53: The Scalar Product of Two Vectors © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core...

53: The Scalar Product 53: The Scalar Product of Two Vectorsof Two Vectors

© Christine Crisp

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

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

The Scalar Product of Two Vectors

Module C4

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In this presentation we are going to look at one of these methods: the scalar product.

Vectors have direction as well as magnitude so it seems strange that we can multiply them together.There are in fact 2 ways of multiplying vectors.


a

bSuppose the angle between two vectors a and b is .

is defined as the angle which is between the vectors when both point towards, or both away from, the point of intersection so, this angle . . .

a

bis NOT . We need to reposition b


a


is defined as the angle which is between the vectors when both point towards, or both away from, the point of intersection so, this angle . . .

a

bis NOT . We need to reposition b

b

then we see that is an obtuse angle.


cos. abba

a


The scalar product is written as a . b and is defined as

The dot must NEVER be missed out.

The scalar product is sometimes called the “dot” product.

The result of the scalar product is a scalar quantity not a vector !


Notice that the scalar product uses the magnitudes, a and b, of the vectors as well as the angle between them, so, we get a different answer for:

30cos)10)(8(cos ab

• different size vectors at the same angle,

)2(69 s.f. 30cos)10)(6(cos ab

)2(52 s.f.

cos. abba

8

10

30

630

10


150cos)10)(6(cos ab

• the same size vectors at different angles.

We also have different answers for

30cos)10)(6(cos ab

)2(52 s.f. )2(52 s.f.

630

10

6 150

10

The scalar product was defined so that the answer is unique to any 2 vectors. It will

enable us easily to find the angle between the vectors, even in 3 dimensions.


Perpendicular Vectors

If 0cos ab

either a = 0 or 0cos

a = 0 or b = 0 are trivial cases as they mean the vector doesn’t exist. So, we must have

0cos 90

The vectors are perpendicular.

b = 0 or

The Scalar Product of Two VectorsSUMMAR

Y

cos. abba


is defined as the angle which is between the vectors when both point towards, or both away from, the point of intersection.

If and0cos ab ,00 ba and

0cos then and the vectors are perpendicular.


Exercise1. Find the scalar product of the following

pairs of vectors.(a)

660

4(b)

5

3

2. Find the value of a . a for any non-zero vector a.

a

b

a

b (c)

120a

b4

4

3. Find the value of (a) i . i and (b) i . k


1.

(a)

660

4(b)

5

3

a

b

a

b (c)

120a

b4

4

12

0. ba

60cos)6)(4(cos ab(a)

0cos90 (b)

(c) 60 ( not )120 860cos)4)(4(. ba

Solutions:


2. Find the value of a . a for any non-zero vector a.

22 0cos. aaaa

Solution: The angle between a vector and itself is 0

(a) i . i = 1 ( since the magnitude of i is one )

90(b) i . k = 0 ( since )

3. i and k are the unit vectors ( magnitude 1 ) along the x- and z- axes respectively, so

The Scalar Product of Two VectorsThe Scalar Product of 2 Column

Vectors

Then to form , there are quantities to multiply

933 ba .

However, six of these are perpendicular.e.g. is along the x-axis and is along the y-axis.

2 4

a

b

Supposeand

2 34-1

-3 -1


Vectors

Supposeand

Then to form , there are quantities to multiply

933 ba .


2 4The scalar product of these components is zero.

a

band

2 34-1

-3 -1


Vectors

Supposeand

Then to form , there are quantities to multiply.

933 ba .

The other three multiplications e.g 2 and 3 . . .



a

band

2 34-1

-3 -1


Vectors

Supposeand

Then to form , there are quantities to multiply.

933 ba .

The other three multiplications e.g 2 and 3 . . .



involve parallel components so the angle between them is zero.

a

band

2 34-1

-3 -1


Vectors

Supposeand

a

band

2 34-1

-3 -1

Since ,10cos


Vectors

Supposeand

Since ,10cos

a

band

2 34-1

-3 -1

)3)(2(ba .


Vectors

Supposeand

10cos

a

band

2 34-1

-3 -1

)4)(1(Since ,10cos

)3)(2(ba .


Vectors

Supposeand

10cos

a

band

2 34-1

-3 -1

)4)(1( )1)(3( Since ,10cos

)3)(2(ba .


Vectors

Supposeand

10cos

a

band

2 34-1

-3 -1

)4)(1( )1)(3( Since ,10cos

)3)(2(ba .

346

5


Y For the scalar product of 2 column

vectors,

3

2

1

aaa

a

3

2

1

bbb

be.g. and

we multiply the “tops”,and add the results. So,

332211. babababa

“middles”

and “bottoms”


e.g.1 Find the scalar product of the vectors

31

2a

224

band

332211. babababa Solutio

n:

224

31

2. .ba

)2)(3()2)(1()4)(2( 12628


332211. babababa Solution:

143

11

1. .ba

143

e.g.2 Show that the vectors

kjia kjib 43

andare perpendicular.

00cos0. abba

0,0 ba 0cos 90


Exercise1. Find the scalar product of the following

pairs of vectors.

(a)

14

3a

222

band

(b)

23

1a

31

3band

What can you say about the vectors in part (b) ?


(b)

31

3

23

1. .ba


332211. babababa

)2)(1()2)(4()2)(3(. ba 4

)3)(2()1)(3()3)(1( 0

Solutions:

222

14

3. .ba

(a)


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.


cos. abba

a



The dot must NEVER be missed out.

The scalar product is sometimes called the “dot” product.

The result of the scalar product is a scalar quantity not a vector !


Perpendicular Vectors

If 0cos ab

either a = 0 or 0cos

a = 0 or b = 0 are trivial cases as they mean the vector doesn’t exist. So, we must have

0cos 90


b = 0 or


Y

cos. abba


is defined as the angle which is between the vectors when both point towards, or both away from, the point of intersection.

If and0cos ab ,00 ba and

0cos then and the vectors are perpendicular.


Y For the scalar product of 2 column

vectors,

3

2

1

aaa

a

3

2

1

bbb

be.g. and

we multiply the “tops”,and add the results. So,

332211. babababa

“middles”

and “bottoms”


e.g.1 Find the scalar product of the vectors

31

2a

224

band

332211. babababa Solutio

n:

224

31

2. .ba

)2)(3()2)(1()4)(2( 12628


332211. babababa Solution:

143

11

1. .ba

143

e.g.2 Show that the vectors

kjia kjib 43

andare perpendicular.

00cos0. abba

0,0 ba 0cos 90

53: The Scalar Product of Two Vectors © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core...

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