Question

How do I find the intersection of two lines in three-dimensional space?

Answer 1

I'll show how to solve this problem using an example using the vector based form of a two straight lines. Bear in mind that there will be one of the following outcomes:

• a single unique point.
• no solution (if the lines do not intersect).
• infinitely many solutions (if the lines coincide).

Suppose we have;

`L_1: \ \ \ \ vec(r_1) = ((5),(2),(-1)) +lamda((1),(-2),(-3))`

`L_2: \ \ \ \ vec(r_2) = ((2),(0),(4)) +mu((1),(2),(-1))`

It is vital that the variable parameters (`lamda` and `mu`) have different symbols, as they are different parameters relative to each line.

If the lines do meet then for some specific values of `lamda` and `mu` then:

`vec(r_1) = vec(r_2)`

In which case we would have:

`((5),(2),(-1)) +lamda((1),(-2),(-3)) = ((2),(0),(4)) +mu((1),(2),(-1))`
`:. \ \ \ \ \ \ ((5+lamda),(2-2lamda),(-1-3lamda)) = ((2+mu),(2mu),(4-mu))`

By comparing the coefficients of `hat(i)`, `hat(j)` and `hat(k)` we have three equations in two unknowns, so we can solve for `lamda` and `mu` using the first two equations and check if the third is satisfied:

`hat(i) : \ \ \ \ \ \ \ \ \ \ \ 5+lamda = 2+mu " " ....... [1]`
`hat(j) : \ \ \ \ \ \ \ \ \ 2-2lamda = 2mu " " ....... [2]`
`hat(k) : \ \ \ \ -1-3lamda = 4-mu \" " ....... [3]`

`Eq [1] + 1/2Eq[2]` gives `=> 6=2+2mu => mu=2`
Subs `mu=2` into `Eq[1] => 5+lamda=4=>lamda=-1`
Subs `mu=2` and `lamda=-1` into `Eq [3]=>-1+3=2=4-2`

So we have established that if we choose `lamda=-1` and `mu=2` then we get a unique solution simultaneously satisfying all three equations.

We can then substitute `lamda=-1` into `L_1`, (equally `mu=2` into `L_2` would work) to determine the actual coordinate of intersection.

`vec(r_1) = ((5),(2),(-1)) -((1),(-2),(-3))`
`\ \ \ = ((5-1),(2+2),(-1+3))`
`\ \ \ = ((4),(4),(2))`

so in this example, the lines `L_1` and `L_2` intersect at the coordinate `(4,4,2)`