Dot Products
There are two ways to multiply two vectors
•The dot product produces a scalar quantity
•It has no direction
•It can be pretty easily computed from geometry
•It can be easily computed from components
v  w  vw cos   vx wx  v y wy  vz wz
w

•The dot product of two unit vectors is easy to memorize
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  1
ˆi  ˆj  ˆj  ˆi  0
ˆi  kˆ  kˆ  ˆi  0
ˆj  kˆ  kˆ  ˆj  0
•The dot product is commutative
vw  w v
v
Cross Products
The cross product produces a vector quantity
•It is perpendicular to both vectors
•Requires the right-hand rule
•Its magnitude can be easily computed from
geometry
•It is a bit of a pain to compute from components
v  w  vw sin 
 ˆi

v  w  det  vx
w
 x
ˆj
vy
wy
vw
w

v
kˆ 

vz    v y wz  vz wy  ˆi   vz wx  vx wz  ˆj
wz 
  vx wy  v y wx  kˆ
Determinants
•Finding the cross product requires that you memorize the
formula, or know how to compute determinants
Computing a 33 determinant:
1. Multiply on the diagonal down-right
2. Add the other two down-rights, wrapping as needed
3. Subtract the diagonal down-left
4. Subtract the other two down-lefts, wrapping as needed
5. Simplify
1

det  4
7

2
5
8
3

6   1 5  9 2  6  7 3  4  8 3  5  7 2  4  9 1 6  8
9 
 45  84  96 105  72  48  0
Simple Rules for Cross-Products
•Vectors that are parallel or anti-parallel have zero cross product
ab  ac  bc  0
•Cross products are anti-symmetric
b
a
c
v  w  w  v
Basis vectors:
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
•Any vector with itself gives zero
•Think of ijk as a circle: any two in order gives the third
î
•Any two in reverse order gives minus the third
ˆi  ˆj  kˆ
ˆj  kˆ  ˆi
ˆj  ˆi  kˆ
kˆ  ˆj   ˆi
kˆ  ˆi  ˆj
ˆi  kˆ   ˆj
k̂
ĵ
Sample Nasty Problem
An electron moving in the xy-plane at a speed of 4.00 m/s at an angle
of 37 below the x-axis enters a region where the magnetic field is 312
mT in the xz-plane and pointed at a 60 angle above the x-axis. What
is the acceleration of the electron?
v   3.19, 2.41, 0 m/s
y
Bsin60
x
vsin37
vcos37
37
B
z
B   0.156, 0, 0.270  T
v
 ˆi

v  B  det  3.19
 0.156

60
Bcos60
ˆj
kˆ 

2.41
0  T  m/s
0
0.270 
x
Sample Nasty Problem (cont.)
 ˆi

v  B  det  3.19
 0.156

ˆj
kˆ 

2.41
0 
0
0.270 
 ˆi  2.41 0.270   kˆ  2.41 0.156   ˆj  3.19  0.270 
 .651ˆi  0.861ˆj  0.375kˆ
F  q  v  B
F
a
m
q
a   v  B
m
1.602 1019 C
a
 .651, .861,.375 T  m/s
31
9.109 10 kg
N m m
11 C
a  1.145,1.514, .660  10
kg A  m s s 2