# What is the unit vector that is orthogonal to the plane containing  <3, -6, 2>  and  <1, 1, 1> ?

Jan 7, 2016

$\left(- \frac{8}{\sqrt{146}} , - \frac{1}{\sqrt{146}} , \frac{9}{\sqrt{146}}\right)$

#### Explanation:

First take the cross product of both the vectors. Let's say $u = \left(3 , - 6 , 2\right)$ and $v = \left(1 , 1 , 1\right)$.

The cross product of 2 vectors can be seen as the calculus of 3 determinants, and the vector you will create with the cross product is always orthogonal to the 2 first vectors, that's why we have to begin with it.

$u \wedge v = \left(1 \cdot \left(- 6\right) - 2 \cdot 1 , 2 \cdot 1 - 3 \cdot 1 , 3 \cdot 1 - 1 \cdot \left(- 6\right)\right) = \left(- 8 , - 1 , 9\right)$

Now that we have $u \wedge v = w$, we now need to normalize it, it means we have to multiply the vector $w$ by $\frac{1}{| | w | |}$ in order to make it a unit vector.

$| | w | | = \sqrt{{\left(- 8\right)}^{2} + {\left(- 1\right)}^{2} + {9}^{2}} = \sqrt{146}$. So the unit vector orthogonal to the plan containing $\left(3 , - 6 , 2\right)$ and $\left(1 , 1 , 1\right)$ is $\frac{w}{|} | w | | = \left(- \frac{8}{\sqrt{146}} , - \frac{1}{\sqrt{146}} , \frac{9}{\sqrt{146}}\right)$.