Question

If \(\vec V_1\) and \(\vec V_2\) are the vectors joining the fixed points \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\) respectively to a variable point \((x,y,z)\), then \(\operatorname{curl}(\vec V_1\times \vec V_2)=\)

• A. \(0\)
• B. \(\vec V_1+\vec V_2\)
• C. \(\vec V_1-\vec V_2\)
• D. \(2(\vec V_1-\vec V_2)\)

Hint:

Use the vector identity for \(\nabla\times(\vec A\times\vec B)\) carefully in curl-cross product questions.

Answer 1

The Correct Option is D

Concept:
Let the variable point be \[ P(x,y,z) \] and the fixed points be \[ A(x_1,y_1,z_1),\qquad B(x_2,y_2,z_2) \] The vectors from fixed points to \(P\) are position-type vectors.

Step 1: Write \(\vec V_1\) and \(\vec V_2\). \[ \vec V_1=(x-x_1)\hat i+(y-y_1)\hat j+(z-z_1)\hat k \] \[ \vec V_2=(x-x_2)\hat i+(y-y_2)\hat j+(z-z_2)\hat k \] For such vectors, \[ \nabla\cdot \vec V_1=3,\qquad \nabla\cdot \vec V_2=3 \] and \[ \nabla\times \vec V_1=\vec 0,\qquad \nabla\times \vec V_2=\vec 0 \]

Step 2: Use the identity. \[ \nabla\times(\vec A\times \vec B) = \vec A(\nabla\cdot \vec B)-\vec B(\nabla\cdot \vec A) +(\vec B\cdot\nabla)\vec A-(\vec A\cdot\nabla)\vec B \] Put \[ \vec A=\vec V_1,\qquad \vec B=\vec V_2 \]

Step 3: Substitute known results. Since \[ \nabla\cdot\vec V_1=3,\qquad \nabla\cdot\vec V_2=3 \] and for position-type vectors, \[ (\vec V_2\cdot\nabla)\vec V_1=\vec V_2 \] \[ (\vec V_1\cdot\nabla)\vec V_2=\vec V_1 \] we get \[ \nabla\times(\vec V_1\times\vec V_2) = 3\vec V_1-3\vec V_2+\vec V_2-\vec V_1 \] \[ =2\vec V_1-2\vec V_2 \] \[ =2(\vec V_1-\vec V_2) \]

Step 4: Final answer. \[ \boxed{2(\vec V_1-\vec V_2)} \]