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

We know that, given two vectors, say $\stackrel{\to }{x}&\stackrel{\to }{y}$, their Vector
or Outer Product , denoted by $\stackrel{\to }{x}×\stackrel{\to }{y},$ is a vector that is
perpendicular to the plane that contains them.
The given pts. $A\left(3,-1,2\right),B\left(1,-1,-3\right)\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}C\left(4,-3,1\right)$ lie in the plane $ABC$
Accordingly, the vectors $\stackrel{\to }{AB}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\stackrel{\to }{AC}\in \phantom{\rule{1ex}{0ex}}\text{the plane}\phantom{\rule{1ex}{0ex}}ABC.$
Thus, $\stackrel{\to }{AB}×\stackrel{\to }{AC}\perp \text{plane}\phantom{\rule{1ex}{0ex}}ABC$
Therefore, the reqd. unit vector will be
$\frac{\stackrel{\to }{AB}×\stackrel{\to }{AC}}{||\stackrel{\to }{AB}×\stackrel{\to }{AC}||}$
We have, $\stackrel{\to }{AB}=\left(1-3,-1+1,-3-2\right)=\left(-2,0,-5\right)$,
$\stackrel{\to }{AC}=\left(1,-2,-1\right)$, so that,
$\stackrel{\to }{AB}×\stackrel{\to }{AC}=\mathrm{det}|\begin{array}{ccc}i& j& k\\ -2& 0& -5\\ 1& -2& -1\end{array}|$
$=-10i-7j-4k=\left(-10,-7,-4\right)$
$⇒||\stackrel{\to }{AB}×\stackrel{\to }{AC}||=\sqrt{100+49+16}=\sqrt{165}$
Finally, the desired unit vector is
$\left(-\frac{10}{\sqrt{165}},-\frac{7}{\sqrt{165}},-\frac{4}{\sqrt{165}}\right).$

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