# 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)?

Jul 4, 2016

$\hat{n} = \left\{- \frac{3}{\sqrt{14}} , - \sqrt{\frac{2}{7}} , \frac{1}{\sqrt{14}}\right\}$

#### Explanation:

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

${p}_{1} = \left\{0 , 1 , 1\right\}$
${p}_{2} = \left\{1 , - 1 , 0\right\}$
${p}_{3} = \left\{1 , 0 , 2\right\}$

${p}_{1} , {p}_{2} , {p}_{2}$ 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

$\hat{n} = \frac{\left({p}_{2} - {p}_{1}\right) \times \left({p}_{3} - {p}_{1}\right)}{\left\mid \left({p}_{2} - {p}_{1}\right) \times \left({p}_{3} - {p}_{1}\right) \right\mid} = \left\{- \frac{3}{\sqrt{14}} , - \sqrt{\frac{2}{7}} , \frac{1}{\sqrt{14}}\right\}$