Let s be the surface defined by x 2 + 2y 3 + 3z 4 = 6. let t be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−2, uv+1). these two surfaces intersect in a curve c which passes through the point (x, y, z) = (1, 1, 1). find the tangent line to c through (1, 1, 1).

The tangent to through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces and at that point.

Let . Then is the level curve . Recall that the gradient vector is perpendicular to level curves; we have

so that the gradient of at (1, 1, 1) is

For the surface , we have

so that . We can obtain a vector normal to by taking the cross product of the partial derivatives of , and evaluating that product for :

Now take the cross product of the two normal vectors to and :

The direction of vector (24, 8, -8) is the direction of the tangent line to at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by . Then adding (1, 1, 1) shifts this line to the point of tangency on . So the tangent line has equation

c is the final answer! : )