Question

If \[ \vec{v}=(x+y+1)\hat{i}+\hat{j}+(-x-y)\hat{k}, \] then the value of \[ \vec{v}\cdot(\nabla\times\vec{v}) \] is

• A. \(0\)
• B. \(0\)
• C. \(-\hat{i}+\hat{j}+\hat{k}\)
• D. \(x-x-y\)

Hint:

To solve questions involving curl and dot product:
• First identify the components \(P\), \(Q\), and \(R\).
• Compute the curl using the determinant formula.
• Finally, evaluate the dot product using \[ \vec{A}\cdot\vec{B}=A_xB_x+A_yB_y+A_zB_z. \]
• Always simplify completely before selecting the option.

Answer 1

The Correct Option is A

Concept: For a vector field \[ \vec{F}=P\hat{i}+Q\hat{j}+R\hat{k}, \] its curl is defined as \[ \nabla\times\vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \dfrac{\partial}{\partial x} &\\ \dfrac{\partial}{\partial y} &\\ \dfrac{\partial}{\partial z} P & Q & R \end{vmatrix}. \] The curl measures the rotational tendency of the vector field. After finding the curl, we compute the dot product \[ \vec{F}\cdot(\nabla\times\vec{F}), \] using \[ \vec{A}\cdot\vec{B} =A_xB_x+A_yB_y+A_zB_z. \] If the curl is the zero vector, then its dot product with any vector field is also zero.

Step 1: Identify the components of the given vector field.
Given, \[ \vec{v} = (x+y+1)\hat{i} +\hat{j} +(-x-y)\hat{k}. \] Comparing with \[ \vec{v}=P\hat{i}+Q\hat{j}+R\hat{k}, \] we get \[ P=x+y+1, \] \[ Q=1, \] \[ R=-x-y. \]

Step 2: Compute the curl of the vector field.
Using \[ \nabla\times\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \dfrac{\partial}{\partial x} &\\ \dfrac{\partial}{\partial y} &\\ \dfrac{\partial}{\partial z}\\ x+y+1 & 1 & -x-y \end{vmatrix}, \] we evaluate each component separately. For the \(\hat{i}\)-component, \[ \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = \frac{\partial(-x-y)}{\partial y} - \frac{\partial(1)}{\partial z} = -1-0 = -1. \] For the \(\hat{j}\)-component, \[ \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 0-(-1) = 1. \] For the \(\hat{k}\)-component, \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0-1 = -1. \] Thus, \[ \nabla\times\vec{v} = -\hat{i} +\hat{j} -\hat{k}. \]

Step 3: Find the dot product.
Now, \[ \vec{v}\cdot(\nabla\times\vec{v}) = (x+y+1)(-1) + (1)(1) + (-x-y)(-1). \] Expanding, \[ = -(x+y+1)+1+x+y. \] Simplifying, \[ =-x-y-1+1+x+y. \] \[ =0. \] Hence, \[ \boxed{\vec{v}\cdot(\nabla\times\vec{v})=0.} \] Therefore, the correct option is \[ \boxed{\text{(A) }0.} \]