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Question Number 85865 by subhankar10 last updated on 25/Mar/20
provethatcurl(rnc→×r→)=(n+2)rnc→−nrn−2(r→.c→).wherecistheconstantvector.
Answered by TANMAY PANACEA. last updated on 25/Mar/20
r→=ix+jy+kz→r2=x2+y2+z2rn=(x2+y2+z2)n2c→=ia+ib+kdcurlA→=▽→×A→(i∂∂x+j∂∂y+k∂∂z)×A→now▽→×(c→×rnr→)=(▽→.rnr→)c→−(▽→.c→)rnr→=c→(i∂∂x+j∂∂y+k∂∂z).(x2+y2+z2)n2(ix+jy+kz)=c→[∂∂x{x.(x2+y2+z2)n2}+∂∂y{y(x2+y2+z2)}+∂∂z{z(x2+y2+z2)}]calculationof∂∂x{x(x2+y2+z2)n2}=(x2+y2+z2)n2×1+x×n2(x2+y2+z2)n2−1×2x=rn+x2×n(r2)n2−1=rn+x2×n×rn−2addingthree3rn+nrn−2(x2+y2+z2)=3rn+nrn=rn(n+3)soanswerisrn(n+3)c→timorrowishallsolveinpaper.=..c→×r→∣ijk∣∣abd∣∣xyz∣=i(bz−yd)−j(az−xd)+k(ay−bx)now∣ij
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