Calculus 3, dealing with tangent planes and surfaces.

[yopy]

okay i came up with doing the gradient of the ellipsoid. Then set that equal to the vector, <4,-4,6>. I solved and got x,y,z = 1,-2,1

I looked at the answer key and it said (1/3) (1,-2,1)

Does anyone know where the 1/3 came from?

 

[tiny-tim]

Hi yopy!

i came up with doing the gradient of the ellipsoid. Then set that equal to the vector, <4,-4,6>. I solved and got x,y,z = 1,-2,1

You'll only get a proportion out of that.

2*1² +(-2)² +3*1 = 9,

so you have to divide everything by 3.

 

[HallsofIvy]

One obvious point should be that (1, -2, 1) is not even on the ellipse! [itex]2(1)^2+ (-2)^2+ 3(1)^2= 9[/itex] not 1!

In order that two planes be parallel, their normal vectors must be parallel, but not necessarily equal. As tiny-tim said, you should have [itex]<4x, 2y, 6z>= \lambda<4, -4, 6>[/itex] for some number [itex]\lambda[/itex]. That together with [itex]2x^2+ y^2+ 3z^2= 1[/itex] gives x, y, and z.