Question

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

• A. \(6\hat{i} - 12\hat{j} - 7\hat{k}\)
• B. \(-12\hat{i} - 10\hat{j} - 12\hat{k}\)
• C. \(-34\)
• D. \(\begin{bmatrix} -12 & 12 & -12 \\ 5 & -10 & 0 \\ 0 & 6 & -12 \end{bmatrix}\)

Hint:

Use \(\text{curl}\,\vec{V}=\nabla\times\vec{V}\) as a \(3\times3\) determinant, find the six needed partial derivatives, then plug in \((2,-1,1)\).

Answer 1

The Correct Option is A

Step 1: Write the three curl components directly, without the determinant notation.
The curl of $\vec{V}=V_x\hat i+V_y\hat j+V_z\hat k$ can be written as three separate formulas, one per axis, without ever setting up the $3\times3$ array:
\[ (\text{curl}\,\vec V)_x = \partial_y V_z - \partial_z V_y,\quad (\text{curl}\,\vec V)_y = \partial_z V_x - \partial_x V_z,\quad (\text{curl}\,\vec V)_z = \partial_x V_y - \partial_y V_x \] Given $V_x=3x^2yz$, $V_y=-5xy$, $V_z=6yz^2$.

Step 2: Work out the x-component on its own.
$\partial_y V_z = \partial_y(6yz^2) = 6z^2$.
$\partial_z V_y = \partial_z(-5xy) = 0$, because $-5xy$ has no $z$ in it at all.
So $(\text{curl}\,\vec V)_x = 6z^2 - 0 = 6z^2$. At $(2,-1,1)$: $6(1)^2=6$.

Step 3: Work out the y-component on its own.
$\partial_z V_x = \partial_z(3x^2yz) = 3x^2y$.
$\partial_x V_z = \partial_x(6yz^2) = 0$, because $6yz^2$ has no $x$ in it.
So $(\text{curl}\,\vec V)_y = 3x^2y - 0 = 3x^2y$. At $(2,-1,1)$: $3(4)(-1)=-12$.

Step 4: Work out the z-component on its own.
$\partial_x V_y = \partial_x(-5xy) = -5y$.
$\partial_y V_x = \partial_y(3x^2yz) = 3x^2z$.
So $(\text{curl}\,\vec V)_z = -5y - 3x^2z$. At $(2,-1,1)$: $-5(-1)-3(4)(1) = 5-12=-7$.

Step 5: Collect the three numbers into one vector.
Putting the three components side by side, in the order (x, y, z), gives the same result as the determinant method, confirming it independently:
\[ \boxed{\text{curl}\,\vec V\big|_{(2,-1,1)} = 6\hat i - 12\hat j - 7\hat k} \]