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Given surface is \phi (x,y,z)=xyz=4\,

Let r=xi+yj+zk\, be the position vector of any point P(x,y,z) of the required tangent plane and let r_{{0}}=i+2j+2k\, be the position vector of the given point (1,2,2).

Here r-r_{{0}}=(xi+yj+zk)-(i+2j+2k)=(x-1)i+(y-2)j+(z-2)k\,

Now,\nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](xyz)=yzi+zxj+xyk\,

Therefore,(\nabla \phi )_{{0}}\,=value of \nabla \phi \, at point (1,2,2)=4i+2j+2k.

Hence the required equation of the tangent plane is (r-r_{{0}})\cdot (\nabla \phi )_{{0}}=0\,

Therefore,[(x-1)i+(y-2)j+(z-2)k]\cdot (4i+2j+2k)=0\,

=4(x-1)+2(y-2)+2(z-2)=0\, or 2x+y+z=6\,</math>

Again,the required equation of normal is (r-r_{{0}})\times (\nabla \phi )_{{0}}=0\,

or [(x-1)i+(y-2)j+(z-2)k]\times (4i+2j+2k)=0\,

or {\begin{vmatrix}i&j&k\\x-1&y-2&z-2\\4&2&2\end{vmatrix}}=0\,

or [2(y-2)-2(z-2)]i-[2(x-1)-4(z-2)]j+[2(x-1)-4(y-2)]k=0\,

which holds if the coefficients of i,j,k vanish simultaneously.

Therefore, 2(y-2)-2(z-2)=0,{\frac  {y-2}{1}}={\frac  {z-2}{1}}\,

2(x-1)-4(z-2)=0,{\frac  {x-1}{2}}={\frac  {z-2}{1}}\,

2(x-1)-4(y-2)=0,{\frac  {x-1}{2}}={\frac  {y-2}{1}}\,

Hence from the above,the equations of the normal in cartesian form is

{\frac  {x-1}{2}}={\frac  {y-2}{1}}={\frac  {z-2}{1}}\,

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