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

##### 2 Answers

Jul 6, 2018

#### Answer:

#### Explanation:

First write as a level surface

#phi(bbx) = z-2+2/3x+y = 0#

The normal vector at any point is:

Plane is in form:

#(bbr - bbr_o)*bbn = 0 #

NB in case you are not familiar with the **directional derivative** , another way to find the normal vector is to take partial derivatives of

So

Then take the vector product of these tangent vectors:

#{((:1,0,- 2/3:), (larr f_x)),((:0, 1,-1:),(larr f_y)):}#

Jul 8, 2018

The tangent plane to the plane