# What is the unit vector that is normal to the plane containing  ( i - 2 j + 3 k)  and  ( i + 7 j + 4 k) ?

Sep 18, 2017

$\frac{1}{\sqrt{923}} \left(- 29 i - j + 9 k\right)$

#### Explanation:

The cross product of these two vectors will be in a suitable direction, so to find a unit vector we can take the cross product then divide by the length...

$\left(i - 2 j + 3 k\right) \times \left(i + 7 j + 4 k\right) = \left\mid \begin{matrix}i & j & k \\ 1 & - 2 & 3 \\ 1 & 7 & 4\end{matrix} \right\mid$

$\textcolor{w h i t e}{\left(i - 2 j + 3 k\right) \times \left(i + 7 j + 4 k\right)} = \left\mid \begin{matrix}- 2 & 3 \\ 7 & 4\end{matrix} \right\mid i + \left\mid \begin{matrix}3 & 1 \\ 4 & 1\end{matrix} \right\mid j + \left\mid \begin{matrix}1 & - 2 \\ 1 & 7\end{matrix} \right\mid k$

$\textcolor{w h i t e}{\left(i - 2 j + 3 k\right) \times \left(i + 7 j + 4 k\right)} = - 29 i - j + 9 k$

Then:

$\left\mid \left\mid - 29 i - j + 9 k \right\mid \right\mid = \sqrt{{29}^{2} + {1}^{2} + {9}^{2}} = \sqrt{841 + 1 + 81} = \sqrt{923}$

So a suitable unit vector is:

$\frac{1}{\sqrt{923}} \left(- 29 i - j + 9 k\right)$