Determining if two lines intersect in 3D

Question

If I have for example two lines in $3d$, presented in parametric form,

$L1: x= 3t+2, ~y=t+6, ~z=2t-9$

$L2: x= t+5, ~y=3t+7, ~z=t+3$

and I want to find if they intersect, do I set the t variable for $L2$ to s and solve simultaneously or is there a different approach?

Answer 1

Clearly the lines are not parallel. Now finding the shortest distance between the two lines. If they intersect, shortest distance will be zero.

You may find the shortest distance using by finding the projection of a vector joining the two lines along the common normal. The S given below is not the shortest distance, but it will be zero if the lines intersect.

$$S = [<3,1,12> <3,1,2> <1,3,1>]$$

Note: $[\rm a\ b\ c]$ represents the scalar triple product.

Answer 2

Write the system of three equations in two unknowns

$$\begin{cases} 3t+2=\ \ t'+5\\ \ \ t+6=3t'+7\\ 2t-9=\ \ t'+3\\ \end{cases}$$

Either solve two $2\times2$ systems and compare the solutions, or put it in a standard form and check that the determinant is zero (i.e. rank $2$).

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In this particular case, a quick solution is obtained by subtracting the first equation from the second(three times) and from the third.

$$\begin{cases} -8t=-8\\ -t-11=-2 \end{cases}$$

This doesn't seem very compatible...