Transcript The Vector Equation of a Line

“Teach A Level Maths”
Vol. 2: A2 Core Modules
51: The Vector Equation of
a Line
© Christine Crisp
The Vector Equation of a Line

The vector AB and the vector equation of the line
through AB are very different things.
xA
A
x

The vector AB
xB
The line through
A and B.
xB

The vector AB has a definite length ( magnitude ).
The line AB is a line passing through the points A
and B and has infinite length.
The Vector Equation of a Line
Finding the Equation of a Line
In coordinate geometry, the equation of a line is
y  mx  c
e.g.
y  2 x  3
The equation gives the value (coordinate) of y for
any point which lies on the line.
The vector equation of a line must give us the
position vector of any point on the line.
We start with fixing a line in space.
We can do this by fixing 2 points, A and B. There
is only one line passing through these points.
The Vector Equation of a Line
A and B are fixed points.
We consider several
more points on the line.
We need an equation
for r, the position
vector of any point R on
x
the line.
O
Starting with R1:
R3
x
xA
x R1
a
x
r1
B
x
r1  a 
1
2

AB
R2
The Vector Equation of a Line
A and B are fixed points.
We consider several
more points on the line.
We need an equation
for r, the position
vector of any point R on
x
the line.
O
Starting with R1:
R3
x
xA
x R1
a
r1  a 
x
1
2

AB

r 2  a  2 AB
B
r2
x
R2
The Vector Equation of a Line
R3
A and B are fixed points.
We consider several
more points on the line.
x
r3
We need an equation
for r, the position
vector of any point R on
x
the line.
O
Starting with R1:
xA
x R1
a
x
B
x
r1  a 
1
2

AB

r 2  a  2 AB
r3  a 

( 14 ) AB
R2
The Vector Equation of a Line
So for R1, R2 and R3
r1  a 
1
2
R3

x
AB

x R1
r 2  a  2 AB
r 3  a  (
1
)
4
xA
a

AB
x
x
B
x
O
For any position of R, we have

r  a  t AB
t is called a parameter and can have any real value.
It is a scalar not a vector.
R2
The Vector Equation of a Line

R3
r  a  t AB
x

xA
x R1
We can substitute for AB
a
x
r  a  t (b  a )
x
B
b
O
Instead of using a here . . . we could use b.
The value of t would then be different to get to
any particular point.
We say the equation is not unique.
x
R2
The Vector Equation of a Line

R3
r  a  t AB
x

p
xA
x R1
We can substitute for AB
a
x
r  a  t (b  a )
x

B
b
O
x
R2
Also, instead of AB we can equally well use any vector
p which is parallel to AB.
If p is not the same length as AB, t will have a
different value for any particular R.
The Vector Equation of a Line
xA
p
x R1
a
x
O
e.g.
r1  a 
1
2
or
r1  a 
1
3

AB
p
r1
x
B
The Vector Equation of a Line
SUMMARY
 The vector equation of the line through 2 fixed
points A and B is given by

r  a  t AB

r  a  t (b  a )
 The vector equation of the line through 1 fixed
point A and parallel to the vector p is given by
r  a  tp
position vector . . .
of a known point
on the line
direction vector . . .
of the line
The Vector Equation of a Line
e.g. 1 Find the equation of the line passing through
the points A and B with position vectors a
and b where
2
a   1
 
2
and
  3
b 4 
 
 1 
Solution: r  a  t ( b  a )
  3  2 
 5
direction  b  a   4     1   5 
   
 
1
2
   
  1
 2 
  5
So, r    1  t  5 
 
 
 2 
  1
The Vector Equation of a Line
Diagram
The Vector Equation of a Line
In this example we had
2
a   1
 
2
giving
We can replace r
with  x
 y
 
z
  3
and b   4 
 
 1 
2
  5
r    1  t  5 
 
 
2
  1
and/or replace
a with b
 x    3
  5
e.g.  y    4   t  5 
   
 
z  1 
  1
The Vector Equation of a Line
So,
2
  5
r    1  t  5  is the same
 
  line as
2
  1
 x    3
  5
 y   4   t  5 
   
 
z  1 
  1
However, the value of t for any particular point has
now changed.
Which point is given by t = 2 in
the 1st version?
 8
ANS:  9 
 
 0
What value of t in the 2nd
version gives the same point?
ANS: t = 1
The Vector Equation of a Line
e.g. 2 Find the equation of the line passing through
the point A (  1 , 3 , 4 ) , parallel to the vector
Solution:
1
b6
 
 1
r = a + t (b-a)=position vector+t(direction v ector)
  1
1
 x    1
1
So, r   3   t  6  or  y    3   t  6 
 
 
   
 
4
  1
z  4 
  1
Demo
The Vector Equation of a Line
 7 
e.g. 3 Show that C with position vector c    6
 
2
  5  3 
lies on the line r    1  t  5 
 
 
2
 
  1
Solution: If C lies on the line, there is a value of
t that makes r = c.
So,
 7   2
 5
  6    1  t  5 
   
 
3
2
   
  1
The Vector Equation of a Line
We have
 7  2 
 -5 
 -6   -1   t  5 
   
 
 3 2 
 -1 
The top row of the vectors gives the x-components,
so,
The Vector Equation of a Line
We have
 7  2 
 -5 
 -6   -1   t  5 
   
 
 3 2 
 -1 
The top row of the vectors gives the x-components,
so,
7  2  t(5)

5t  2  7 
t  1
However, for the point to lie on the line, this value
of t must also give the y- and z- components.
The Vector Equation of a Line
We have
 7  2 
 -5 
 -6   -1   t  5 
   
 
 3 2 
 -1 
The top row of the vectors gives the x-components,
so,

7  2  t(5)
5t  2  7 
t  1
However, for the point to lie on the line, this value
of t must also give the y- and z- components.
y:
The Vector Equation of a Line
We have
 7  2 
 -5 
 -6   -1   t  5 
   
 
 3 2 
 -1 
The top row of the vectors gives the x-components,
so,

7  2  t(5)
5t  2  7 
t  1
However, for the point to lie on the line, this value
of t must also give the y- and z- components.
y:  1  5 t
  1  5( 1)
The Vector Equation of a Line
We have
 7  2 
 -5 
 -6   -1   t  5 
   
 
 3 2 
 -1 
The top row of the vectors gives the x-components,
so,

7  2  t(5)
5t  2  7 
t  1
However, for the point to lie on the line, this value
of t must also give the y- and z- components.
y:  1  5 t
  1  5( 1)
 6
Noticez:how we write this. We
mustn’t start with  1  5 t   6
as we are trying to show that
this is true.
The Vector Equation of a Line
We have
 7  2 
 -5 
 -6   -1   t  5 
   
 
 3 2 
 -1 
The top row of the vectors gives the x-components,
so,

7  2  t(5)
5t  2  7 
t  1
However, for the point to lie on the line, this value
of t must also give the y- and z- components.
y:  1  5 t
  1  5( 1)
 6
z: 2  t
 2  ( 1)
The Vector Equation of a Line
We have
 7  2 
 -5 
 -6   -1   t  5 
   
 
 3 2 
 -1 
The top row of the vectors gives the x-components,
so,

7  2  t(5)
5t  2  7 
t  1
However, for the point to lie on the line, this value
of t must also give the y- and z- components.
y:  1  5 t
  1  5( 1)
 6
z: 2  t
 2  ( 1)
 3
The Vector Equation of a Line
We have
 7  2 
 -5 
 -6   -1   t  5 
   
 
 3 2 
 -1 
The top row of the vectors gives the x-components,
so,

7  2  t(5)
5t  2  7 
t  1
However, for the point to lie on the line, this value
of t must also give the y- and z- components.
y:  1  5 t
  1  5( 1)
 6
z: 2  t
 2  ( 1)
 3
Since all 3 equations are satisfied, C lies on the line.
The Vector Equation of a Line
Exercise
1. Find a vector equation for the line AB for
each of the following:
(a) A ( 2 ,  3 ,  1 ) , B (  2 , 3 ,  3 )
  3
(b) a   1  ,
 
 4 
  1
b    2
 
 0 
1
 2
(c) a  0 and AB is parallel to the vector 2
 
 
3
1 
2. Does the point ( 0 , 0 , 2 ) lie on the line AB in 1(a)?
Solutions
The Vector Equation of a Line
(a) A ( 2 ,  3 ,  1 ) ,
B (2, 3,  3)
2
  4
 r    3  t  6 
r  a  t (b  a )
 
 
  1
  2
  3
  1
  3
 2 
(b) a   1  , b    2  r   1   t   3 
 
 




 4 
 0 
 4 
  4
 2
1
 2
1 
(c) a  0 parallel to 2  r   0   t  2 
 
 




1 
3
1 
 3
Solutions
The Vector Equation of a Line
2. Does the point ( 0 , 0 , 2 ) lie on the line AB in 1(a)?
2
  4
From 1(a), r    3  t  6 
 
 
  1
  2
x:
 4t  2  t 
0  2  4t
y:
z:  1  2t
 3  6t
 3  612 
 1  212 
 1  1
 3  3
 2  2
0
The point does not lie on AB.
1
2
The Cartesian Equation of a Line
Stop here for C4
Carry on for Further Maths
The Cartesian Equation of a Line
The Cartesian Form of the Equation of a Line
The equation y  mx  c is the Cartesian equation
of a line but only if it lies in the x-y plane.
The more general form can be easily found from the
vector form.
The Cartesian form does not contain a parameter.
The Cartesian Equation of a Line
e.g.
 x    3
 2
 y    1   t  3
   
 
 z    4
4
We can extract the 3 components from this equation:
x  3  2t
y  1  3t
z  4  4t
To eliminate the parameter, t, we rearrange to find t :
x3
t
2
So,
z4
y 1
t
t
4
3
x3
z4
y 1

(t )

2
4
3
This 3 part equation is the Cartesian equation.
The Cartesian Equation of a Line
e.g. Find the Cartesian equation of the line
Solution:
  3
r 2
 
 1 
x  3
y  2t
z  1  2t
0
 t   1
 
2
      (1)
      (2)
      (3)
(1) doesn’t contain t so just gives x = 3
z 1
y2
( 3)  t 
( 2)  t 
2
1
y  2 z 1
So the line is x  3 ;

1
2
The Cartesian Equation of a Line
Exercise
1. Write the following lines in Cartesian form:
2
  4
(a) r    3  t  6 
 
 
  1
  2
Answers: (a)
(b)
1
  2
(b) r   1   t  0 
 
 
  2
 3 
y3
x2
z1


4
2
6
x 1
z2

; y1
2
3
The Vector Equation of a Line
The Vector Equation of a Line
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.
The Vector Equation of a Line
SUMMARY
 The vector equation of the line through 2 fixed
points A and B is given by

r  a  t AB

r  a  t (b  a )
 The vector equation of the line through 1 fixed
point A and parallel to the vector p is given by
r  a  tp
position vector
of a known point
on the line
direction vector
of the line
The Vector Equation of a Line
e.g. 1 Find the equation of the line passing through
the points A and B with position vectors a
and b where
2
a    1
 
2
Solution:
and
  3
b4
 
1
r  a  tp
p  b  a

  3  2 
 5
p   4     1   5 
   
 
1
2
   
  1
2
  5
So, r    1  t  5 
 
 
2
 
  1
The Vector Equation of a Line
In this example we had
2
a    1
 
2
giving
We can replace r
with  x
 y
 
z
  3
and b   4 
 
1
2
  5
r    1  t  5 
 
 
2
  1
and/or replace
a with b
 x    3
  5
e.g.  y    4   t  5 
   
 
z  1 
  1
The Vector Equation of a Line
So,
2
  5
r    1  t  5  is the same
 
  line as
2
  1
 x    3
  5
 y   4   t  5 
   
 
z  1 
  1
However, the value of t for any particular point has
now changed.
 8
In the 1st version t = 2 gives  9  .
 
0
The same point is given by t =1 in the 2nd version.
The Vector Equation of a Line
e.g. 2 Find the equation of the line passing through
the point A (  1 , 3 , 4 ) , parallel to the vector
1
b6
 
  1
Solution:
r  a  tp
p  b
  1
1
 x    1
1
So, r   3   t  6  or  y    3   t  6 
 
 
   
 
4

1
z
4
 
 
   
  1
The Vector Equation of a Line
7
e.g. 3 Show that C with position vector c    6
 
3
2
  5
lies on the line r    1  t  5 
 
 
2
 
  1
Solution: If C lies on the line, there is a value of
t that makes r = c.
So,
 7  2
  5
  6     1  t  5 
   
 
3
2
   
  1
The Vector Equation of a Line
The top row of the vectors gives the x-components,
so,
7  2  t(5)
5t  2  7 
t  1
However, for the point to lie on the line, this value
of t must also give the y- and z- components.
y:  1  5 t
  1  5( 1)
 6
z: 2  t
 2  ( 1)
 3
Since all 3 equations are satisfied, C lies on the line.
The Vector Equation of a Line
OCR/MEI only
The Cartesian Equation of a Line
e.g.
 x    3
 2
 y    1   t  3
   
 
 z    4
4
We can extract the 3 components from this equation:
x  3  2t
y  1  3t
z  4  4t
To eliminate the parameter, t, we rearrange to find t :
x3
t
2
So,
z4
y 1
t
t
4
3
x3
z4
y 1

(t )

2
4
3
This 3 part equation is the Cartesian equation.
The Vector Equation of a Line
We can generalise the result by comparing the 2
forms:
 x   3
2
 y    1   t 3 and
   
 

4
z
   
4
y1
x3
z4


2
4
3
The denominators of the Cartesian form are the
elements of the direction vector.
The numerators of the Cartesian form are
x  a1 ,
y  a2 ,
z  a3
where the position vector of the point on the line is
a1
a2
a 
 3
The Vector Equation of a Line
SUMMARY
The Cartesian equation of a line is given by
x  a1 y  a 2 z  a 3


p2
p1
p3
p1 , p 2 , p 3  0
where the position vector of the point on the line is
a1
a2
a 
 3
 p1
and the direction vector is  p2
 p
 3
The Vector Equation of a Line
It is easy to make a mistake when writing the
Cartesian equation of a line.
Also, it isn’t obvious what to do if an element of
zero.
p is
For both these reasons you may prefer to rearrange
the vector equation to find the parameter rather then
quote the formula.
The Vector Equation of a Line
e.g. Find the Cartesian equation of the line
Solution:
  3
r 2
 
 1 
x  3
y  2t
z  1  2t
0
 t   1
 
2
      (1)
      (2)
      (3)
(1) doesn’t contain t so just gives x = 3
z 1
y2
( 3)  t 
( 2)  t 
2
1
y  2 z 1
So the line is x  3 ;

1
2