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:

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

Explanation:

First write as a level surface #phi(bbx)#, with #z = f(x,y):#

  • #phi(bbx) = z-2+2/3x+y = 0#

The normal vector at any point is:

#bbn = nabla phi = (: phi_x, phi_y, phi_z :) = (: 2/3,1,1:)#

Plane is in form:

  • #(bbr - bbr_o)*bbn = 0 #

#implies (:x-3 ,y+1, z-1 :) * (:2/3,1,1:) = 0 #

#2/3 x - 2 + y + 1 + z - 1 = 0#

#:. 2 x +3 y + 3z = 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(x,y)#

So #f_x = - 2/3 qquad f_y = - 1#

Then take the vector product of these tangent vectors:

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

#bbn = det [(hat x, haty, hatz),( 1,0,- 2/3),(0, 1,-1)]#

#= hat x(2/3) - hat y (- 1) + hat z (1) = (: 2/3,1,1:)#

Jul 8, 2018

The tangent plane to the plane #z = 2 - (2/3)x - y# is itself.