VC3.42

From Exampleproblems

Here $r=xi+yj+zk,r^2=x^2+y^2+z^2\,$

Therefore, $\mathrm{curl}[r\times(a\times r)]=\mathrm{curl}[(r\cdot r)a-(r\cdot a)r]=\mathrm{curl}[r^2a-(r\cdot a)r]=\mathrm{curl}(r^2a)-\mathrm{curl}[(r\cdot a)r]\,$

= $[r^2\mathrm{curl}a+\mathrm{grad}(r^2\times a)]-[(r\cdot a)\mathrm{curl}r+\mathrm{grad}(r\cdot a)\times r]\,$ [by using vector identity $\mathrm{curl}(fg)=f\mathrm{curl}g+\mathrm{grad}f\times g]\,$ ]

Now a is a constant vector implies $\mathrm{curl}a=0\,$

$\mathrm{grad}r^2=\sum i\frac{\partial}{\partial x}r^2=\sum i(2r)\frac{\partial r}{\partial x}=\sum i(2r)(\frac{x}{r})=2\sum xi=2(xi+yj+zk)=2r\,$ --(1)

$\mathrm{curl}r=\sum i\times\frac{\partial r}{\partial x}=\sum (i\times r)=(i\times i)+(j\times j)+(k\times k)=0\,$

$\mathrm{grad}(r\cdot a)=\sum i\frac{\partial}{\partial x}(r\cdot a)=\sum i[\frac{\partial r}{\partial x}\cdot a+r\cdot\frac{\partial a}{\partial x}]=\sum i[i\cdot a)+0]=(i\cdot a)i+(j\cdot a)j+(k\cdot a)k=a\,$