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 $(1, 0, 2)$ $(1,0,2)$ ?

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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 $(1, 0, 2)$ $(1,0,2)$ ?

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Answer 1 · 2016-07-04T11:13:59

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

Explanation:

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

$p_{1} = {0, 1, 1}$ $p_1={0,1,1}$
$p_{2} = {1, - 1, 0}$ $p_2={1,-1,0}$
$p_{3} = {1, 0, 2}$ $p_3 ={1,0,2}$

$p_{1}, p_{2}, p_{2}$ $p_1,p_2,p_2$ define two segments

$p_{2} - p_{1}$ $p_2-p_1$ and $p_{3} - p_{1}$ $p_3-p_1$ parallel to the plane which contains $p_{1}, p_{2}, p_{3}$ $p_1,p_2,p_3$ .

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

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

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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 $(1, 0, 2)$ $(1,0,2)$ ?

1 Answer

Explanation:

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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)?

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Explanation:

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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 $(1, 0, 2)$ $(1,0,2)$ ?