### Vector product

```The Vector or x Product
In C4 the scalar product was covered
Pronounced a dot b
a . b = |a||b|cos
Suppose the angle between two
vectors a and b is 
b

a
The scalar product is written as a . b and is defined as
a . b  a  b cos 
The dot must NEVER be missed out.
The result of the scalar product is a scalar quantity not a vector !
The scalar product is sometimes called the “dot” product.
a . b  a  b cos 
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.i=0
j.k=0
k.j=0
i.k=0
k.i=0
So an answer is obtained when the components of the vector
are in the same direction.
SUMMARY
 For the scalar product of 2 column vectors,
e.g.
a1 
a  a 2 
 
a 3 
we multiply the “tops”,
and
 b1 
b  b2 
 
b3 
“middles” and “bottoms”
a . b  a1 b1  a 2 b2  a 3 b3
e.g.1 Find the scalar product of the vectors
 2 
a    1


 3
and
 4 
b   2 



2


a . b  a1 b1  a 2 b2  a 3 b3
Solution:

 2   4 
a .b    1 .  2 

 

 3  2
 ( 2 )( 4 )  (  1 )( 2 )  (  3 )(  2 )
 8  2  6  12
Perpendicular Vectors
If a.b = 0
a b cos   0
Then
either a = 0 or
b = 0 or
cosθ= 0
a = 0 or b = 0 are trivial cases as they mean the vector
doesn’t exist. So, we must have
co sθ = 0

θ  90
The vectors are perpendicular.
Finding Angles between Vectors
The scalar product can be rearranged to find the angle
between the vectors. 
a . b  a b cos  
a .b
cos  
ab
Notice how careful we must be with the lines under the vectors.
The r.h.s. is
the product of 2 vectors divided by
the product of the 2 magnitudes of the vectors
e.g. Find the angle between
b  2i  j  k
Solution:
a  i  j  k and
a .b
cos  
ab
a . b  ( 1 )( 2 )  (  1 )( 1 )  (  1 )(  1 )  2
Tip: If at
2 this2 stage
2 you get zero, STOP. 2
2
2
a The
 vectors
1  1 are
 1perpendicular.
 3 , b  2 1 1 

cos  
2
3 6

  61  9
( 3s.f. )

6
When solving problems, we have to be careful to use the correct
vectors.
e.g. The triangle ABC is given by
A ( 2 ,  1, 3 ) ,
B (1 , 1 , 0 ) ,
C ( 2 ,  1,  3 )
Find the
cosine of angle ABC.
Solution: We always sketch and label a triangle
B
( any shape triangle will do )

A
a .b
cos  
ab
C BUT the a and b of the formula are not
the a and b of the question.
We need the vectors
and
 
AB CB
A ( 2 ,  1, 3 ) ,
B (1 , 1 , 0 ) ,
= ` − `
C ( 2 ,  1,  3 )

1 
 2 
  1

AB  b  a   1     1    2 
 




0
 3 
 3
1 
 2 
  1

CB  b  c   1     1    2 
 




0
 3
 3 
 1
  
AB 
2


 3
  1
  
CB  2


 3 
  1    1
     
 1  4  9  4
AB . CB 
2 . 2

 

 3  3 
AB 
12  22  32 
14 , CB 
 
AB . CB
cos  
 cos  
AB  CB
 cos   
12  2 2  3 2 
4
14
2
7
14
2
4
 cos  
14
7
14
Finding Angles between Lines
e.g.
a
With lines instead of vectors, we have 2 possible angles. We
usually give the acute angle.
( If the obtuse angle is found, subtract from 180 . )
We use the 2 direction vectors only since these define the angle.
e.g. Find the acute angle, a, between the lines
 1 
 0
r    3   s  1


 
 2 
 2
Solution:
a .b
cos  
ab
and
where
2
 1 
r   0   t  1
 
 
1 
 2
e.g. Find the acute angle, a, between the lines
 1 
 0
r    3   s  1


 
 2 
 2
Solution:
a .b
cos  
ab
and
where
2
 1 
r   0   t  1
 
 
1 
 2
 0
a   1 and
 
 2
e.g. Find the acute angle, a, between the lines
 1 
 0
r    3   s  1


 
 2 
 2
Solution:
a .b
cos  
ab
and
where
2
 1
r   0   t  1
 
 
1 
 2
 0
 1
a   1 and b    1
 
 
 2
 2
e.g. Find the acute angle, a, between the lines
 1 
 0
r    3   s  1


 
 2 
 2
Solution:
and
2
 1 
r   0   t  1
 
 
1 
 2
 0
 1 
a .b
cos  
where a   1 and b    1
 
 
ab
 2
 2
a . b  ( 0 )( 1 )  ( 1 )(  1 )  ( 2 )(  2 )   5
a 

1
2
 2
cos  
2

5
5 6
5,

b 
1
2
  156
1

2

(nearest degree)
 2
2

a  24
Autograph
6

The Vector Product
The 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
a  b = |a||b|sin
Using this definition:
ii=jj=kk= 0
as the angle between these unit vectors is zero and sin0 = 0
ij = k
jk = i
ki = j
ik = –j
ji = –k
kj = –i
a  b = |a||b|sin
a  b = (a1i + a2j + a3k)  (b1i + b2j + b3k)
= a1b2k – a1b3j – a2b1k + a2b3i + a3b1j – a3b2i
= (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)k
 a 2b 3 - a 3b 2 


 a 3 b 1 - a 1b 3


ab - a b 
2 1 
 1 2
This result is on pg 4 of the formula book written as a column vector
Ex
 a1 


a2 


a 
 3
 2 


1 


 2 


 b 1   a 2b 3 - a 3b 2 
  

b 2  a 3 b 1 - a 1b 3
  

b   a b - a b 
2 1 
 3  1 2
0
 1  3  (  2)  1 
 


1  (  2)  0  2  3 
 


3
 2  1 1 0 
 


So the vector
 5 


6


 2 


 5 


 6  5i  6 j  2k


 2 


is perpendicular to
 2 


1


 2 


and
0
 
1
 
3
 
This fact will be essential in understanding the equation of a
plane