# Question #807f7

Mar 17, 2017

$m a t h b f {u}_{1} = \left(\begin{matrix}\frac{4}{5} \\ \frac{3}{5}\end{matrix}\right)$

$m a t h b f {u}_{2} = \left(\begin{matrix}- \frac{4}{5} \\ - \frac{3}{5}\end{matrix}\right)$

#### Explanation:

I think what you're saying is that $\left(4 , 3\right)$ is the normal vector to the plane you mention.

If that is so, the first unit vector is:

$m a t h b f {u}_{1} = \frac{1}{\sqrt{{4}^{2} + {3}^{2}}} \left(\begin{matrix}4 \\ 3\end{matrix}\right)$

$= \left(\begin{matrix}\frac{4}{5} \\ \frac{3}{5}\end{matrix}\right)$

The other unit vector will point in the opposite direction, ie:

$m a t h b f {u}_{2} = \left(\begin{matrix}- \frac{4}{5} \\ - \frac{3}{5}\end{matrix}\right)$