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September 30, 2014

September 30, 2014

Posted by **Becky** on Thursday, September 27, 2012 at 8:48pm.

z = x2 + y2

and the ellipsoid

6x2 + 5y2 + 7z2 = 39

at the point

(−1, 1, 2)

- Calc 3 -
**Steve**, Friday, September 28, 2012 at 10:37amif u and v are functions of x,y,z, then the normals are

Nu = ∇u

Nv = ∇v

The tangent line of the intersection is perpendicular to both normals

T(x,y,z) = ∇u × ∇v

∇u = 2xi + 2yj - k

∇v = 12xi + 10yj + 14zk

T(-1,1,2) = 33i + 34j + 2k

so that would be

x = -1 + 33t

y = 1 + 34t

z = 2 + 2t

- Calc 3 -
**Becky**, Friday, September 28, 2012 at 4:46pmcan you please explain to me where you got the numbers for T(-1,1,2) I know you did the cross product but I can not seem to get the same numbers as you. Please explain

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