0 0 votes 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} \\ $ $\hat{k} \\$ $\dfrac{1}{\sqrt{3}} \hat{i} +\dfrac{1}{\sqrt{3}} \hat{j} +\dfrac{1}{\sqrt{3}} \hat{k}$ Calculus gateme-2012 calculus vector-identities + – Andrijana3306 325 views answer comment Share Follow Add Sync Questions 0 reply Please log in or register to add a comment.