How do you find all unit vectors normal to the plane which contains the points $(0,1,1),(1,−1,0)$ , and $(1,0,2)$ ?

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Answer 1 · 2016-07-04T11:13:59

$hat n = {-3/sqrt[14], -sqrt[2/7], 1/sqrt[14]}$

Given three non aligned points there is an unique plane which contains them.

$p_1={0,1,1}$
$p_2={1,-1,0}$
$p_3 ={1,0,2}$

$p_1,p_2,p_2$ define two segments

$p_2-p_1$ and $p_3-p_1$ parallel to the plane which contains $p_1,p_2,p_3$ .

The normal to them is also the normal to the plane so

$hat n = ((p_2-p_1) xx (p_3-p_1))/abs( (p_2-p_1) xx (p_3-p_1)) = {-3/sqrt[14], -sqrt[2/7], 1/sqrt[14]}$

How do you find all unit vectors normal to the plane which contains the points (0,1,1),(1,−1,0)(0,1,1),(1,−1,0), and (1,0,2)(1,0,2)?