Finding the Tangent Plane to a Surface at a Given Point

[tylerc1991]

Homework Statement

Suppose you need to know an equation of the tangent plane to a surface S at the point P(2,1,3). You don't have an equation for S but you know that the curves:

r_1(t) = <2 + 3t, 1 - t^2, 3 - 4t + t^2>
r_2(u) = <1 + u^2, 2u^3 - 1, 2u + 1>

both lie on S. Find an equation of the tangent plane at P

Homework Equations

The general form of a plane at a point P(x_0, y_0, z_0) with normal vector n = <a, b, c> is

a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

The Attempt at a Solution

t = 0 for P to lie on r_1(t), and u = 1 for P to lie on r_2(u)
taking the first derivatives of r_1(t) and r_2(u) and evaluating them at the specified t and u values gives:
r'_1(0) = <3, 0, -4> = A
r'_2(u) = <2, 6, 2> = B

Since both of these vectors are tangent to the surface at P, I took the cross product of them, A x B and got <24, 14, 18>, which simplifies to <12, 7, 9>.

so the normal vector to the tangent plane to the surface is <12, 7, 9> and the point is P.
plugging this into the general form gives:

12(x - 2) + 7(y - 1) + 9(z - 3) = 0.

This can be simplified but I wanted to know if this was the correct answer. Thank you very much anyone for your help!

 

[Dick]

Certainly nothing wrong with the approach. Are you sure <3, 0, -4>X<2, 6, 2>=<24, 14, 18>. I get something a little different. Can you check it?

 

[tylerc1991]

Certainly nothing wrong with the approach. Are you sure <3, 0, -4>X<2, 6, 2>=<24, 14, 18>. I get something a little different. Can you check it?

Whoops, forgot the minus in the determinant, I am getting <24, -14, 18> which reduces to <12, -7, 9>. Thank you.