Vector equation for the tangent line of the intersection of $x^2 + y^2 = 25$ and $y^2 + z^2 = 20$ What is the vector equation for the tangent line of the intersection of $x^2 + y^2 = 25$ and $y^2 + z^2 = 20$ at the point $(3,4,2)$?
I think I should find a vector
$$
\gamma(t) = (x(t),y(t),z(t))
$$
that represents the intersection. Because then, I can find easily the tangent line.
But I do not know a general technique to find such a $\gamma$.
Any suggestion please?
 A: You can define functions $f(x, y, z) = x^2 + y^2 - 25$ and $g(x, y, z) = y^2 + z^2 - 20$. Then by taking the gradient vectors $\nabla f(x, y, z) = (2x, 2y, 0)$ and $\nabla g(x, y, z) = (0, 2y, 2z)$ you get vectors that are perpendicular to each of the level surfaces defined by $f(x, y, z) = 0$ and $g(x, y, z) = 0$. 
Therefore if you take the cross product of the two gradients, you will get a directional vector for the tangent line.
$$
\vec{n} = \nabla f \times \nabla g = 
\begin{vmatrix}
i & j & k\\
2x &2y &0\\
0 &2y &2z
\end{vmatrix}
= (4yz, -4xz, 4xy)
$$
Thus evaluating at the point $(3, 4, 2)$ gives the vector $\vec{n} = (32, -24, 48) = 8(4, -3, 6)$ in the direction of the tangent line. Finally, a vector equation for the tanget line at the given point would just be
$$
(x, y, z) = (3, 4, 2) + t(32, -24, 48)
$$
You can see the two surfaces and the tangent line just computed in the plot below.

A: Hint: Maybe use $y=t,\ x=(25-t^2)^{1/2},\ z=(20-t^2)^{1/2}$ and restrict $t$ to make the radicals defined. I think the tangent line would then be possible except at the endpoints, for specific values of $t$ chosen.
ADDED: Actually $x$ and/or $z$ could be chosen as the negatives of the above radicals, and that would also be part of the intersection. So one has to choose specific signs for these at each given point. (This means for a fixed $t_0$ to choose either positive or negative radicals for the $x,z$ coordinates to get both the point on the intersection and the value of the derivative vector at that point.)
