Begin2.DVI

or

|A
+B
|^2 =|A
|^2 +2(A
·B
)+ |B
 |^2 ≤| A
|^2 +2 |A
·B
 |+|B
|^2 (6 .19)

Using the Schwarz inequality |A
·B
|≤| A
|| B
|the equation (6.19) can be expressed

|A
+B
 |^2 ≤|A
|^2 +2 |A
|| B
 |+|B
 |^2 = (|A
|+|B
 |)^2 (6 .20)

Taking the square root of both sides of the equation (6.20) gives the triangle inequal-

ity |A
+B
|≤|A
|+|B
|.

The Cross Product or Outer Product

The cross or outer product of two nonzero vectors A
and B
is denoted using the

notation A
×B
and represents the construction of a new vector C
defined as

C
=A
×B
 =|A
||B
|sin θˆen, (6 .21)

where θis the smaller angle between the two nonzero vectors

A
and B
when their origins coincide, and eˆn is a unit vector

perpendicular to the plane containing the vectors A
and B

when their origins are made to coincide. The direction of ˆen

is determined by the right-hand rule. Place the fingers of your

right-hand in the direction of A
and rotate the fingers toward

the vector B
, then the thumb of the right-hand points in the

direction C
.

The vectors A,
B,
C
then form a right-handed system.^2 Note that the cross product

A
×B
is a vector which will always be perpendicular to the vectors A
and B
, whenever

A
and B
are linearly independent.

A special case of the above definition occurs when A
×B
=
0 and in this case one

can state that either θ= 0 , which implies the vectors A
and B
are parallel or A
=
0

or B
=
0.

Use the above definition of a cross product and show that the orthogonal

unit vectors ˆe 1 , eˆ 2 , ˆe 3 satisfy the relations

ˆe 1 ׈e 1 =
 0

ˆe 1 ׈e 2 = ˆe 3

ˆe 1 ׈e 3 =−ˆe 2

ˆe 2 ׈e 1 =−ˆe 3

ˆe 2 ׈e 2 =
 0

ˆe 2 ׈e 3 = ˆe 1

ˆe 3 ׈e 1 = ˆe 2

ˆe 3 ׈e 2 =−ˆe 1

ˆe 3 ׈e 3 =
 0

(6 .22)

(^2) Note many European technical books use left-handed coordinate systems which produces results different from
using a right-handed coordinate system.