Discussion
Consider the following vector \(\vec{V} = x\,y\,z\,\hat{k}\).
Which of the following statements is correct for the resultant vector \(\vec{\nabla} \times \vec{\nabla} \times \vec{V}\)?
Answer: A
The resultant vector lies in the \(xy\)-plane.
Use the identity below with \(\vec{V} = xyz\,\hat{k}\):
\[ \begin{aligned} & \nabla \times (\nabla \times \vec{V}) \\ &= \nabla(\nabla \cdot \vec{V}) - \nabla^{2}\vec{V} \end{aligned} \]
\[ \begin{aligned} \nabla \cdot \vec{V} &= \frac{\partial (xyz)}{\partial z} = xy \\ \nabla(xy) &= y\,\hat{\imath} + x\,\hat{\jmath} \\ \nabla^{2}(xyz) &= 0 \end{aligned} \]
So \(\nabla \times \nabla \times \vec{V} = y\,\hat{\imath} + x\,\hat{\jmath}\), which has no \(\hat{k}\) component and lies in the \(xy\)-plane. Working it directly gives the same: \(\nabla \times \vec{V} = xz\,\hat{\imath} - yz\,\hat{\jmath}\), whose curl is \(y\,\hat{\imath} + x\,\hat{\jmath}\).
It has both \(\hat{\imath}\) and \(\hat{\jmath}\) parts, so it is not along \(x\) alone, and it has no \(z\) component, so it cannot lie in the \(yz\)- or \(xz\)-plane in general.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development