Jaelyn Mueller

2023-02-18

How to find a unit vector normal to the surface ${x}^{3}+{y}^{3}+3xyz=3$ ay the point(1,2,-1)?

Jayden Landry

$f\left(x,y,z\right)={x}^{3}+{y}^{3}+3xyz-3=0$
The gradient of $f\left(x,y,z\right)$ at point $x,y,z$ is a vector normal to the surface at this point.
Following is how the gradient is obtained
$\nabla f\left(x,y,z\right)=\left({f}_{x},{f}_{y},{f}_{z}\right)=3\left({x}^{2}+yz,{y}^{2}+xz,xy\right)$ at point
$\left(1,2,-1\right)$ has the value
$3\left(-1,3,2\right)$ and the unit vector is
$\frac{\left\{-1,3,2\right\}}{\sqrt{1+{3}^{2}+{2}^{2}}}=\left\{-\frac{1}{\sqrt{14}},\frac{3}{\sqrt{14}},\sqrt{\frac{2}{7}}\right\}$

Do you have a similar question?