GATE 2026 EC Q.12: solution and answer

Official answer: DSource: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Write the surface as a level surface: $\varphi (x,y,z)=2x^2-y^2-z^2=0$. The point is $P=(1,0,\sqrt{2})$. It lies on the surface because $2(1{)}^2-0^2-(\sqrt{2}{)}^2=2-0-2=0$.

2. The gradient of $\varphi$ is normal to the level surface $\varphi =c$: $\nabla \varphi =\frac{\partial \varphi }{\partial x}\hat{ı}+\frac{\partial \varphi }{\partial y}\hat{ȷ}+\frac{\partial \varphi }{\partial z}\hat{k}=4x\hat{ı}-2y\hat{ȷ}-2z\hat{k}$.

3. At $P=(1,0,\sqrt{2})$: $\nabla \varphi =4\hat{ı}+0\hat{ȷ}-2\sqrt{2}\hat{k}$.

4. Its magnitude is $|\nabla \varphi |=\sqrt{4^2+0^2+(2\sqrt{2}{)}^2}=\sqrt{16+8}=\sqrt{24}=2\sqrt{6}$.

5. Divide by the magnitude to get the unit normal: $$\vec{n}=\frac{\nabla \varphi }{|\nabla \varphi |}=\frac{4\hat{ı}-2\sqrt{2}\hat{k}}{2\sqrt{6}}=\frac{2}{\sqrt{6}}\hat{ı}-\frac{\sqrt{2}}{\sqrt{6}}\hat{k}=\frac{\sqrt{2}}{\sqrt{3}}\hat{ı}-\frac{1}{\sqrt{3}}\hat{k}=\frac{\sqrt{2}\hat{ı}-\hat{k}}{\sqrt{3}}$$

6. Check: $|\vec{n}{|}^2=\frac{(\sqrt{2}{)}^2+1^2}{3}=\frac{3}{3}=1$, so this is a unit vector. It is option (D), and $-\vec{n}$ is the other unit normal.

7. Rule out the others. (A) has magnitude $\sqrt{1+2}=\sqrt{3}$ and (C) has magnitude $\sqrt{2+3}=\sqrt{5}$, so neither is a unit vector. (B) has magnitude $\sqrt{\frac{4}{9}+\frac{1}{9}}=\frac{\sqrt{5}}{3}\ne 1$, and it points the wrong way too: its $\hat{ı}:\hat{k}$ ratio is $2:-1$, but the normal needs $\sqrt{2}:-1$. The answer is (D).