Question Details

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

Options

$\begin{matrix} \frac{}{} \hat{i} - \sqrt{2} \hat{k} \end{matrix}$

$\begin{matrix} \frac{2}{3} \hat{i} - \frac{1}{3} \hat{k} \end{matrix}$

$\begin{matrix} \sqrt{2} \hat{i} - \sqrt{3} \hat{k} \end{matrix}$

$\frac{\sqrt{2} \hat{i} - \hat{k}}{\sqrt{3}}$

Show Answer

Correct Answer :

Option D

$\frac{\sqrt{2} \hat{i} - \hat{k}}{\sqrt{3}}$

Solution :

The correct option is:

$\frac{\sqrt{2} \hat{i} - \hat{k}}{\sqrt{3}}$

Step-by-step Explanation:

1. Define the surface function:
We are given the equation of the surface:
$z^{2} = 2 x^{2} - y^{2}$
Let us rewrite this in the form of a level surface $F (x, y, z) = 0$ :
$F (x, y, z) = 2 x^{2} - y^{2} - z^{2} = 0$

2. Identify the coordinates of point P:
The given position vector of point $\vec{P}$ is:
$\vec{P} = \hat{i} + \sqrt{2} \hat{k}$
This corresponds to the coordinates:
$(x, y, z) = (1, 0, \sqrt{2})$

3. Calculate the gradient vector:
The normal vector to any level surface at a given point is in the direction of the gradient of the surface function. The gradient vector $\nabla F$ is defined as:
$\nabla F = \frac{\partial F}{\partial x} \hat{i} + \frac{\partial F}{\partial y} \hat{j} + \frac{\partial F}{\partial z} \hat{k}$
We compute the partial derivatives of $F$ :
$\frac{\partial F}{\partial x} = 4 x$
$\frac{\partial F}{\partial y} = - 2 y$
$\frac{\partial F}{\partial z} = - 2 z$
Substituting these back, we get:
$\nabla F = 4 x \hat{i} - 2 y \hat{j} - 2 z \hat{k}$

4. Evaluate the gradient at point P:
Substitute $x = 1$ , $y = 0$ , and $z = \sqrt{2}$ into the gradient expression:
$\nabla F_{P} = 4 (1) \hat{i} - 2 (0) \hat{j} - 2 (\sqrt{2}) \hat{k} = 4 \hat{i} - 2 \sqrt{2} k$

5. Normalize the gradient vector to find the unit normals:
To find the unit normal vector, we divide the normal vector by its magnitude:
$| \nabla F_{P} | = \sqrt{4^{2} + {(- 2 \sqrt{2})}^{2}} = \sqrt{16 + 8} = \sqrt{24} = 2 \sqrt{6}$
Therefore, the unit normal vectors $\pm \vec{n}$ are:
$\vec{n} = \pm \frac{4 \hat{i} - 2 \sqrt{2} \hat{k}}{2 \sqrt{6}} = \pm \frac{2 \hat{i} - \sqrt{2} \hat{k}}{\sqrt{6}}$

6. Simplify the expression:
Divide the numerator and denominator by $\sqrt{2}$ :
$\vec{n} = \pm \frac{\sqrt{2} \hat{i} - \hat{k}}{\sqrt{3}}$
Among the options provided, one of these unit normal vectors is:
$\frac{\sqrt{2} \hat{i} - \hat{k}}{\sqrt{3}}$