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**My Little Pony** The curve r(t) passing through the point (1,1,1) is defined implicitly by the equations:

x^2y + y^2x + xyz + z^2 = 4

x + y + z = 3

Find the equation of the tangent line to the curve at the given point.

__Attempt at a solution__

GRAD f(x,y,z) = (2xy + y^2)i + (x^2 + 2xy)j + (xy + 2z)k

At (1,1,1):

GRAD f(1,1,1) = 3i + 3j + 3k

Tangent line has normal of (3,3,3).

Therefore:

3(x - 1) + 3(y - 1) + 3(z - 1) = 0

3x + 3y + 3z = 9

x + y + z = 3

I'm not sure what I'm doing wrong. Am I supposed to use both equations given in the question?