# VC3.46

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\,[/itex]

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

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