GATE EC 2026 Set 1 — Question 12

The surface is given by the equation $z^2 = 2x^2 - y^2$. We can rewrite this as a level surface of a function $F(x, y, z) = 2x^2 - y^2 - z^2 = 0$.The normal vector to the surface at any point $(x, y, z)$ is given by the gradient of $F$, $\nabla F$.
$\nabla F = \frac{\partial F}{\partial x} \hat{\mathbf{i}} + \frac{\partial F}{\partial y} \hat{\mathbf{j}} + \frac{\partial F}{\partial z} \hat{\mathbf{k}}$Let's compute the partial derivatives:
$\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}(2x^2 - y^2 - z^2) = 4x$
$\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(2x^2 - y^2 - z^2) = -2y$
$\frac{\partial F}{\partial z} = \frac{\partial}{\partial z}(2x^2 - y^2 - z^2) = -2z$So, the gradient vector is $\nabla F = 4x \hat{\mathbf{i}} - 2y \hat{\mathbf{j}} - 2z \hat{\mathbf{k}}$.We need to find the normal vector at the point $P = \hat{\mathbf{i}} + \sqrt{2} \hat{\mathbf{k}}$. This corresponds to $(x, y, z) = (1, 0, \sqrt{2})$.
First, let's verify that this point lies on the surface:
$z^2 = (\sqrt{2})^2 = 2$
$2x^2 - y^2 = 2(1)^2 - (0)^2 = 2 - 0 = 2$
Since $2 = 2$, the point $P$ is indeed on the surface.Now, substitute the coordinates of $P$ into the gradient vector:
$\nabla F \Big|_{P} = 4(1) \hat{\mathbf{i}} - 2(0) \hat{\mathbf{j}} - 2(\sqrt{2}) \hat{\mathbf{k}}$
$\nabla F \Big|_{P} = 4 \hat{\mathbf{i}} - 2\sqrt{2} \hat{\mathbf{k}}$This is a normal vector to the surface at $P$. To find a unit normal vector $\hat{\mathbf{n}}$, we normalize this vector:
Magnitude of the normal vector: $|\nabla F \Big|_{P}| = \sqrt{(4)^2 + (0)^2 + (-2\sqrt{2})^2} = \sqrt{16 + 0 + 8} = \sqrt{24} = 2\sqrt{6}$.So, a unit normal vector is $\hat{\mathbf{n}} = \frac{4 \hat{\mathbf{i}} - 2\sqrt{2} \hat{\mathbf{k}}}{2\sqrt{6}} = \frac{2 \hat{\mathbf{i}} - \sqrt{2} \hat{\mathbf{k}}}{\sqrt{6}}$.Now, let's compare this with the given options:
(A) $\hat{\mathbf{i}} - \sqrt{2} \hat{\mathbf{k}}$: This vector is not a unit vector (magnitude is $\sqrt{1^2 + (-\sqrt{2})^2} = \sqrt{1+2} = \sqrt{3}$) and its direction is not proportional to $2 \hat{\mathbf{i}} - \sqrt{2} \hat{\mathbf{k}}$.
(B) $\frac{2}{3} \hat{\mathbf{j}} - \frac{1}{3} \hat{\mathbf{k}}$: This vector has a $\hat{\mathbf{j}}$ component and no $\hat{\mathbf{i}}$ component, which is incorrect.
(C) $\sqrt{2} \hat{\mathbf{i}} - \sqrt{3} \hat{\mathbf{k}}$: This vector is not proportional to $2 \hat{\mathbf{i}} - \sqrt{2} \hat{\mathbf{k}}$.
(D) $\frac{\sqrt{2} \hat{\mathbf{i}} - \hat{\mathbf{k}}}{\sqrt{3}}$: Let's check its direction and magnitude.
The direction of our calculated normal vector $4 \hat{\mathbf{i}} - 2\sqrt{2} \hat{\mathbf{k}}$ can be simplified by factoring out $2\sqrt{2}$: $2\sqrt{2}(\sqrt{2} \hat{\mathbf{i}} - \hat{\mathbf{k}})$.
So, the direction is indeed $(\sqrt{2} \hat{\mathbf{i}} - \hat{\mathbf{k}})$.
Now, let's check the magnitude of the vector in option (D):
Magnitude $= \left|\frac{\sqrt{2} \hat{\mathbf{i}} - \hat{\mathbf{k}}}{\sqrt{3}}\right| = \frac{1}{\sqrt{3}} \sqrt{(\sqrt{2})^2 + (-1)^2} = \frac{1}{\sqrt{3}} \sqrt{2 + 1} = \frac{1}{\sqrt{3}} \sqrt{3} = 1$.Thus, option (D) represents a unit vector in the correct direction of the normal to the surface at point $P$.The final answer is $\boxed{\text{D}}$.