# How do you find all unit vectors that is orthogonal to the plane through the points P = (3, -3, 0), Q = (5, -1, 2), and R = (5, -1, 6)?

Jul 27, 2016

$= \left(\frac{1}{\sqrt{2}}\right) \left(1 , - 1 , 0\right)$

#### Explanation:

Vector $P Q = \left(5 , - 1 , 2\right) - \left(3 , - 3 , 0\right) = \left(2 , 2 , 2\right)$.

Vector $Q R = \left(5 , - 1 , 6\right) - \left(5 , - 1 , 2\right) = \left(0 , 0 , 4\right)$

Vector $\pm P Q X Q R = \left(8 , - 8 , 0\right)$ is normal to the plane PQR.

So, the unit orthogonal vectors are

$\pm \left(\left(8 , - 8 , 0\right)\right)$/ | ( 8, -8, 0 ) |

$= \left(\frac{1}{\sqrt{2}}\right) \left(1 , - 1 , 0\right)$