Find intersection of 2 parameterized planes I have two parameterized planes, for example, {u, 0, v} and {u-1, v-1, 1}. And I have to find the parametric equation of the line that intersects both planes. By setting both planes equal to each other I get {u=u-1, 0=v-1, v=1} = {0=-1, v=1, v=1}. Then by substituting v into the first equation {u, 0, v} I get the answer {u, 0, 1} which is correct but this approach doesn't seem correct. Is there another way to solve this?
 A: I believe you meant that your planes are described by the parameterizations
$$
S_1(u, v) = (u, 0, v) \\
S_2(u, v) = (u-1, v-1, 1)
$$
That makes it easy to find the normal vector (vector perpendicular) to each. For $S_1$, take a derivative with respect to $u$ and to $v$ to get that two vectors in the plane are $(1, 0,0)$ and $(0, 0, 1)$. Their cross-product is the normal vector, i.e.
$$
n_1 = (0, -1, 0).
$$
In fact, you can write an implicit equation for the plane at this point: it has the form 
$$
n_1 \cdot (x, y, z) = \text{constant}.
$$
Plugging in a known point of the plane --- say $S(0,0) = (0,0,0)$ for $(x,y, z)$, you get that the constant, in this case, must be 0, so the plane equation is 
$$
n_1  \cdot (x,y,z) = 0. 
$$
Computing, in the same way, the normal for $S_2$, we get $(1, 0, 0)$ and $(0, 1, 0)$ as vectors in the plane, with normal vector 
$$
n_2 = (0, 0, 1). 
$$
The cross product of these two normals, $w = (-1, 0, 0)$, is a direction perpendicular to both, hence it lies in each of the two planes. That means your intersection line runs in this direction. So your line has the form
$$
P(t) = P_0 + t w,
$$
where $P_0$ is any point of the intersection between the planes. 
To find a point of intersection (there are many!), you need a point on the second plane that also lies on the first, i.e., a point $S_2(u, v)$ that satisfies $n_1 \cdot (x, y, z) = 0$. Plugging in, we get
$$
(u-1, v-1, 1) \cdot (0, -1, 0) = 0 \\
1-v = 0
$$
so any point with $v = 1$ lies on both planes. Let's take $(u, v) = (0, 1)$ to get 
$P_0 = S_2(u, v) = S_2(0, 1) = (-1, 0, 1)$. So now your intersection line is
$$
P(t) = (-1, 0, 1) + (-1, 0, 0) t.
$$
The method described applies to any two planes:
1.Find normals $n_1, n_2$ to both planes. 


*

*Write a parametric equation, $n_1 \cdot (x, y,z) = C$ for the first, using the normal and $S_1(0,0)$ to find the constant $C$. 

*Let $w = n_1 \times n_2$.

*Solve $n_1 \cdot S_2(u, v) = C$ to find some pair $(u, v)$ such that $P_0 = S_2(u, v)$ is a point on both planes. (You'll have to make an arbitrary choice of $u$ or $v$ here).

*Write down the parametric equation $P(t)= P_0 + t w$. 
