SOLUTION: Two lines have equations L1 :[x,y,z] = [1,3,2] + t[3,0,-1] L2 :[x,y,z] = [3,1,-2] + s[2,-1,3] Determine points P(on L1) and Q(on L2) such that PQ is perpendicular to both l

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Question 961162: Two lines have equations
L1 :[x,y,z] = [1,3,2] + t[3,0,-1]
L2 :[x,y,z] = [3,1,-2] + s[2,-1,3]
Determine points P(on L1) and Q(on L2) such that PQ is perpendicular to both lines.
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
L1 :[x,y,z] = [1,3,2] + t[3,0,-1]
L2 :[x,y,z] = [3,1,-2] + s[2,-1,3]
A direction vector of L1 is < 3,0,-1 >
A direction vector of L2 is < 2,-1,3 > 

A general point of L1 is
P = (1+3t, 3, 2-t)

A general point of L2 is 
Q = (3+2s, 1-s, -2+3s) 

The components of a vector from P to Q is
found by subtracting their coordinates:

< -2+3t-2s,2+s,4-t-3s >
 
We want to make this vector be perpendicular
to the direction vectors of L1 and L2 

< 3,0,-1 > and < 2,-1,3 >. 

So we dot it with each of those and set
each equal to 0: 

< -2+3t-2s,2+s,4-t-3s > • < 3,0,-1 > = 0
< -2+3t-2s,2+s,4-t-3s > • < 2,-1,3 > = 0

-6+9t-6s+0+0-4+t+3s = 0
-4+6t-4s-2-s+12-3t-9s = 0

10t- 3s = 10
 3t-14s =  6

Solve that system and get 

t=122/131, s=-30/131

What god-awful fractions!  Hope I didn't
make a mistake.

So to see what those points are we
have to substitute those god-awful 
fractions in

P = (1+3t, 3, 2-t) = (497/131,3,140/131)

and

Q = (3+2s, 1-s, -2+3s) = (333/131,161/131,-352/131)

Edwin
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