Vector field V is defined as V = 3x2yz\,i - 5xy\,j + 6yz2\,k The curl of V at point (2, -1, 1) is

$-12\hat{i} - 10\hat{j} - 12\hat{k}$

$\begin{bmatrix} -12 & 12 & -12 \\ 5 & -10 & 0 \\ 0 & 6 & -12 \end{bmatrix}$

GATE 2026 CE (CE2) – Question 37

Engineering Mathematics · Calculus: Integrals, partial and total derivatives, gradient, divergence, curl, line, surface and volume integrals · 2 marks · Multiple choice

Vector field $\vec{V}$ is defined as

$$\vec{V} = 3x^2yz\,\hat{i} - 5xy\,\hat{j} + 6yz^2\,\hat{k}$$

The curl of $\vec{V}$ at point $(2, -1, 1)$ is

$6\hat{i} - 12\hat{j} - 7\hat{k}$
$-12\hat{i} - 10\hat{j} - 12\hat{k}$
$-34$
$\begin{bmatrix} -12 & 12 & -12 \\ 5 & -10 & 0 \\ 0 & 6 & -12 \end{bmatrix}$

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Correct answer: (A) $6\hat{i} - 12\hat{j} - 7\hat{k}$

Explanation

The curl is $\left(\frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z}\right)\hat{i} + \left(\frac{\partial V_x}{\partial z} - \frac{\partial V_z}{\partial x}\right)\hat{j} + \left(\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y}\right)\hat{k} = 6z^2\,\hat{i} + 3x^2y\,\hat{j} + (-5y - 3x^2z)\,\hat{k}$. At $(2, -1, 1)$ this is $6\hat{i} + 3(4)(-1)\hat{j} + (5 - 12)\hat{k} = 6\hat{i} - 12\hat{j} - 7\hat{k}$.