Bruno Schneider

2023-03-11

How would you find a unit vector perpendicular to plane ABC where points are A(3,-1,2), B(1,-1,-3) and C(4,-3,1)?

Jaidyn Velez

Beginner2023-03-12Added 2 answers

We know that, given two vectors, say $\overrightarrow{x}\&\overrightarrow{y}$, their Vector

or Outer Product , denoted by $\overrightarrow{x}\times \overrightarrow{y},$ is a vector that is

perpendicular to the plane that contains them.

The given pts. $A(3,-1,2),B(1,-1,-3)\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}C(4,-3,1)$ lie in the plane $ABC$

Accordingly, the vectors $\overrightarrow{AB}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\overrightarrow{AC}\in \phantom{\rule{1ex}{0ex}}\text{the plane}\phantom{\rule{1ex}{0ex}}ABC.$

Thus, $\overrightarrow{AB}\times \overrightarrow{AC}\perp \text{plane}\phantom{\rule{1ex}{0ex}}ABC$

Therefore, the reqd. unit vector will be

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

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

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

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

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

$\Rightarrow \left||\overrightarrow{AB}\times \overrightarrow{AC}|\right|=\sqrt{100+49+16}=\sqrt{165}$

Finally, the desired unit vector is

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

or Outer Product , denoted by $\overrightarrow{x}\times \overrightarrow{y},$ is a vector that is

perpendicular to the plane that contains them.

The given pts. $A(3,-1,2),B(1,-1,-3)\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}C(4,-3,1)$ lie in the plane $ABC$

Accordingly, the vectors $\overrightarrow{AB}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\overrightarrow{AC}\in \phantom{\rule{1ex}{0ex}}\text{the plane}\phantom{\rule{1ex}{0ex}}ABC.$

Thus, $\overrightarrow{AB}\times \overrightarrow{AC}\perp \text{plane}\phantom{\rule{1ex}{0ex}}ABC$

Therefore, the reqd. unit vector will be

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

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

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

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

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

$\Rightarrow \left||\overrightarrow{AB}\times \overrightarrow{AC}|\right|=\sqrt{100+49+16}=\sqrt{165}$

Finally, the desired unit vector is

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