“Teach A Level Maths” Vol. 2: A2 Core Modules 52: The Intersection of Lines © Christine Crisp.
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Slide 1
“Teach A Level Maths”
Vol. 2: A2 Core Modules
52: The Intersection of
Lines
© Christine Crisp
Slide 2
The Intersection of Lines
In 2 dimensions, 2 lines are either parallel . . .
or they intersect
Slide 3
The Intersection of Lines
However, in 3 dimensions the lines may be skew ,
they neither intersect nor are parallel.
e.g.
y
E
O
A
x
C
z
If you find it hard to visualise this, look at the
room you are in . . . Imagine a line on the wall on
your right from the bottom corner in front of you
to the top corner behind you . . . Now imagine a
line diagonally across the floor from the left-hand
corner in front of you. These lines are skew.
Slide 4
The Intersection of Lines
Diagram
Slide 5
The Intersection of Lines
e.g. Determine whether the lines given below intersect.
If they do, find the coordinates of the point of
intersection.
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
6
0
3
3
( Notice that different parameters are used. )
Solution:
We notice first that the lines aren’t parallel.
The direction vector of the 2nd is not a multiple
of the direction vector of the 1st.
Slide 6
The Intersection of Lines
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
6
0
3
3
If the lines intersect, there is a set of values for x, y
and z that satisfy both equations.
With the left-hand sides of the equations equal, the
right-hand sides will also be equal. This gives three
equations, one for each component.
Slide 7
The Intersection of Lines
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
6
0
3
3
x:
1t1 s
(1 )
y:
t2
(2)
z:
6 3 3s
(3)
There are 3 equations but we only need 2 of them
( the easiest ) to solve for the 2 unknowns.
(2) gives t = 2 and (3) gives s = 1
Check in (1): l.h.s. 1 t
r.h.s. 1 s
1 21
11 0
So, the 3rd equation ( number (1) ) is not satisfied.
The equation are said to be inconsistent.
The lines do not intersect.
Slide 8
The Intersection of Lines
Autograph tab 3
Slide 9
The Intersection of Lines
If the equation of the 3rd line is changed:
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
3
z
z
6
0
3
3
x:
1t1 s
(1 )
y:
t2
(2)
z:
We now get
(3)
3 6 3 3s
(2) gives t = 2 and (3) now gives s = 0
Check in (1): l.h.s. 1 t
r.h.s. 1 s
1 21
10 1
All 3 equations are now satisfied so the lines intersect.
The point of intersection is found by substituting for s
or t. s 0 in 2 nd line Point is ( 1 , 2 , 3 )
( t 2 in the 1 st line gives the sam e. )
Slide 10
The Intersection of Lines
Autograph tab 3
Slide 11
SUMMARY
The Intersection of Lines
Since a point of intersection (x, y, z) would lie on
both lines:
• Equate the right-hand sides of the equations
of the lines.
•
•
Write down the 3 component equations.
Solve any 2 equations. Try to pick the easiest.
( Never use all 3. )
• Check whether the values of s and t satisfy
the unused equation. If not, the lines are
skew.
If all equations are satisfied, substitute into
either of the lines to find the position vector or
coordinates of the point of intersection.
Slide 12
The Intersection of Lines
Exercise
1. Determine whether the following pairs of lines
intersect. If they do, find the coordinates
of the point of intersection.
2
0
x
1
2
(a) x
y 3 t 1 and y 6 s 2
z
1
2
z
1
2
(b)
r 2i 2 j k t ( i j 2k )
r i 6 j 23 k s ( 2 i 3 j 2 k )
(c) The line AB and the line CD where
A( 0, 2, 3 ) ,
B ( 1, 1, 0 ) ,
C ( 1, 5 , 3 ) ,
D ( 4, 4, 6 )
Slide 13
The Intersection of Lines
Solutions:
1(a) The lines are not parallel.
2
0 1
2
For intersection, 3 t 1 6 s 2
1
2
1
2
x:
2 1 2 s
(1 )
y:
3 t 6 2s
(2)
z:
1 2t 1 2 s
(3)
I’ll use (1) and (2): (1) s 3
2
Subs. in (2): t 0
Check in (3): l.h.s. 1,
r.h.s. 2
The equations are inconsistent. The lines are skew.
Slide 14
The Intersection of Lines
Diagram
Slide 15
The Intersection of Lines
(b)
r 2i 2 j k t ( i j 2k )
r i 6 j 23 k s ( 2 i 3 j 2 k )
Solution: The lines are not parallel.
For intersection, x :
2 t 1 2 s (1 )
y:
2 t 6 3s
(2)
z : 1 2 t 23 2 s ( 3 )
(1 ) ( 2 ) : 0 5 s s 5
2 t 6 15 t 7
Subs. in (2):
Check in (3): l.h.s. = 13,
Lines meet.
r.h.s. = 13
Subs. for s in the 2nd line or t in the 1st line:
point of int. is ( 9, 9, 13 )
Slide 16
The Intersection of Lines
Diagram
Slide 17
The Intersection of Lines
(c) The line AB and the line CD where
A( 0, 2, 3 ) ,
Solution:
B ( 1, 1, 0 ) , C ( 1, 5 , 3 ) ,
D ( 4, 4, 6 )
r a t p
1
For AB,
p AB 3
3
For CD,
3
p CD 9
9
CD 3 AB
The lines are parallel.
However, before we assume they don’t intersect, we
should check whether they are collinear.
Slide 18
The Intersection of Lines
We could have
either
Ax
or
Bx
x
x
C
no points of
intersection
D
Bx
A
Cx
Dx
x
an infinite number of
points of intersection
We can check by seeing whether C, or D, lies on AB.
r a t p
x
0
1
y 2 t 3
equation of AB is
z
3
3
C ( 1 , 5 , 3 ) does not lie on this line.
The lines don’t intersect.
Slide 19
The Intersection of Lines
Slide 20
The Intersection of Lines
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Slide 21
SUMMARY
Since a point of intersection (x, y, z) would lie on
both lines:
• Equate the right-hand sides of the equations
of the lines.
•
Write down the 3 component equations.
•
Solve any 2 equations. Try to pick the easiest.
( Never use all 3. )
Check whether the values of s and t satisfy
the unused equation. If not, the lines are
skew.
•
If all equations are satisfied, substitute into one
of the lines to find the position vector or
coordinates of the point of intersection.
Slide 22
e.g. Determine whether the lines given below intersect.
If they do, find the coordinates of the point of
intersection.
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
1
0
3
3
( Notice that different parameters are used. )
Solution:
We notice first that the lines aren’t parallel.
The direction vector of the 2nd is not a multiple
of the direction vector of the 1st.
Slide 23
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
6
0
3
3
x:
1t1 s
(1 )
y:
t2
(2)
z:
6 3 3s
(3)
There are 3 equations but we only need 2 of them
( the easiest ) to solve for the 2 unknowns.
(2) gives t = 2 and (3) gives s = 1
Check in (1): l.h.s. 1 t
r.h.s. 1 s
1 21
11 0
So, the 3rd equation ( number (1) ) is not satisfied.
The equation are said to be inconsistent.
The lines do not intersect.
Slide 24
If the equation of the 1st line is changed:
x
1
1
x
y 0 t 1 and y
3
z
6
0
z
x:
1t1 s
y:
t2
z:
We now get
3 6 3 3s
(2) gives t = 2 and (3) now gives s =
1
1
2 s 0
3
3
(1 )
(2)
(3)
0
Check in (1): l.h.s. 1 t
r.h.s. 1 s
1 21
10 1
All 3 equations are now satisfied so the lines intersect.
The point of intersection is found by substituting for s
or t. s 0 in 2 nd line Point is ( 1 , 2 , 3 )
( t 2 in the 1 st line gives the sam e. )
“Teach A Level Maths”
Vol. 2: A2 Core Modules
52: The Intersection of
Lines
© Christine Crisp
Slide 2
The Intersection of Lines
In 2 dimensions, 2 lines are either parallel . . .
or they intersect
Slide 3
The Intersection of Lines
However, in 3 dimensions the lines may be skew ,
they neither intersect nor are parallel.
e.g.
y
E
O
A
x
C
z
If you find it hard to visualise this, look at the
room you are in . . . Imagine a line on the wall on
your right from the bottom corner in front of you
to the top corner behind you . . . Now imagine a
line diagonally across the floor from the left-hand
corner in front of you. These lines are skew.
Slide 4
The Intersection of Lines
Diagram
Slide 5
The Intersection of Lines
e.g. Determine whether the lines given below intersect.
If they do, find the coordinates of the point of
intersection.
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
6
0
3
3
( Notice that different parameters are used. )
Solution:
We notice first that the lines aren’t parallel.
The direction vector of the 2nd is not a multiple
of the direction vector of the 1st.
Slide 6
The Intersection of Lines
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
6
0
3
3
If the lines intersect, there is a set of values for x, y
and z that satisfy both equations.
With the left-hand sides of the equations equal, the
right-hand sides will also be equal. This gives three
equations, one for each component.
Slide 7
The Intersection of Lines
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
6
0
3
3
x:
1t1 s
(1 )
y:
t2
(2)
z:
6 3 3s
(3)
There are 3 equations but we only need 2 of them
( the easiest ) to solve for the 2 unknowns.
(2) gives t = 2 and (3) gives s = 1
Check in (1): l.h.s. 1 t
r.h.s. 1 s
1 21
11 0
So, the 3rd equation ( number (1) ) is not satisfied.
The equation are said to be inconsistent.
The lines do not intersect.
Slide 8
The Intersection of Lines
Autograph tab 3
Slide 9
The Intersection of Lines
If the equation of the 3rd line is changed:
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
3
z
z
6
0
3
3
x:
1t1 s
(1 )
y:
t2
(2)
z:
We now get
(3)
3 6 3 3s
(2) gives t = 2 and (3) now gives s = 0
Check in (1): l.h.s. 1 t
r.h.s. 1 s
1 21
10 1
All 3 equations are now satisfied so the lines intersect.
The point of intersection is found by substituting for s
or t. s 0 in 2 nd line Point is ( 1 , 2 , 3 )
( t 2 in the 1 st line gives the sam e. )
Slide 10
The Intersection of Lines
Autograph tab 3
Slide 11
SUMMARY
The Intersection of Lines
Since a point of intersection (x, y, z) would lie on
both lines:
• Equate the right-hand sides of the equations
of the lines.
•
•
Write down the 3 component equations.
Solve any 2 equations. Try to pick the easiest.
( Never use all 3. )
• Check whether the values of s and t satisfy
the unused equation. If not, the lines are
skew.
If all equations are satisfied, substitute into
either of the lines to find the position vector or
coordinates of the point of intersection.
Slide 12
The Intersection of Lines
Exercise
1. Determine whether the following pairs of lines
intersect. If they do, find the coordinates
of the point of intersection.
2
0
x
1
2
(a) x
y 3 t 1 and y 6 s 2
z
1
2
z
1
2
(b)
r 2i 2 j k t ( i j 2k )
r i 6 j 23 k s ( 2 i 3 j 2 k )
(c) The line AB and the line CD where
A( 0, 2, 3 ) ,
B ( 1, 1, 0 ) ,
C ( 1, 5 , 3 ) ,
D ( 4, 4, 6 )
Slide 13
The Intersection of Lines
Solutions:
1(a) The lines are not parallel.
2
0 1
2
For intersection, 3 t 1 6 s 2
1
2
1
2
x:
2 1 2 s
(1 )
y:
3 t 6 2s
(2)
z:
1 2t 1 2 s
(3)
I’ll use (1) and (2): (1) s 3
2
Subs. in (2): t 0
Check in (3): l.h.s. 1,
r.h.s. 2
The equations are inconsistent. The lines are skew.
Slide 14
The Intersection of Lines
Diagram
Slide 15
The Intersection of Lines
(b)
r 2i 2 j k t ( i j 2k )
r i 6 j 23 k s ( 2 i 3 j 2 k )
Solution: The lines are not parallel.
For intersection, x :
2 t 1 2 s (1 )
y:
2 t 6 3s
(2)
z : 1 2 t 23 2 s ( 3 )
(1 ) ( 2 ) : 0 5 s s 5
2 t 6 15 t 7
Subs. in (2):
Check in (3): l.h.s. = 13,
Lines meet.
r.h.s. = 13
Subs. for s in the 2nd line or t in the 1st line:
point of int. is ( 9, 9, 13 )
Slide 16
The Intersection of Lines
Diagram
Slide 17
The Intersection of Lines
(c) The line AB and the line CD where
A( 0, 2, 3 ) ,
Solution:
B ( 1, 1, 0 ) , C ( 1, 5 , 3 ) ,
D ( 4, 4, 6 )
r a t p
1
For AB,
p AB 3
3
For CD,
3
p CD 9
9
CD 3 AB
The lines are parallel.
However, before we assume they don’t intersect, we
should check whether they are collinear.
Slide 18
The Intersection of Lines
We could have
either
Ax
or
Bx
x
x
C
no points of
intersection
D
Bx
A
Cx
Dx
x
an infinite number of
points of intersection
We can check by seeing whether C, or D, lies on AB.
r a t p
x
0
1
y 2 t 3
equation of AB is
z
3
3
C ( 1 , 5 , 3 ) does not lie on this line.
The lines don’t intersect.
Slide 19
The Intersection of Lines
Slide 20
The Intersection of Lines
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Slide 21
SUMMARY
Since a point of intersection (x, y, z) would lie on
both lines:
• Equate the right-hand sides of the equations
of the lines.
•
Write down the 3 component equations.
•
Solve any 2 equations. Try to pick the easiest.
( Never use all 3. )
Check whether the values of s and t satisfy
the unused equation. If not, the lines are
skew.
•
If all equations are satisfied, substitute into one
of the lines to find the position vector or
coordinates of the point of intersection.
Slide 22
e.g. Determine whether the lines given below intersect.
If they do, find the coordinates of the point of
intersection.
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
1
0
3
3
( Notice that different parameters are used. )
Solution:
We notice first that the lines aren’t parallel.
The direction vector of the 2nd is not a multiple
of the direction vector of the 1st.
Slide 23
x
1
1
x
1
1
y 0 t 1 and y 2 s 0
z
z
6
0
3
3
x:
1t1 s
(1 )
y:
t2
(2)
z:
6 3 3s
(3)
There are 3 equations but we only need 2 of them
( the easiest ) to solve for the 2 unknowns.
(2) gives t = 2 and (3) gives s = 1
Check in (1): l.h.s. 1 t
r.h.s. 1 s
1 21
11 0
So, the 3rd equation ( number (1) ) is not satisfied.
The equation are said to be inconsistent.
The lines do not intersect.
Slide 24
If the equation of the 1st line is changed:
x
1
1
x
y 0 t 1 and y
3
z
6
0
z
x:
1t1 s
y:
t2
z:
We now get
3 6 3 3s
(2) gives t = 2 and (3) now gives s =
1
1
2 s 0
3
3
(1 )
(2)
(3)
0
Check in (1): l.h.s. 1 t
r.h.s. 1 s
1 21
10 1
All 3 equations are now satisfied so the lines intersect.
The point of intersection is found by substituting for s
or t. s 0 in 2 nd line Point is ( 1 , 2 , 3 )
( t 2 in the 1 st line gives the sam e. )