# Calculus 3, dealing with tangent planes and surfaces. 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?

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tiny-tim
Homework Helper
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! $2(1)^2+ (-2)^2+ 3(1)^2= 9$ 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 $<4x, 2y, 6z>= \lambda<4, -4, 6>$ for some number $\lambda$. That together with $2x^2+ y^2+ 3z^2= 1$ gives x, y, and z.