Question

Find the length and foot of the perpendicular from the point $(7,14,5)$ to the plane $2x+4y-z=2$.

Answer 1

Let the foot of perpendicular from the point $P$ to the given plane $2x+4y-z=2$ be $Q\left(a,b,c\right)$. So,

$\overrightarrow{PQ}=\left(a-7\right)\hat{i}+\left(b-14\right)\hat{j}+\left(c-5\right)\hat{k}$

The normal vector of the given plane is,

$\overrightarrow{N}=2\hat{i}+4\hat{j}-\hat{k}$

As $\overrightarrow{PQ}\parallel \overrightarrow{N}$, so,

$\frac{a-7}{2}=\frac{b-14}{4}=\frac{c-5}{-1}=\lambda$

$a=2\lambda +7$

$b=4\lambda +14$

$c=-\lambda +5$

Since $Q\left(a,b,c\right)$ lies on the given plane, so,

$2a+4b-c=2$

$2\left(2\lambda +7\right)+4\left(4\lambda +14\right)-\left(-\lambda +5\right)=2$

$\lambda =-3$

The coordinates of the foot of the perpendicular is $Q\left(\left(2\left(-3\right)+7\right),\left(4\left(-3\right)+14\right),\left(-\left(-3\right)+5\right)\right)=Q\left(1,2,8\right)$.The length of the perpendicular is,

$PQ=\sqrt{{\left(7-1\right)}^2+{\left(14-2\right)}^2+{\left(5-8\right)}^2}$

$=\sqrt{189}$

$=3\sqrt{21}$