VC3.48

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Given,\phi (x,y,z)=x^{2}+y^{2}+z^{2}=9,\psi (x,y,z)=x^{2}+y^{2}-z=3\,

Therefore,\nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](x^{2}+y^{2}+z^{2})=2xi+2yj+2zk=4i-2j+4k\, at(2,-1,2).

\nabla \psi ==[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](x^{2}+y^{2}-z)=2xi+2yj-k=4i-2j-k\, at(2,-1,2).

The angle between the curves is \cos \theta ={\frac  {\nabla \phi \cdot \nabla \psi }{|\nabla \phi ||\nabla \psi |}}\,

\cos \theta ={\frac  {(4i-2j+4k)\cdot (4i-2j-k)}{{\sqrt  {16+4+16}}{\sqrt  {16+4+1}}}}={\frac  {16}{6{\sqrt  {21}}}}\,

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