Question

Find a unit vector perpendicular to the plane $ABC$, where the coordinates of $A,B$ and $C$ is $A(3,-1,2),B(1,-1,-3)$ and $C(4,-3,1)$

Answer 1

We know that, given two vectors say $\overrightarrow{x}$ and $\overrightarrow{y}$, their vector product denoted by $\overrightarrow{x}\times \overrightarrow{y}$ is a vector that is perpendicular to the plane containing them.

The given points, $A(3,-1,2),B(1,-1,-3)$ and $C(4,-3,1)$ lie in the plane $ABC$.

Accordingly, the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}ϵ$ the plane $ABC$.

Hence, $\overrightarrow{AB}\times \overrightarrow{AC}$ is perpendicular to the plane $ABC$.

Finally, the required, unit vector will be

$\frac{\overrightarrow{AB}\times \overrightarrow{AC}}{||\overrightarrow{AB}\times \overrightarrow{AC}||}$

We have, $\overrightarrow{AB}=(1-3,-1+1,-3-2)=(-2,0,-5)$.

$\overrightarrow{AC}=(1,-2,-1)$, so that,

$\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}i& j& k\\ -2& 0& -5\\ 1& -2& -1\end{array}\right|$

$=-10i-7j+4k=(-10,-7,4)$

$\Rightarrow ||\overrightarrow{AB}\times \overrightarrow{AC}||=\sqrt{(-10{)}^2+(-7{)}^2+(4{)}^2}=\sqrt{100+49+16}=\sqrt{165}$.

Finally, the desired unit vector is

$(-\frac{10}{\sqrt{165}},-\frac{7}{\sqrt{165}},\frac{4}{\sqrt{165}})$.