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on the diagram for Method 1. To determine the coordinates of these points, we
must use the fact that the vector found by joining the two points is perpendicular
to the direction vector of each line.
We start by writing each line in parametric form.
For
For
The point with coordinates U12 s, 1 s, s2 represents a general point on
L and V1t, 1 t, 2t2 represents a general point on L We next calculate UV ! 1 ,
2 .
.
UV ! 1t 12 s2, 11 t2 11 s2, 2t s2 1t s 2, t s, 2t s2
UV !
represents a general vector with its tail on L 1 and its head on L 2 .
To find the points on each of the two lines ! that produce the ! minimal
distance, we must use equations and m since UV !
m1! # UV 0 2! # UV 0, must be
perpendicular to each of the two planes.
Therefore, 11, 1, 121t s 2, t s, 2t s2 0
11t s 22 11t s2 112t s2 0
or 2t 3s 2
(Equation 1)
and 11, 1, 22 # 1t s 2, t s, 2t s2 0
11t s 22 11t s2 212t s2 0
or 3t s 1
(Equation 2)
This gives the following system of equations:
1
2
L1 ,
L2 ,
! x 2 s, y 1 s, z s, m 1 11, 1, 12.
! x t, y 1 t, z 2t, m 2 11, 1, 22.
2t 3s 2
3t s 1
7t 1
t 1 7
If we substitute t 1 into equation 2 , or s 4 7
3 Q1 7 R s 1 7 .
We now substitute s 4 and t 1 into the equations for each line to find
7
7
the required points.
For L 1, x 2 4 y 1 4 z 4 7 3 7 10 7 ,
7 , 7 .
Therefore, the required point on is Q 10 7 , 3 7 , 4 7 R.
For L 2, x 1 7 ,
3
y 1 Q1 7 R 6 7 ,
z 2Q 1
7 R 2 7 .
Therefore, the required point on L is Q 1 7 , 6 2 7 , 2 7 R.
2
1
L 1
NEL
CHAPTER 9 547