51: The Vector Equation of a Line © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core...

51: The Vector 51: The Vector Equation of a LineEquation of a Line

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

The Vector Equation of a Line

The vector AB and the vector equation of the line through AB are very different things.

A

B

x

xB

Ax

x

The line AB is a line passing through the points A and B and has infinite length.

The vector AB

The line through A and B.

The vector AB has a definite length ( magnitude ).

The Vector Equation of a LineFinding the Equation of a LineIn coordinate geometry, the equation of a line is cmxy e.g. 32 xy

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.


x

B

A

x

Ox

We consider several more points on the line.

2Rx

3Rx

1Rx

A and B are fixed points.

We need an equation for r, the position vector of any point R on the line.

r1a

Starting with R1: 1r aAB2

1


x

B

A

x

Ox

B

AWe consider several more points on the line.

x

1R

2R

3R

x

x

xa

r2


Starting with R1:

2r

1r aAB2

1

a AB2



x

B

A

x

Ox

B

AWe consider several more points on the line.

Ox

1R

2R

3R

x

x

xa

r3


Starting with R1:

2r

1r aAB2

1

a AB2

3r a AB)( 41



x

B

A

x

Ox

B

A

x

1R

2R

3R

x

x

xa

O

So for R1, R2 and R3

2r

1r a

21

a 2

3r a )( 41

AB

AB

AB

For any position of R, we have

r a

t ABt is called a parameter and can have any real value.It is a scalar not a vector.


x

B

A

x

Ox

B

A

x

1R

2R

3R

x

x

xa

b

r a

t AB

We can substitute for AB

r a )( abt

O

The value of t would then be different to get to any particular point.

Instead of using a here . . .

we could use b.

We say the equation is not unique.


x

B

A

x

Ox

B

A

x

1R

2R

3R

x

x

xa

b

r a

t AB

We can substitute for AB

r a )( abt

O

If is not the same length as AB, t will have

a different value for any particular R.

p

Also, instead of we can equally well use any vector which is parallel to AB.

AB

p

p


x

B

A

x

Ox

B

A

x

1Rxa

O

p

e.g. 1r a

21 AB

or 1r a p31

r1


direction vector . . .

SUMMARY

The vector equation of the line through 2 fixed points A and B is given by

)( abtarA Btar

The vector equation of the line through 1 fixed point A and parallel to the vector is given by

p

ptar

position vector . . .of a known point on the line

of the line


)ab(tar

e.g. 1 Find the equation of the line passing through the points A and B with position vectors a and b where

21

2a

143

band

Solution:

abdirection

21

2

143

155

So,

155

21

2tr


Diagram


155

21

2trWe can replace

r

zyx

with

In this example we had

21

2a

143

band

giving

and/or replace a

with b

e.g.

155

143

tzyx


155

21

2tr

However, the value of t for any particular point has now changed.

So,

155

143

tzyxis the

same line as

Which point is given by t = 2 in the 1st version?

ANS:

098

What value of t in the 2nd version gives the same point?

ANS: t = 1


r = a + t (b-a)=position vector+t(d irection v ector)

Solution:

So,

1

61

431

tr

e.g. 2 Find the equation of the line passing through the point , parallel to the vector

161

b

)4,3,1( A

or

1

61

431

tzyx

Demo


Solution: If C lies on the line, there is a value of t that makes r = c.

155

21

2tr

e.g. 3 Show that C with position vector

36

7c

lies on the line

So,

15

21

367 2 5

t

The Vector Equation of a LineWe have

The top row of the vectors gives the x-components, so,

t

7-63

2-12

-55-1



1 t

t

However, for the point to lie on the line, this value of t must also give the y- and z- components.

7-63

2-12

-55-1

)5(27 t

We have

725 t


The top row of the vectors gives the x-components, so, )5(27 t

725 t 1 t

t


y:

7-63

2-12

-55-1

We have



725 t 1 t

t


y: t51

7-63

2-12

-55-1

)1(51

We have



725 t 1 t

t


y: t51

7-63

2-12

-55-1

)1(51 6

z:Notice how we write this. We mustn’t start withas we are trying to show that this is true.

651 t

We have



725 t 1 t

t


z:

)1(2

7-63

2-12

-55-1

y: t51 )1(51

t2

6

We have



725 t 1 t

t


z:

)1(2

7-63

2-12

-55-1

y: t51 )1(51

t2

6 3

We have



725 t 1 t

t


z:

)1(2

7-63

2-12

-55-1

y: t51 )1(51

t2

6 3Since all 3 equations are satisfied, C lies on the line.

We have

The Vector Equation of a LineExercise1. Find a vector equation for the line AB

for each of the following: (a)

,)1,3,2( A )3,3,2( B

(b)

,

413

a

021

b

(c)

102

aand AB is parallel to the

vector

321

2. Does the point lie on the line AB in 1(a)?

)2,0,0(


2

6

4

1

3

2

tr

Solutions (a)

,)1,3,2( A )3,3,2( B

(c)

102

a parallel to

321

)ab(tar

43

2

413

tr

321

102

tr

(b)

,

413

a

021

b

The Vector Equation of a LineSolutions2. Does the point lie on the line AB in

1(a)? )2,0,0(

2

6

4

1

3

2

trFrom

1(a),

x: t420 24 t21 t

y: t63

33 0

z: t21

211

The point does not lie on AB.

2

2163

2121

The Cartesian Equation of a Line

Stop here for C4Carry on for Further Maths


The Cartesian Form of the Equation of a Line

The more general form can be easily found from the vector form.The Cartesian form does not contain a parameter.

The equation is the Cartesian equation of a line but only if it lies in the x-y plane.

cmxy


432

413

tzyxe.g.

We can extract the 3 components from this equation:

tx 23 ty 31 tz 44 To eliminate the parameter, t, we rearrange to find t :

tx

2

3t

y

3

1t

z

4

4

So, 2

3x

3

1y)(

4

4t

z

This 3 part equation is the Cartesian equation.


e.g. Find the Cartesian equation of the line

21

0

123

tr

Solution:

3x )1(ty 2 )2(tz 21 )3(

1

2)2(

y

t2

1)3(

zt

(1) doesn’t contain t so just gives x = 3

So the line is 2

1

1

2;3

zy

x

The Cartesian Equation of a LineExercise1. Write the following lines in Cartesian

form:

(a)

(b)

2

6

4

1

3

2

tr

302

11

2

tr

(a)

4 6 22x 3y 1z

(b) ;

2 3

1x 2z 1y

Answers:


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.


direction vector

SUMMARY

The vector equation of the line through 2 fixed points A and B is given by

)( abtarA Btar

The vector equation of the line through 1 fixed point A and parallel to the vector is given by

p

ptar

position vectorof a known point on the line

of the line


ptar

e.g. 1 Find the equation of the line passing through the points A and B with position vectors a and b where

2

1

2

a

1

4

3

band

Solution:

pabp

2

1

2

1

4

3

1

5

5

So,

1

5

5

2

1

2

tr


1

5

5

2

1

2

trWe can replace r

zyx

with

In this example we had

2

1

2

a

1

4

3

band

giving

and/or replace a

with b

e.g.

1

5

5

143

tzyx


1

5

5

2

1

2

tr

However, the value of t for any particular point has now changed.

So,

1

5

5

143

tzyxis the

same line as

In the 1st version t = 2 gives .

0

9

8

The same point is given by t =1 in the 2nd version.


ptar

e.g. 2 Find the equation of the line passing through the point , parallel to the vector

1

6

1

b

Solution:

bp

So,

1

61

431

tr

)4,3,1( A

or

1

61

431

tzyx


Solution: If C lies on the line, there is a value of t that makes r = c.

1

5

5

2

1

2

tr

e.g. 3 Show that C with position vector

3

6

7

c

lies on the line

So,

1

5

2

1

3

67 2 5

t


)5(27 t725 t 1 t


z:

)1(2 y: t51

)1(51 t2

6 3Since all 3 equations are satisfied, C lies on the line.



432

413

tzyxe.g.

We can extract the 3 components from this equation:

tx 23 ty 31 tz 44 To eliminate the parameter, t, we rearrange to find t :

tx

2

3t

y

3

1t

z

4

4

So, 2

3x

3

1y)(

4

4t

z

This 3 part equation is the Cartesian equation.


OCR/MEI only


are the elements of the direction vector.

We can generalise the result by comparing the 2 forms:

and

The denominators of the Cartesian form

The numerators of the Cartesian form are

321 ,, azayax

t

zyx

432

2 3 4413

3x 1y 4z

where the position vector of the point on the line is

3

2

1

aaa


The Cartesian equation of a line is given by

SUMMARY

1p 2p 3p

1ax 2ay 3az

where the position vector of the point on the line is

3

2

1

aaa

and the direction vector is

3

2

1

ppp

0,, 321 ppp


It is easy to make a mistake when writing the Cartesian equation of a line.

For both these reasons you may prefer to rearrange the vector equation to find the parameter rather then quote the formula.

Also, it isn’t obvious what to do if an element of is zero.

p


e.g. Find the Cartesian equation of the line

21

0

123

tr

Solution:

3x )1(ty 2 )2(tz 21 )3(

1

2)2(

y

t2

1)3(

zt

(1) doesn’t contain t so just gives x = 3

So the line is 2

1

1

2;3

zy

x

51: The Vector Equation of a Line © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core...

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Transcript of 51: The Vector Equation of a Line © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core...