VC3.26

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Here r=xi+yj+zk,r^{2}=|r|^{2}=x^{2}+y^{2}+z^{2}\, --(1)

Since a is a constant vector,{\frac  {\partial a}{\partial x}}=0\, --(2)

From (1),{\frac  {\partial r}{\partial x}}=i,{\frac  {\partial r}{\partial x}}={\frac  {x}{r}}\, --(3)

Now,{\mathrm  {curl}}({\frac  {a\times r}{r^{3}}})=\sum {i\times {\frac  {\partial }{\partial x}}[{\frac  {a\times r}{r^{3}}}]}\, by definition of curl --(4)

But {\frac  {\partial }{\partial x}}[{\frac  {a\times r}{r^{3}}}]={\frac  {\partial }{\partial x}}(r^{{-3}}(a\times r)\,

=-3r^{{-4}}{\frac  {\partial r}{\partial x}}(a\times r)+r^{{-3}}{\frac  {\partial }{\partial x}}(a\times r)\,

=-{\frac  {3r^{{-4}}x}{r}}(a\times r)+r^{{-3}}[{\frac  {\partial a}{\partial x}}\times r+a\times {\frac  {\partial r}{\partial x}}]\, by (3)

=-3r^{{-5}}x(a\times r)+r^{{-3}}(a\times i)\, by (2),(3)

Therefore,i\times {\frac  {\partial }{\partial x}}[{\frac  {a\times r}{r^{3}}}]\,

=i\times [-3r^{{-5}}x(a\times r)+r^{{-3}}(a\times i)]\,

=-3r^{{-5}}xi\times (a\times r)+r^{{-3}}i\times (a\times i)\,

=-3r^{{-5}}x[(i\cdot r)a-(i\cdot a)r]+r^{{-3}}[(i\times i)a-(i\cdot a)i]\, --(5)

Let a=a_{1}i+a_{2}j+a_{3}k\, so that i\cdot a=a_{1}\, etc. --(6)

Also i\cdot r=i\cdot (xi+yj+zk)=x\, --(7)

Using (6) and (7),(5)reduces to

i\times {\frac  {\partial }{\partial x}}[{\frac  {a\times r}{r^{3}}}]\,

=3r^{{-5}}x[xa-a_{1}r]+r^{{-3}}(a-a_{1}i)\,

=-{\frac  {3x^{2}}{r^{5}}}a+{\frac  {3a_{1}x}{r^{5}}}r+{\frac  {1}{r^{3}}}a-{\frac  {1}{r^{3}}}a_{1}i\,

=\sum i\times {\frac  {\partial }{\partial x}}[{\frac  {a\times r}{r^{3}}}]\,

=\sum [-{\frac  {3x^{2}}{r^{5}}}a+{\frac  {3a_{1}x}{r^{5}}}r+{\frac  {1}{r^{3}}}a-{\frac  {a_{1}}{r^{3}}}i]\,

=-{\frac  {3}{r^{5}}}[\sum x^{2}]a+{\frac  {3}{r^{5}}}(\sum a_{1}x)r+{\frac  {1}{r^{3}}}(\sum 1)a-{\frac  {1}{r^{3}}}\sum a_{1}i\, --(8)

But \sum x^{2}=x^{2}+y^{2}+z^{2}=r^{2},\sum a_{1}x=a_{1}x+a_{2}y+a_{3}z=r\cdot a\,

and \sum 1=3,\sum a_{1}i+a_{2}j+a_{3}k=a\,

Therefore \sum i\times {\frac  {\partial }{\partial x}}[{\frac  {a\times r}{r^{3}}}]\,

=-{\frac  {3}{r^{5}}}r^{2}a+{\frac  {3}{r^{5}}}(r\cdot a)r+{\frac  {3}{r^{3}}}a-{\frac  {1}{r^{3}}}a\, by (8)

or {\mathrm  {curl}}({\frac  {a\times r}{r^{3}}})=-{\frac  {a}{r^{{-3}}}}+{\frac  {3}{r^{5}}}(a\cdot r)r\, by (4)

Hence the required is proved.

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