Definite method for finding the intersection of two cartesian lines in 3D.

Question

I have the following problem:

Determine if these lines intersect. If so, find their point of intersection.

$L1 = (4,5,-1)+t(1,1,2)$

$L2 = (6,11,-3)+s(2,4,1)$

I managed to solve this problem by 1) Doing a cross product, seeing that the result is not 0, and 2) plugging in a bunch of $t$'s and $s$'s by trial-and-error until I ended up finding the point $(2,3,-5)$, as the answers to many other similar homework questions say to do (like this one).

This process took me a very long time. As opposed to trial and error, is there a method that will always give me the point of intersection of two lines in either cartesian, parametric or symmetric form?

Answer 1

Simply write $$\left\{ \begin{array}{rcl} 4+t&=&6+2s\\ 5+t&=&11+4s\\ -1+2t&=&-3+s \end{array} \right.$$

And then solve the system formed by the two fisrt equations.
If the solution satisfies the third one, then the lines intersect and its intersection can be found substituting the found value for $t$ in the first line, or the value for $s$ in the second line.
If the solution doesn't satisfies the third equation, the lines do not intersect.

Answer 2

To find the intersection of two lines you have to solve a system of linear equations. In your case the system is $$ (4,5,-1) + t (1,1,2) = (6,11,-3) + s (2,4,1) $$ which is a system of three equations in two unknowns.

Answer 3

As Emanuele suggests, you will have a system of three linear equations for two unknowns. You can create an augmented matrix and then row reduce. We have the equations $t-2s=2,t-4s=6,2t-s=-2$ which implies the following augmented matrix: $$\begin{pmatrix} 1 & -2 & 2 \\ 1 & -4 & 6 \\ 2 & -1 & -2 \end{pmatrix}\Rightarrow \begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{pmatrix}.$$

This gives $t=-2,s=-2$. If the system has a solution, the 3rd row of the reduced matrix must be all zeros, which in this case it does.

We can easily generalize this process. For $L_1=(c1x,c1y,c1z)+t(p1x,p1y,p1z),L_2=(c2x,c2y,c2z)+s(p2x,p2y,p2z)$ we have the following augmented matrix: $$\begin{pmatrix} p1x & -p2x & c2x-c1x \\ p1y & -p2y & c2y-c1y \\ p1z & -p2z & c2z-c1z \end{pmatrix}.$$

Now row reduce this after putting in the corresponding numbers.