# Consider the surface xyz = 45. How do you find the unit normal vector to the surface at the point (3, 5, 3) with positive first coordinate?

Jul 16, 2016

$\hat{n} = \frac{1}{\sqrt{59}} \left(\begin{matrix}5 \\ 3 \\ 5\end{matrix}\right)$

#### Explanation:

the gradient will give you the normal vector

so for level surface $f \left(x , y , z\right) = x , y , z - 45$

$\nabla f = \vec{n} = \left(\begin{matrix}y z \\ x z \\ x y\end{matrix}\right)$

$= \left(\begin{matrix}5 \cdot 3 \\ 3 \cdot 3 \\ 3 \cdot 5\end{matrix}\right) = \left(\begin{matrix}15 \\ 9 \\ 15\end{matrix}\right)$

so $\vec{n} = \left(\begin{matrix}5 \\ 3 \\ 5\end{matrix}\right)$ is a normal vector

and $\hat{n} = \frac{1}{\sqrt{59}} \left(\begin{matrix}5 \\ 3 \\ 5\end{matrix}\right)$ is a normal unit vector