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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