MCV4U1-UNIT NINE-LESSON ONE
Lesson One: Intersection of Lines in  3
Recall that in  2 there are three ways that two lines can intersect.
1.
Parallel/non-coincident
2.
Parallel/coincident
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MCV4U1-UNIT NINE-LESSON ONE
3.
One solution
In  3 we have these three cases, plus one more.....
If you have two parallel planes that contain two non-parallel lines, then you have what is
called SKEW LINES.
Parallel
Planes
Line L1 will never intersect L2 . Therefore there will be no solution.
Method for finding the intersection of two lines in  3 .
1.
Check to see if the lines are parallel. If they are, the direction vectors will be the
same, or multiples of each other.
2.
If they are parallel, check to see if they are coincident. If they are coincident then
the point on one line will satisfy the other line.
3.
If the lines are not parallel, write them in parametric form. Equate any two
equations (say, for x and y) and solve for the two parameters. Test the parameters
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MCV4U1-UNIT NINE-LESSON ONE
in the remaining equations (say, z ). If the same value is obtained for each z, the
lines intersect in a point. Find the value of the remaining coordinates by
substituting the parametric value just determined.
If a different values are obtained for each z, then the lines are skew and there is no
solution.
Example 1:
Find the intersection of ....

r1  0,1,1  t 2,6,4

r2  2,3,4  r 3,9,6
Are they parallel?
Yes, b/c 3,9,6 
3
2,6,4
2
Are they coincident?


Substitute (2,3,-4) into r1 .......................... [ or (0,1,-1) into r2 ]
(2,3,-4)=(0,1,-1)+t(2,6,4)
(2,3,-4)=(0,1,-1)+(2t,6t,4t)
(2,3,-4)=(2t,1+6t,-1+4t)
2=2t
3=1+6t
-4=-1+4t
1
3
t=1
t=
t=
3
4
Since at least ONE of the t values is different, the lines are parallel, but non-coincident.
There is no solution.
Example 2:
L1 :
What is the intersection of L1 and L2 below?
x=3–r
y=-2+r
z = 3 + 2r
L2 :
x=1+t
y=-1+t
z = 4 + 4t
Are they parallel?
No.
 They either intersect, or they are skew. So we look for a solution even though it may
not exist.
1. 3 – r = 1 + t
2. - 2 + r = - 1 + t
1. t + r = 2
2. t – r = - 1
1. - 2. 2r = 3
r = 3/2
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MCV4U1-UNIT NINE-LESSON ONE
 t=1/2
Check r=3/2 and t=1/2 in the equations for z.....
6
3
L1 : z  3  2r  3  2   3   6
2
2
1
L2 : z  4  4t  4  4   4  2  6
2
Since the z values are the same, there is an intersection point.
3
1
Substitute r 
in for the equations for x and y in L1 or substitute t  in for the
2
2
equations for x and y in L2 .
Using L1 .......
3 3

2 2
3 1
y  2  r  2  
2
2
z  3  2r  6 (already done)
x  3r  3
Using L2 .......
1 3
x  1 t  1 
2 2
1 1
y  1  t  1  
2
2
z  4  4t  6 (already done)
 3 1 
 L1 and L2 int er sec t in the po int  , ,6 
2 2 
Example 3: Find the intersection of...
L1 :
x 1 y  3 z


3
1
2
L2 :
x  2 y z 1
 
2
2
4
Are they parallel?
No.
Introduce parameters for the lines. Let the parameter for L1 be t, and for L2 be r.
Then in parametric form, the lines become.....
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MCV4U1-UNIT NINE-LESSON ONE
L1 : x  3t  1
L2 : x  2r  2
y  t  3
z  2t
y  2r
z  4r  1
Equating x and y.....
1.
3t+1=-2r+2
2.
-t-3=2r
1.-2.
1.
2.
3t+2r=1
t+2r=-3
2t=4
t=2
substituting t=2 into equation 2.
2+2r=-3
2r=-5
5
r
2
Check t and r in the equations for z.
 5
L2 : z  4r  1  4
  1  10  1  9
 2 
L1 : z  2t  2(2)  4
Since the z’s are different, the lines do not intersect (no solution), and since we already
know they are not parallel, so the lines must be skew.
.
DISTANCE BETWEEN SKEW LINES:

d1

n
P1
D

n
P2

d2
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MCV4U1-UNIT NINE-LESSON ONE



D  projP1 P2 onto n   P1 P2




n  P1 P2
 n  P1 P2
cos   P1 P2   

n
n P1 P2
 
n  P1P2   
D
 , n  d1  d 2
n
Example: Find the distance between the skew lines:

r  5  a,5  2a,3  4a 

r  3  4k ,10  k ,1  2k 
Let P1  5,5,3 and P2  3,10,1

n  1,2,4   4,1,2 

n  0,18,9  or 0,2,1
P1 P2   2,5,4 
D
D
D
(0,2,1)   2,5,4 
2 2  12
0  10  4
5
14
5

14 5
5
6