Expert: Paul Klarreich - 4/24/2007

Question
1. Show that the line with equation (x,y,z)=(4,-1,3)+ t(-3,3,0) intersects the line with equation (x,y,z)=(5,4,1)+s(2,1,-1). Find the coordinates of the point of intersection, and determine the angle between the two lines.

2. Find the minimum distance between the point w=(3,4,5) and any point on the line(x,y,z)=(1,3,-2) +t(0,-2,1), tER.

Answer
Questioner:   scott
Category:  Advanced Math
 
Subject:  algebra-vectors
Question:  1. Show that the line with equation (x,y,z)=(4,-1,3)+ t(-3,3,0) intersects the line with equation (x,y,z)=(5,4,1)+s(2,1,-1). Find the coordinates of the point of intersection, and determine the angle between the two lines.

2. Find the minimum distance between the point w=(3,4,5) and any point on the line(x,y,z)=(1,3,-2) +t(0,-2,1), tER.

.................................................
Hi, Scott,

For the first one, you just have to find the proper values of s and t, satisfying these equations:

4 - 3t = 5 + 2s    X
-1 + 3t = 4 + s    Y
3 = 1 - s          Z

Since you have three equations in two variables, there won't always be a solution, and the lines won't always intersect.  (that's what the problem is about)

You get  s = -2 from equation Z.  Substitute into X:

4 - 3t = 5 + 2(-2)
4 - 3t = 5 - 4
4 - 3t = 1
 - 3t = -3
    t = 1.

Now try that into Y to see if it checks:

-1 + 3(1) = 4 + (-2)
-1 + 3 = 4 - 2
    2 = 2  
Yes, it checks. Therefore the lines do intersect.

******** USE COURIER FONT TO VIEW THIS ************

Now as to the angle between the lines, use the standard formula for the angle between the two direction vectors:
         A dot B
cos @ = ------------
       | A | | B |,

where  A and B are the direction vectors for your lines:

A = (-3,3,0)
B = (2,1,-1)

A dot B = -6 + 6 + 0

Hmmm.  It looks as if A and B are orthogonal.  The angle between them is 90 degrees.

..........................................................

2. Find the minimum distance between:

the point w=(3,4,5) and [any point on]
the line: L: (x,y,z) = (1,3,-2) +t(0,-2,1)

The vector D = (0,-2,1) is the direction vector for the line L.
p(1,3,-2) is a point on the line L.

First, determine the angle between vector pw and vector D.

pw = (3,4,5) - (1,3,-2) = (2,1,7)
D = (0,-2,1)

pw dot D = 0 - 2 + 7 = 5
|pw| = sqrt(4 + 1 + 49) = sqrt(54)
| D | = sqrt(4 + 1) = sqrt(5)
            5
cos @ = ---------------
       sqrt(5) sqrt(54)

       sqrt(5)
cos @ = --------
       sqrt(54)

Now make a right triangle with points p (which is on L), w (which is not), and the altitude from w to the line L:

          w
         /|
        / |
       /  |
      /   |
     /    |
    /@    |
----------------------
  p       q

Now  qw is your desired distance.  It is equal to  |pw| sin @

| pw | = sqrt(54)

sin @ = sqrt( 1 - cos^2(@))

sin @ = sqrt( 1 - 5/54)

sin @ = sqrt((54 - 5)/54)

sin @ = sqrt(49)/sqrt(54)
                    sqrt(49)
Distance = sqrt(54) -----------
                    sqrt(54)

Distance = sqrt(49)