# How do you find the equations for the tangent plane to the surface f(x,y)=2-2/3x-y through (3,-1,1)?

Jul 6, 2018

$2 x + 3 y + 3 z = 6$

#### Explanation:

First write as a level surface $\phi \left(\boldsymbol{x}\right)$, with $z = f \left(x , y\right) :$

• $\phi \left(\boldsymbol{x}\right) = z - 2 + \frac{2}{3} x + y = 0$

The normal vector at any point is:

$\boldsymbol{n} = \nabla \phi = \left\langle{\phi}_{x} , {\phi}_{y} , {\phi}_{z}\right\rangle = \left\langle\frac{2}{3} , 1 , 1\right\rangle$

Plane is in form:

• $\left(\boldsymbol{r} - {\boldsymbol{r}}_{o}\right) \cdot \boldsymbol{n} = 0$

$\implies \left\langlex - 3 , y + 1 , z - 1\right\rangle \cdot \left\langle\frac{2}{3} , 1 , 1\right\rangle = 0$

$\frac{2}{3} x - 2 + y + 1 + z - 1 = 0$

$\therefore 2 x + 3 y + 3 z = 6$

NB in case you are not familiar with the directional derivative , another way to find the normal vector is to take partial derivatives of $f \left(x , y\right)$

So ${f}_{x} = - \frac{2}{3} q \quad {f}_{y} = - 1$

Then take the vector product of these tangent vectors:

• $\left\{\begin{matrix}\left\langle1 & 0 & - \frac{2}{3}\right\rangle & \left(\leftarrow {f}_{x}\right) \\ \left\langle0 & 1 & - 1\right\rangle & \left(\leftarrow {f}_{y}\right)\end{matrix}\right.$

$\boldsymbol{n} = \det \left[\begin{matrix}\hat{x} & \hat{y} & \hat{z} \\ 1 & 0 & - \frac{2}{3} \\ 0 & 1 & - 1\end{matrix}\right]$

$= \hat{x} \left(\frac{2}{3}\right) - \hat{y} \left(- 1\right) + \hat{z} \left(1\right) = \left\langle\frac{2}{3} , 1 , 1\right\rangle$

Jul 8, 2018

The tangent plane to the plane $z = 2 - \left(\frac{2}{3}\right) x - y$ is itself.