Begin2.DVI

Properties of the Cross Product

A
×B
=−B
×A
(noncommutative)

A
×(B
+C
) = A
×B
+A
×C
(distributive law)

m(A
×B
) = (mA
)×B
=A
×(mB
) ma scalar

A
×A
=
0 since A
is parallel to itself.

Let A
=A 1 ˆe 1 +A 2 eˆ 2 +A 3 ˆe 3 and B
=B 1 ˆe 1 +B 2 ˆe 2 +B 3 ˆe 3 be two nonzero vectors

in component form and form the cross product A
×B
to obtain

A
×B
 = (A 1 ˆe 1 +A 2 ˆe 2 +A 3 ˆe 3 )×(B 1 ˆe 1 +B 2 ˆe 2 +B 3 ˆe 3 ). (6 .23)

The cross product can be expanded by using the distributive law to obtain

A
×B
=A 1 B 1 eˆ 1 ×eˆ 1 +A 1 B 2 ˆe 1 ׈e 2 +A 1 B 3 ˆe 1 ׈e 3

+A 2 B 1 eˆ 2 ׈e 1 +A 2 B 2 ˆe 2 ׈e 2 +A 2 B 3 ˆe 2 ׈e 3

+A 3 B 1 eˆ 3 ׈e 1 +A 3 B 2 ˆe 3 ׈e 2 +A 3 B 3 ˆe 3 ׈e 3.

(6 .24)

Simplification by using the previous results from equation (6.22) produces the im-

portant cross product formula

A
×B
 = (A 2 B 3 −A 3 B 2 )ˆe 1 + (A 3 B 1 −A 1 B 3 )ˆe 2 + (A 1 B 2 −A 2 B 1 )ˆe 3 , (6 .25)

This result that can be expressed in the determinant form^3

A
×B
 =

∣∣

∣∣

∣∣

eˆ 1 ˆe 2 ˆe 3

A 1 A 2 A 3

B 1 B 2 B 3

∣∣

∣∣

∣∣=

∣∣

∣∣A^2 A^3

B 2 B 3

∣∣

∣∣ˆe 1 −

∣∣

∣∣A^1 A^3

B 1 B 3

∣∣

∣∣ˆe 2 +

∣∣

∣∣A^1 A^2

B 1 B 2

∣∣

∣∣ˆe 3. (6 .26)

In summary, the cross product of two vectors A
and B
is a new vector C ,
where

C
=A
×B
 =C 1 ˆe 1 +C 2 eˆ 2 +C 3 ˆe 3 =

∣∣

∣∣

∣∣

ˆe 1 ˆe 2 ˆe 3

A 1 A 2 A 3

B 1 B 2 B 3

∣∣

∣∣

∣∣

with components

C 1 =A 2 B 3 −A 3 B 2 , C 2 =A 3 B 1 −A 1 B 3 , C 3 =A 1 B 2 −A 2 B 1 (6 .27)

(^3) For more information on determinants see chapter 10.