# How do you find a vector perpendicular to the plane containing (0,0,1), (1,0,1) and (-1,-1,-1)?

Aug 12, 2016

$\pm < 0 , 2 , - 1 >$.

#### Explanation:

The vector $\pm P Q X P R$ is perpendicular to the plane containing

$P \left(0. 0. 1\right) , Q \left(1. 0. 1\right) \mathmr{and} R \left(- 1 , - 1 , - 1\right)$.

If O is the origin the vectors $O P . O Q , O R$ are <0, 0, 1>, <1, 0, 1>

and <-1, -1, -1>.

Now, pQ=OQ-OP= <1. 0, 0>

and PR = OR-OP = <-1. -1, -2>

$\pm P Q X P R = \pm < \left(0\right) \left(- 2\right) - \left(0\right) \left(- 1\right) , \left(0\right) \left(- 1\right) - \left(1\right) \left(- 2\right) , \left(1\right) \left(- 1\right) - \left(0\right) \left(- 1\right) >$

$= \pm < 0 , 2 , - 1 >$..