Question

A surface is given by \(z^{2}=2x^{2}-y^{2}\) and \(\vec{n}\) and \(-\vec{n}\) are unit normal vectors to the surface at the point \(\vec{P}=\hat{\imath}+\sqrt{2}\,\hat{k}\). Which of the following vectors can be \(\vec{n}\), where \(\hat{\imath}\), \(\hat{\jmath}\) and \(\hat{k}\) are the unit vectors along \(x\), \(y\) and \(z\) axes, respectively?

• A. \(\hat{\imath}-\sqrt{2}\,\hat{k}\)
• B. \(\dfrac{2}{3}\hat{\imath}-\dfrac{1}{3}\hat{k}\)
• C. \(\sqrt{2}\hat{\imath}-\sqrt{3}\,\hat{k}\)
• D. \(\dfrac{\sqrt{2}\hat{\imath}-\hat{k}}{\sqrt{3}}\)

Hint:

The gradient of F(x,y,z)=2x^2-y^2-z^2 at a point gives the normal direction; divide it by its own magnitude to get a unit vector.

Answer 1

The Correct Option is D

Step 1: Write the surface as a level set.
Rewrite the surface equation as \(F(x,y,z)=2x^{2}-y^{2}-z^{2}=0\). For a surface \(F(x,y,z)=0\), the gradient \(\nabla F\) always points along the normal direction to the surface.

Step 2: Compute the gradient.
\[ \nabla F=\left(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z}\right)=(4x,\,-2y,\,-2z) \]

Step 3: Evaluate the gradient at the given point.
The point is \(\vec{P}=\hat{\imath}+\sqrt{2}\,\hat{k}\), so \(x=1\), \(y=0\), \(z=\sqrt{2}\).
\[ \nabla F\big|_P=(4(1),\,-2(0),\,-2\sqrt{2})=(4,\,0,\,-2\sqrt{2}) \]

Step 4: Find the length of this normal vector.
\[ |\nabla F|=\sqrt{4^{2}+0^{2}+(2\sqrt{2})^{2}}=\sqrt{16+8}=\sqrt{24}=2\sqrt{6} \]

Step 5: Divide by the length to get a unit normal.
\[ \vec{n}=\frac{(4,\,0,\,-2\sqrt{2})}{2\sqrt{6}}=\left(\frac{2}{\sqrt{6}},\,0,\,-\frac{\sqrt{2}}{\sqrt{6}}\right) \] Simplify each component: \(\dfrac{2}{\sqrt{6}}=\dfrac{\sqrt{2}}{\sqrt{3}}\) and \(\dfrac{\sqrt{2}}{\sqrt{6}}=\dfrac{1}{\sqrt{3}}\). So
\[ \vec{n}=\frac{\sqrt{2}}{\sqrt{3}}\hat{\imath}-\frac{1}{\sqrt{3}}\hat{k}=\frac{\sqrt{2}\hat{\imath}-\hat{k}}{\sqrt{3}} \]

Step 6: Check the other options.
Option (A) is the raw gradient direction with \(z\)-component \(-\sqrt{2}\) but is not divided by its own length, since its magnitude is \(\sqrt{1+2}=\sqrt{3}\neq1\). Option (B) uses the wrong ratio \(2/3\) to \(1/3\) instead of \(\sqrt{2}/\sqrt{3}\) to \(1/\sqrt{3}\). Option (C) is not a unit vector either, since \(\sqrt{(\sqrt{2})^{2}+(\sqrt{3})^{2}}=\sqrt{5}\neq1\).

Step 7: Final conclusion.
\[ \boxed{\vec{n}=\dfrac{\sqrt{2}\hat{\imath}-\hat{k}}{\sqrt{3}}} \]