For the spherical surface $x^2+y^2+z^2=1$, the unit outward normal vector at th point $\left( \dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}, 0\right)$ is given by $\dfrac{1}{\sqrt{2}} \hat{i} +\dfrac{1}{\sqrt{2}} \hat{j} \\$ $\dfrac{1}{\sqrt{2}} \hat{i} -\dfrac{1}{\sqrt{2}} \hat{j} \\ $ $\hat{k} \\$ $\dfrac{1}{\sqrt{3}} \hat{i} +\dfrac{1}{\sqrt{3}} \hat{j} +\dfrac{1}{\sqrt{3}} \hat{k}$

asked
Mar 20, 2018
in Calculus
Andrijana3306
1.5k points