VC3.46

From Exampleproblems

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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