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

1 Answer
Jan 4, 2017

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)#