VC3.42

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

Therefore,{\mathrm  {curl}}[r\times (a\times r)]=0+2r\times a-0-a\times r=2r\times a+r\times a=3r\times a\,

Hence the required is proved.

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