VC1.1

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{\frac  {\partial A}{\partial y}}={\frac  {\partial }{\partial y}}(xyz)i+{\frac  {\partial }{\partial y}}(xz^{2})j-{\frac  {\partial }{\partial y}}(y^{3})k

Therefore, {\frac  {\partial ^{2}A}{\partial y^{2}}}={\frac  {\partial }{\partial y}}(xz)i-{\frac  {\partial }{\partial y}}(3y^{2})k=-6yk

Again, {\frac  {\partial B}{\partial x}}={\frac  {\partial }{\partial x}}(x^{3})i-{\frac  {\partial }{\partial x}}(xyz)j+{\frac  {\partial }{\partial x}}(x^{2}z)k=2x^{3}i-yzj+2xzk

Therefore, {\frac  {\partial ^{2}B}{\partial x^{2}}}={\frac  {\partial }{\partial x}}(3x^{2})i-{\frac  {\partial }{\partial x}}(yz)j+{\frac  {\partial }{\partial x}}(2xz)k

Now, {\frac  {\partial ^{2}A}{\partial y^{2}}}\times {\frac  {\partial ^{2}B}{\partial x^{2}}}=-6yk\times (6xi+2zk)=-36xyk\times i (As k\times k=0 )

Hence {\frac  {\partial ^{2}A}{\partial y^{2}}}\times {\frac  {\partial ^{2}B}{\partial x^{2}}}=-36xyj (As k\times i=j )

Which is equal to -36j at the point (1,1,0).

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