If the vector field $\vec{F} = (ay + 2z)\hat{i} + (3x - bz)\hat{j} + (cx - y)\hat{k}$ is irrotational, the value of $(a + b + c)$ is ________ (answer in integer).
Answer Key:
6
Step-by-Step Solution & Explanation
For $\vec{F}$ to be irrotational, $\nabla \times \vec{F} = \vec{0}$:
$\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ ay + 2z & 3x - bz & cx - y \end{vmatrix} = \vec{0}$
Computing components:
$\hat{i}\left(\frac{\partial}{\partial y}(cx - y) - \frac{\partial}{\partial z}(3x - bz)\right) = \hat{i}(-1 - (-b)) = (b - 1)\hat{i}$
$\hat{j}\left(\frac{\partial}{\partial z}(ay + 2z) - \frac{\partial}{\partial x}(cx - y)\right) = \hat{j}(2 - c)$
$\hat{k}\left(\frac{\partial}{\partial x}(3x - bz) - \frac{\partial}{\partial y}(ay + 2z)\right) = \hat{k}(3 - a)$
Equating each component to zero:
$b - 1 = 0 \implies b = 1$
$2 - c = 0 \implies c = 2$
$3 - a = 0 \implies a = 3$
Therefore:
$a + b + c = 3 + 1 + 2 = 6$