# Find the normal vector of the tangent plane to x^2yz+3y^2-2xz^2+8z=0  at (1,2,-1) ?

May 18, 2017

$- 6 \hat{i} + 12 \hat{j} + 14 \hat{k}$

#### Explanation:

First we rearrange the equation of the surface into the form $f \left(x , y , z\right) = 0$, this is already done for us:

${x}^{2} y z + 3 {y}^{2} - 2 x {z}^{2} + 8 z = 0$

And so we define our surface function, $f$, by:

$f \left(x , y , z\right) = {x}^{2} y z + 3 {y}^{2} - 2 x {z}^{2} + 8 z$

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

$\nabla f \left(x , y , z\right) = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k}$

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

$\nabla f = \left(\frac{\partial}{\partial x} \left({x}^{2} y z + 3 {y}^{2} - 2 x {z}^{2} + 8 z\right)\right) \hat{i} +$
$\text{ } \left(\frac{\partial}{\partial y} \left({x}^{2} y z + 3 {y}^{2} - 2 x {z}^{2} + 8 z\right)\right) \hat{j} +$
$\text{ } \left(\frac{\partial}{\partial z} \left({x}^{2} y z + 3 {y}^{2} - 2 x {z}^{2} + 8 z\right)\right) \hat{k}$
$\text{ } = \left(2 x y z - 2 {z}^{2}\right) \hat{i} + \left({x}^{2} z + 6 y\right) \hat{j} + \left({x}^{2} y - 4 x + 8\right) \hat{k}$

So for the particular point $\left(1 , 2 , - 1\right)$ the normal vector to the surface is given by:

$\nabla f \left(1 , 2 , - 11\right) = \left(- 4 - 2\right) \hat{i} + \left(- 1 + 12\right) \hat{j} + \left(2 - 4 + 8\right) \hat{k}$
$\text{ } = - 6 \hat{i} + 12 \hat{j} + 14 \hat{k}$

We can confirm this graphically: Here is the surface with the normal vector: