Discussion
For the spherical surface \(x^{2}+y^{2}+z^{2}=1\), the unit outward normal vector at the point \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)\) is given by
Answer: A
The unit outward normal is \(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}\).
Take \(\phi = x^{2}+y^{2}+z^{2}\). Its gradient is normal to the surface:
\[ \begin{aligned} \nabla\phi &= 2x\hat{i}+2y\hat{j}+2z\hat{k} \\ &= \sqrt{2}\hat{i}+\sqrt{2}\hat{j} \text{ at the point} \\ |\nabla\phi| &= 2 \end{aligned} \]
Dividing by 2 gives the unit normal. On a sphere centred at the origin, the outward normal is simply the position vector of the point, which is the same answer.
The minus sign in the second option makes the vector tangent to the sphere, \(\hat{k}\) is the normal at the north pole, and the last option has a z-component although z = 0 at this point.
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