cosbab.a

a

bSuppose the angle between two vectors a and b is

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 !

The Vector or x ProductIn C4 the scalar product was covered a . b = |a||b|cosq Pronounced a dot b

Using this definition:i . i = j . j = k . k = 1 as the angle between these unit vectors is zero and cos0 = 1

i . j = 0 j . k = 0 i . k = 0j . i = 0 k . j = 0 k . i = 0

So an answer is obtained when the components of the vector are in the same direction.

cosbab.a

SUMMARY

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 Solution:

224

31

2. .ba

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

Perpendicular Vectors

Then 0cosba

either a = 0 or cosθ=0

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

cosθ = 090θ

The vectors are perpendicular.

b = 0 or

If a.b = 0

the product of 2 vectors divided by

Finding Angles between Vectors

cosbab.a ba

b.acos

The scalar product can be rearranged to find the angle between the vectors.

Notice how careful we must be with the lines under the vectors.

The r.h.s. is

the product of the 2 magnitudes of the vectors

Solution:

)1)(1()1)(1()2)(1(. ba 2

,3111 222 a 6112 222 b

ba

b.acos

e.g. Find the angle between kjia kjib 2

and

63

2cos 961

( 3s.f. )

Tip: If at this stage you get zero, STOP. The vectors are perpendicular.

)3,1,2(,)0,1,1(,)3,1,2( CBA

When solving problems, we have to be careful to use the correct vectors.e.g. The triangle ABC is given by

Solution: ( any shape triangle will do )

We always sketch and label a triangle

A

B

C

ba

b.acos

BUT the a and b of the formula are not the a and b of the question.

We need the vectors and AB CB

Find the cosine of angle ABC.

)3,1,2(,)0,1,1(,)3,1,2( CBA

321

31

2

011

abAB

321

31

2

011

cbCB

�⃗�𝐁=𝐁 𝐬−𝐀 𝐬𝐀𝐁𝐁𝐀𝐑𝐮𝐥𝐞

,14321 222 AB

321

AB

321

CB

14321 222 CB

1414

4cos

321

321

. .CBAB

4941

CBAB

CB.ABcos

7

2cos

14

4cos

2

7

Finding Angles between Lines

e.g.

With lines instead of vectors, we have 2 possible angles. We usually give the acute angle.

a

We use the 2 direction vectors only since these define the angle.

( If the obtuse angle is found, subtract from 180 . )

ba

b.acos where

Solution:

sr23

1

tr

102

and

210

21

1

e.g. Find the acute angle, a, between the lines

sr23

1

tr

102

and

Solution:

ba

b.acos where

a and

210

210

21

1

e.g. Find the acute angle, a, between the lines

b

210

sr23

1

tr

102

and

Solution:

ba

b.acos where

a

21

1

and

21

1

210

e.g. Find the acute angle, a, between the lines

e.g. Find the acute angle, a, between the lines

sr23

1

tr

102

and

Solution:

ba

b.acos where

a

21

1

and

b

21

1

)2)(2()1)(1()1)(0(. ba 5

,521 22 a 6211 222 b

65

5cos

156 24

210

210

(nearest degree)Autograph

The Vector ProductThe vector product is defined as a b = |a| |b|sin is a unit vector perpendicular to both a and b.

To determine the direction of use the right hand rule where finger 1 is a, finger 2 is b and the thumb is a b

a b is out of the paper as the direction of is out

a b is into the paper as the direction of is in

Using this definition:i i = j j = k k = as the angle between these unit vectors is zero and sin0 = 0

ij = jk = ki =

ik = ji = kj =

a b = |a||b|sin

0

k i j

–j –k –i

a b = (a1i + a2j + a3k) (b1i + b2j + b3k)

= (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)k

2 3 3 2

3 1 1 3

1 2 2 1

a b - a b

a b - a b

a b - a b

This result is on pg 4 of the formula book written as a column vector

= a1b2k – a1b3j – a2b1k + a2b3i + a3b1j – a3b2i

a b = |a||b|sin

Link to applet

Ex

2 0

1 1

2 3

1 1 2 3 3 2

2 2 3 1 1 3

3 3 1 2 2 1

a b a b - a b

a b a b - a b

a b a b - a b

Link to demo

5

6 5i 6j 2k

2

1 3 ( 2) 1

( 2) 0 2 3

2 1 1 0

So the vector is perpendicular to 5

6

2

2 0

1 and 1

2 3

This fact will be essential in understanding the equation of a plane

Link to worksheet