VC3.49

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Since (1,-1,2) lies on the given surface ax^{2}-byz=(a+2)x\,

Therefore,a+2b=a+2\, so that b=1.

Thus,the given surfaces can be written as

\phi (x,y,z)=ax^{2}-yz-(a+2)x=0,\psi (x,y,z)=4x^{2}y+z^{3}=4\,

Therefore,\nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](ax^{2}-yz-(a+2)x)=(a-2)i-2j+k\, at (1,-1,2).

\nabla \psi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](4x^{2}y+z^{3})=8xyi+4x^{2}j+3z^{2}k=-8i+4j+12k\, at (1,-1,2).

Since the two surfaces are orthogonal,(\nabla \phi )\cdot (\nabla \psi )=0\,

or [(a-2)i-2j+k]\cdot [-8i+4j+12k]=0\,

or -8(a-2)-8+12=0\, so that a={\frac  {5}{2}}\,

Thus the required values are a={\frac  {5}{2}},b=1\,

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