Begin2.DVI

Solution Use the determinant form for the cross product and express the triple scalar

product as a determinant as follows.

A
·(B
 ×C
)=(A 1 ˆe 1 +A 2 ˆe 2 +A 3 ˆe 3 )·

∣∣

∣∣

∣∣

ˆe 1 eˆ 2 ˆe 3

B 1 B 2 B 3

C 1 C 2 C 3

∣∣

∣∣

∣∣

A
·(B
 ×C
)=(A 1 ˆe 1 +A 2 ˆe 2 +A 3 ˆe 3 )·[(B 2 C 3 −B 3 C 2 )ˆe 1 −(B 1 C 3 −B 3 C 1 )ˆe 2 +(B 1 C 2 −B 2 C 1 )ˆe 3 ]

A
·(B
 ×C
)= A 1 (B 2 C 3 −B 3 C 2 )−A 2 (B 1 C 3 −B 3 C 1 )+ A 3 (B 1 C 2 −B 2 C 1 )

A
·(B
 ×C
)= A 1

∣∣

∣∣B^2 B^3

C 2 C 3

∣∣

∣∣−A 2

∣∣

∣∣B^1 B^3

C 1 C 3

∣∣

∣∣+A 3

∣∣

∣∣B^1 B^2

C 1 C 2

∣∣

∣∣=

∣∣

∣∣

∣∣

A 1 A 2 A 3

B 1 B 2 B 3

C 1 C 2 C 3

∣∣

∣∣

∣∣

Determinants have the property^4 that the interchange of two rows of a determinant

changes its sign. One can then show

∣∣

∣∣

∣∣

A 1 A 2 A 3

B 1 B 2 B 3

C 1 C 2 C 3

∣∣

∣∣

∣∣=

∣∣

∣∣

∣∣

B 1 B 2 B 3

C 1 C 2 C 3

A 1 A 2 A 3

∣∣

∣∣

∣∣=

∣∣

∣∣

∣∣

C 1 C 2 C 3

A 1 A 2 A 3

B 1 B 2 B 2

∣∣

∣∣

∣∣

or

A
·(B
×C
)= B
·(C
×A
)= C
·(A
×B
)

Example6-8. For nonzero vectors A,
B,
C
show that the triple vector product

satisfies A
×(B
×C
)
=(A
×B
)×C

That is, the triple vector product is not associative and the order of execution of

the cross product is important.

Solution Let B
×C
=D
denote the vector

perpendicular to the plane determined by the

vectors B
and C
. The vector A
×D
=E
is a

vector perpendicular to the plane determined by

the vectors A
and D
and therefore must lie in

the plane of the vectors B
and C
. One can then

say the vectors B,
C
and A
×(B
×C
)are coplanar

and consequently there must exist scalars αand

βsuch that

A
×(B
×C
)= αB
+βC
 (6 .35)

(^4) See chapter 10 for properties of determinants.