VC3.23

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a).Here the curl f ={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\\sin y+z&x\cos y-z&x-y\end{vmatrix}}\,

=[{\frac  {\partial }{\partial y}}(x-y)-{\frac  {\partial }{\partial z}}(x\cos y-z)]i-[{\frac  {\partial }{\partial x}}(x-y)-{\frac  {\partial }{\partial z}}(\sin y+z)]j+[{\frac  {\partial }{\partial x}}(x\cos y-z)-{\frac  {\partial }{\partial z}}(\sin y+z)]k\,

=(-1+1)i-(1-1)j+(\cos y-\cos y)k=0\,

Since curl f is equal to zero,it follows that f is irrotational.

b).f is irrotational vector implies curl f=0.

{\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\(x+2y+az)&(bx-3y-z)&(4x+cy+2z)\end{vmatrix}}=0\,

=[{\frac  {\partial }{\partial y}}(4x+cy+2z)-{\frac  {\partial }{\partial z}}(bx-3y-z)]i-[{\frac  {\partial }{\partial x}}(4x+cy+2z)-{\frac  {\partial }{\partial z}}(x+2y+az)]j+[{\frac  {\partial }{\partial x}}(bx-3y-z)-{\frac  {\partial }{\partial y}}(x+2y+az)]k=0\,

=(c+1)i-(4-a)j+(b-2)k=0\,

=c+1=0,-(4-a)=0,b-2=0\,

=c=-1,a=4,b=2\,

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