Question:medium
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 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?
Show 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.
Updated On: Jul 20, 2026
- \(\hat{\imath}-\sqrt{2}\,\hat{k}\)
- \(\dfrac{2}{3}\hat{\imath}-\dfrac{1}{3}\hat{k}\)
- \(\sqrt{2}\hat{\imath}-\sqrt{3}\,\hat{k}\)
- \(\dfrac{\sqrt{2}\hat{\imath}-\hat{k}}{\sqrt{3}}\)
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Get two tangent directions on the surface at P.
Write the surface as $z=\sqrt{2x^2-y^2}$ near $P=(1,0,\sqrt2)$, treating $z$ as a function of $x$ and $y$. The partial derivatives give the slopes needed to build a normal.
$\dfrac{\partial z}{\partial x}=\dfrac{2x}{\sqrt{2x^2-y^2}}$, which at $(1,0)$ gives $\dfrac{2}{\sqrt2}=\sqrt2$.
$\dfrac{\partial z}{\partial y}=\dfrac{-y}{\sqrt{2x^2-y^2}}$, which at $(1,0)$ gives $0$.
Step 2: Build the normal from these slopes.
For a surface $z=f(x,y)$, an unnormalized normal vector is $\left(-\dfrac{\partial z}{\partial x},\,-\dfrac{\partial z}{\partial y},\,1\right)$. Substituting the slopes found above,
\[ \vec{n}_{raw}=(-\sqrt2,\,0,\,1) \]
Step 3: Normalize this vector.
Its length is $\sqrt{(\sqrt2)^2+0^2+1^2}=\sqrt{3}$, so a unit normal is $\dfrac{(-\sqrt2,0,1)}{\sqrt3}$. This points opposite to the direction found by the gradient method, but the question allows either $\vec n$ or $-\vec n$, so flipping the sign gives
\[ \vec n=\frac{\sqrt2\,\hat\imath-\hat k}{\sqrt3} \]
Step 4: Match to the options.
This is exactly the vector in option (D), already scaled to length one, unlike the other three choices which are either unnormalized or use the wrong ratio between components.
\[ \boxed{\dfrac{\sqrt2\hat\imath-\hat k}{\sqrt3}} \]
Write the surface as $z=\sqrt{2x^2-y^2}$ near $P=(1,0,\sqrt2)$, treating $z$ as a function of $x$ and $y$. The partial derivatives give the slopes needed to build a normal.
$\dfrac{\partial z}{\partial x}=\dfrac{2x}{\sqrt{2x^2-y^2}}$, which at $(1,0)$ gives $\dfrac{2}{\sqrt2}=\sqrt2$.
$\dfrac{\partial z}{\partial y}=\dfrac{-y}{\sqrt{2x^2-y^2}}$, which at $(1,0)$ gives $0$.
Step 2: Build the normal from these slopes.
For a surface $z=f(x,y)$, an unnormalized normal vector is $\left(-\dfrac{\partial z}{\partial x},\,-\dfrac{\partial z}{\partial y},\,1\right)$. Substituting the slopes found above,
\[ \vec{n}_{raw}=(-\sqrt2,\,0,\,1) \]
Step 3: Normalize this vector.
Its length is $\sqrt{(\sqrt2)^2+0^2+1^2}=\sqrt{3}$, so a unit normal is $\dfrac{(-\sqrt2,0,1)}{\sqrt3}$. This points opposite to the direction found by the gradient method, but the question allows either $\vec n$ or $-\vec n$, so flipping the sign gives
\[ \vec n=\frac{\sqrt2\,\hat\imath-\hat k}{\sqrt3} \]
Step 4: Match to the options.
This is exactly the vector in option (D), already scaled to length one, unlike the other three choices which are either unnormalized or use the wrong ratio between components.
\[ \boxed{\dfrac{\sqrt2\hat\imath-\hat k}{\sqrt3}} \]
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