Question

If \[ \vec{F}=(x+2y+az)\hat{i}+(bx-3y-z)\hat{j}+(4x+cy+2z)\hat{k} \] is irrotational, then $(a,b,c)$ is:

• A. (1,4,2)
• B. (4,2,1)
• C. (4,2,-1)
• D. (2,4,-1)

Hint:

For irrotational fields, compute curl component-wise and equate to zero.

Answer 1

The Correct Option is C

Concept: A vector field is irrotational if: \[ \nabla \times \vec{F}=0 \] So all components of curl must be zero.

Step 1: Write components.
\[ P=x+2y+az,\quad Q=bx-3y-z,\quad R=4x+cy+2z \]

Step 2: Compute curl components.
\[ \nabla \times \vec{F}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} P & Q & R \end{vmatrix} \] For $\hat{i$ component:} \[ \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}=c-(-1)=c+1=0 \Rightarrow c=-1 \] For $\hat{j$ component:} \[ \frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}=4-a=0 \Rightarrow a=4 \] For $\hat{k$ component:} \[ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=b-2=0 \Rightarrow b=2 \] \[ \Rightarrow (a,b,c)=(4,2,-1) \]