Question

If I know the coordinates for two lines
l:(x, y, z)= (1+t, 1+2t, 1+3t)
and
l':(x, y, z)=(t, 2+t, 13-2t)

How can I find the equation for the plane which these lines span?

Answer 1

If there's a single plane they are both on, then they must intersect.

Where do they intersect?

Answer 2

http://www.physicsforums.com/showthread.php?t=11972

Answer 3

Find a vector representation of each line. Cross the two vectors. The result is a vector normal to the plane. Any point on a line is also in the plane, so you have the normal and a point on the plane...

Answer 4

I find the point where the two points intersect by doing the following:
\[\left[\begin{matrix} 1+s\\1+2s \\1+3s \end{matrix}\right] = \left[\begin{matrix} t \\ 2+t \\13-2t \end{matrix}\right]\]
\[\left[\begin{matrix} 1 = t-s\\ -1 = t-2s \\-12=-2t-3s \end{matrix}\right] R_2-2R_1, R_3-3R_1 \implies \left[\begin{matrix} 1=t-s \\-3=-t \\-15=-5t \end{matrix}\right] R_3+5R_2 \implies \left[\begin{matrix} 1=t-s\\-3 =-t\\0=0\end{matrix}\right]\]
I feel that the last step is unnecessary, but it feels better getting zeroes that work.
This then gives me
\[t=3\]
\[s=t-1 \implies s=3-1=2\]
Soo.. the lines intersect in the point
\[x=1+2=3, y=1+2*2=5, z=1+3*2=7\]

All right. I have an intersection point and two lines. I know that a planes equation can be described through exactly that - I just don't know how to make the leap from lines and point to plane. Hint?