PDA

View Full Version : Find a Point on a 3D surface.


Spectre32
Oct10-04, 03:33 PM
Ok I have a surface withthe eqn of x^2 + 2y^2 +3z^2 = 12. The question tells us that there is a perpendicular plane tangent to the the line is as follows:

x = 1 + 2t
y = 3 + 8t
z = 2 - 6t

Spectre32
Oct10-04, 03:36 PM
Whoops I neglected to see something on his answer key... I see whats going on.. Nevermind

Spectre32
Oct10-04, 03:45 PM
lol .. ok now i have question. After looking at the real solution my teacher used the Para eqn's to make a vector... athen took the derivite of F(x,y,z) and got <2x,4y,6z>. Then he gets this point (1,2,-1)... I'm unclear as to how he got this point.

Spectre32
Oct10-04, 06:34 PM
Yeah I'm been crackin away at this for a while and i sitll don't get it, any one anyone anyone..