# Calculus

posted by .

Find the distance between the lines [x,y,z] = [-1,1,0] + s[3,4,-2] and [x,y,z] = [5,9,-4] + t[2,3,1]. Interpret the result.

• Calculus -

As I recall, if you have two lines

L1: p1 + v1*t
L2: p2 + v2*t

the distance can be obtained by projecting the unit normal between the lines onto the line joining the two points:

v1 × v2 / |v1 × v2| • (p1 - p2)

For a brief example and discussion, see

physicsforums . com / showthread.php?t=276253

• Calculus -

for this question, do i have to solve it? it just says interpret the results, does that mean only explain?

im trying to solve it right now. i have:

L1: [x,y,z]=[-1,1,0]+s[3,4,-2]
L2: [x,y,z]=[5,9,-4]+t[2,3,1]

First, cross product of the two directional vectors from both lines:

[3,4,-2] x [2,3,1] =

That's all I have so far :/
I don't know what to do next because in the example shown on the forum, I don't know how they got the answer [from the forum: (-2,3,-2) x (-2,-3,-1) = (-9,-2, 12)] I know 3 x (-3) is -9 but I don't understand where the -2 and +12 came from since (-2)(-2) = +4 and (-2)(-1)= +2.

Also on the forum there is:

And its unit vector is:
-9x-2y+12z
-----------
229^1/2

I don't know how they got this answer either. They mentioned at the end of it they may not be right so I don't know ....

### Answer This Question

 First Name: School Subject: Answer:

### Related Questions

More Related Questions

Post a New Question