13 B• Lines in 2D and

3D

The vector AB and the vector equation of the line 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 ).

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Vector Equation of a Line

Finding the Equation of a LineIn coordinate geometry, the equation of a line is

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.

y mx b

Vector Equation of a Line

In Other Words:

We can determine the vector equation of a line using its direction and any fixed point on the line.

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

Vector Equation of a Line

x

B

A

x

Ox

B

AWe consider several more points on the line.

x

1R

2R

3R

x

x

xa

r2

A and B are fixed points.

Starting with R1:

2r

1r aAB2

1

a AB2

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

Vector Equation of a Line

x

B

A

x

Ox

B

AWe consider several more points on the line.

Ox

1R

2R

3R

x

x

xa

r3

A and B are fixed points.

Starting with R1:

2r

1r aAB2

1

a AB2

3r a AB)( 41

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

Vector Equation of a Line

x

B

A

x

Ox

B

A

x

1R

2R

3R

x

x

xa

O

So for R1, R2 and R3

For any position of R, we have

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

Vector Equation of a Line

1

2

3

1

2

2

1

4

r a AB

r a AB

r a AB

��������������

��������������

��������������

r a ABt ��������������

x

B

A

x

Ox

B

A

x 1Rx

x

xa

O

This is what I’m Saying:

t is called a parameter and can have any real value.

It is a scalar not a vector.

Fixed Point

Fixed Point

Any Poin

t

b

Say you have a line that passes through a fixed point A with position vector a, and that line is parallel to vector b. Point R is on the line

OR r��������������

r

By vector addition:OR OA AR ������������������������������������������

Vector Equation of a Line

r a ABt ��������������

x

B

A

x

Ox

B

A

x 1Rx

x

xa

O

Fixed Point

Fixed Point

Any Poin

t

b

Since is parallel to b,

r

AR tb�������������� AR��������������

r a tb

Vector Equation of a Line

direction vector . . .

SUMMARY

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

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

r a t AB r a tb ��������������

Vector Equation of a Line

In 2-D, we are dealing with a line in a plane

A

(a1 , a2

)

R

(x , y)

1 1

2 2

a bxt

a by

The vector equation of the line

1

2

bb

b

Is the direction vector

1

2

bb

b

Slope or gradient

is:2

1

bm

b

We can write the parametric equations of the line:

1 1

2 2

a b tx

a b ty

1 1 2 2&x a b t y a b t Each point on the line corresponds

to exactly one value of t.Convert into Cartesian form by equating t

values.1 2

1 2

x a y at

b b

We get the Cartesian equation of

the line as: 2 1 2 1 1 2b x b y b a b a

A line passes through the point A(1,5) and has the direction vector . Describe the line using:3

2

Draw a sketch

A

(1 , 5)

R

(x , y)

3

2b

O

ra

1 3&

5 2a OA b

��������������

A) A vector equation

1 3

5 2

xt

y

A line passes through the point A(1,5) and has the direction vector . Describe the line using:3

2

A

(1 , 5)

R

(x , y)

3

2b

O

ra

B) Parametric Equations

1 3 & 5 2x t y t

1 1

2 2

a b tx

a b ty

A line passes through the point A(1,5) and has the direction vector . Describe the line using:3

2

A

(1 , 5)

R

(x , y)

3

2b

O

ra

C) Cartesian Equation

1 2

1 2

x a y at

b b

1 5

3 2

x yt

2 1 2 1 1 2b x b y b a b a 2 2 3 15

2 3 13

x y

x y

In 3-D, we are dealing with a line in Space

A

(a 1,a2,a3

)

R

(x,y,z)

1 1

2 2

3 3

a bx

a bty

a bz

Is the vector equation of the line

1

2

3

b

bb

b

Is the direction vector

1

2

3

b

bb

b

Slope or gradient NONE. We just describe its direction by its direction vector

We can write the parametric equations of the line:

1 1

2 2

3 3

a b tx

a b ty

a b tz

Each point on the line corresponds to exactly one value of t.

Find a vector equation and the corresponding parametric equations of the line through (1,-2, 3) in the direction of 4i + 5j – 6k

The parametric equations are:

1 4

2 5

3 6

x

ty

z

The vector equation is: r = a + tb

1 1

2 2

3 3

a b tx

a b ty

a b tz

1 4

2 5

3 6

x t

y t

z t

Non-Uniqueness of a line

Consider the line passing through (5,4) and (7,3). When writing the equation of the line, we could use either point to give the position vector a

(5,4)(7,3)

We could also use theDirection vector

2

1

2

1

We could also use the Direction vector

2

1

We could also use ANY Direction vector that is a non-zero scalar multiple of these vectors

Non-Uniqueness of a line

So, we could write the equation of the line as:

(5,4)(7,3)

2

1

5 2

4 1x t

7 2

3 1x s

We could keep going.

Non-Uniqueness of a lineSo, we could write the equation of the line as:

(5,4)(7,3)2

1

5 2

4 1x t

7 2

3 1x s

Non-Uniqueness of a lineFind a parametric equations of the line through A(2,-1,4) and B(-1,0,2)

We need a direction vector for the line, either

AB or BA����������������������������

1 2 3

0 1 1

2 4 2

AB

��������������Using point A, the equations are: 2 3

1

4 2

x t

y t

z t

Using point B, the equations are: 1 3

2 2

x s

y s

z s

Homework Page 325 (1 –

8)