Let's solve the problem step by step: ### Step 1: **Given Vector Field and Point** The vector field is: \[ \vec{F}(x, y, z) = 2x\hat{i} + 2z\hat{j} + y\hat{k} \] The point is \((1, -2, 3)\). ### Step 2: **Curl Formula** The curl of a vector field \(\vec{F} = M\hat{i} + N\hat{j} + P\hat{k}\) is: \[ \text{curl } \vec{F} = \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ M & N & P \end{vmatrix} \] Where \(M = 2x\), \(N = 2z\), \(P = y\). ### Step 3: **Compute Partial Derivatives** Let's fill in the determinant: \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2x & 2z & y \end{vmatrix} \] - For \(\hat{i}\) component: \[ \frac{\partial y}{\partial y} - \frac{\partial 2z}{\partial z} = 1 - 2 = -1 \] - For \(\hat{j}\) component: \[ -\left(\frac{\partial y}{\partial x} - \frac{\partial 2x}{\partial z}\right) = -(0 - 0) = 0 \] - For \(\hat{k}\) component: \[ \frac{\partial 2z}{\partial x} - \frac{\partial 2x}{\partial y} = 0 - 0 = 0 \] So, \[ \text{curl } \vec{F} = -\hat{i} + 0\hat{j} + 0\hat{k} = -\hat{i} \] ### Step 4: **Substitute the Point (1, -2, 3)** Notice, the curl is a constant vector (independent of \(x, y, z\)), so at any point: \[ \text{curl } \vec{F}(1, -2, 3) = -\hat{i} \] --- ## **Final Answer:** \[ \boxed{\text{curl } \vec{F}(1, -2, 3) = -\hat{i}} \] --- **Explanation:** The curl was computed using the determinant formula, filling in the partial derivatives, and substituting the vector field's components. The curl turns out to be a constant vector, independent of the point. Therefore, at the point \((1, -2, 3)\), the curl is \(-\hat{i}\).