2},4\times \left(\frac{-1}{2}\right)+2]i.e.,(0,\frac{5}{2},0)$
$\therefore$ Required length of perpendicular = $\sqrt{(1-0{)}^2+{(\frac{3}{2}-\frac{5}{2})}^2+(2-0{)}^2}$
=$\sqrt{1+1+4}=\sqrt{6}$ units.