what is the Difference between TRIPLE SCALAR PRODUCT, TRIPLE VECTOR PRODUCT ?

 TRIPLE SCALAR PRODUCT, TRIPLE VECTOR PRODUCT

Triple Scalar Product

Sections 1.3 and 1.4 cover the two types of multiplication of interest here. However, there

are combinations of three vectors, A · (B×C) and A×(B×C), that occur with sufficient

frequency to deserve further attention. The combination

A · (B ×C)

is known as the triple scalar product. B × C yields a vector that, dotted into A, gives a

scalar. We note that (A · B)×C represents a scalar crossed into a vector, an operation that

is not defined. Hence, if we agree to exclude this undefined interpretation, the parentheses

may be omitted and the triple scalar product written A · B×C.

Using Eqs. (1.38) for the cross product and Eq. (1.24) for the dot product, we obtain

A · B×C = Ax(ByCz −BzCy )+Ay(BzCx −BxCz)+Az(BxCy − ByCx )

= B · C×A = C · A×B

=−A · C×B=−C · B×A=−B · A×C, and so on. (1.48)

There is a high degree of symmetry in the component expansion. Every term contains the

factors Ai , Bj , and Ck. If i, j , and k are in cyclic order (x, y, z), the sign is positive. If the

order is anticyclic, the sign is negative. Further, the dot and the cross may be interchanged,

A · B×C = A×B · C. (1.49)

26 Chapter 1 Vector Analysis

FIGURE 1.16 Parallelepiped representation of triple scalar product.

A convenient representation of the component expansion of Eq. (1.48) is provided by the

determinant

A · B×C =

Ax Ay Az

Bx By Bz

Cx Cy Cz

. (1.50)

The rules for interchanging rows and columns of a determinant12 provide an immediate

verification of the permutations listed in Eq. (1.48), whereas the symmetry of A, B, and

C in the determinant form suggests the relation given in Eq. (1.49). The triple products

encountered in Section 1.4, which showed that A× B was perpendicular to both A and B,

were special cases of the general result (Eq. (1.48)).

The triple scalar product has a direct geometrical interpretation. The three vectors A, B,

and C may be interpreted as defining a parallelepiped (Fig. 1.16):

|B×C| = BC sin θ

= area of parallelogram base. (1.51)

The direction, of course, is normal to the base. Dotting A into this means multiplying the

base area by the projection of A onto the normal, or base times height. Therefore

A · B×C = volume of parallelepiped defined by A,B, and C.

The triple scalar product finds an interesting and important application in the construction

of a reciprocal crystal lattice. Let a, b, and c (not necessarily mutually perpendicular)

12See Section 3.1 for a summary of the properties of determinants.

1.5 Triple Scalar Product, Triple Vector Product 27

represent the vectors that define a crystal lattice. The displacement from one lattice point

to another may then be written

r = naa+nbb +ncc, (1.52)

with na,nb, and nc taking on integral values. With these vectors we may form

a′ =

b×c

a · b×c

, b′ =

c×a

a · b×c

, c′ =

a×b

a · b×c

. (1.53a)

We see that a′ is perpendicular to the plane containing b and c, and we can readily show

that

a′ · a = b′ · b = c′ · c = 1, (1.53b)

whereas

a′ · b = a′ · c = b′ · a = b′ · c = c′ · a = c′ · b = 0. (1.53c)

It is from Eqs. (1.53b) and (1.53c) that the name reciprocal lattice is associated with the

points r′ = n′aa′ +n′bb′ +n′cc′. The mathematical space in which this reciprocal lattice exists

is sometimes called a Fourier space, on the basis of relations to the Fourier analysis of

Chapters 14 and 15. This reciprocal lattice is useful in problems involving the scattering of

waves from the various planes in a crystal. Further details may be found in R. B. Leighton’s

Principles of Modern Physics, pp. 440–448 [New York: McGraw-Hill (1959)].

Triple Vector Product

The second triple product of interest is A×(B×C), which is a vector. Here the parentheses

must be retained, as may be seen from a special case (ˆx× ˆx)× ˆy = 0, while ˆx×(ˆx× ˆy) =

ˆx× ˆz=−ˆy.

Example 1.5.1 A TRIPLE VECTOR PRODUCT

For the vectors

A = ˆx +2ˆy− ˆz = (1, 2,−1), B = ˆy+ ˆz = (0, 1, 1), C = ˆx − ˆy = (0, 1, 1),

B×C =

ˆx ˆy ˆz

0 1 1

1 −1 0

= ˆx+ ˆy − ˆz,

and

A×(B×C) =

ˆx ˆy ˆz

1 2 −1

1 1 −1

=−ˆx− ˆz=−(ˆy + ˆz)−(ˆx − ˆy)

=−B− C.

By rewriting the result in the last line of Example 1.5.1 as a linear combination of B and

C, we notice that, taking a geometric approach, the triple vector product is perpendicular

28 Chapter 1 Vector Analysis

FIGURE 1.17 B and C are in the xy-plane.

B×C is perpendicular to the xy-plane and

is shown here along the z-axis. Then

A×(B×C) is perpendicular to the z-axis

and therefore is back in the xy-plane.

to A and to B × C. The plane defined by B and C is perpendicular to B × C, and so the

triple product lies in this plane (see Fig. 1.17):

A×(B ×C) = uB+ vC. (1.54)

Taking the scalar product of Eq. (1.54) with A gives zero for the left-hand side, so

uA · B + vA · C = 0. Hence u = wA · C and v = −wA · B for a suitable w. Substituting

these values into Eq. (1.54) gives

A×(B×C) = w

B(A · C)− C(A · B)

; (1.55)

we want to show that

w = 1

in Eq. (1.55), an important relation sometimes known as the BAC–CAB rule. Since

Eq. (1.55) is linear in A, B, and C, w is independent of these magnitudes. That is, we

only need to show that w = 1 for unit vectors Aˆ , Bˆ , Cˆ . Let us denote Bˆ · Cˆ = cosα,

Cˆ · Aˆ = cosβ, Aˆ · Bˆ = cos γ , and square Eq. (1.55) to obtain

Aˆ ×(Bˆ ×Cˆ )

2

= Aˆ 2(Bˆ ×Cˆ )2 −

Aˆ · (Bˆ ×Cˆ )

2

= 1 −cos2 α −

Aˆ · (Bˆ ×Cˆ )

2

= w2

(Aˆ · Cˆ )2 +(Aˆ · Bˆ )2 − 2(Aˆ · Bˆ )(Aˆ · Cˆ )(Bˆ · Cˆ )

= w2

cos2 β +cos2 γ −2 cosα cosβ cos γ

, (1.56)

1.5 Triple Scalar Product, Triple Vector Product 29

using (Aˆ × Bˆ )2 = Aˆ 2 ˆB2 − (Aˆ · Bˆ )2 repeatedly (see Eq. (1.43) for a proof). Consequently,

the (squared) volume spanned by Aˆ , Bˆ , Cˆ that occurs in Eq. (1.56) can be written as

Aˆ · (Bˆ ×Cˆ )

2

= 1− cos2 α − w2

cos2 β + cos2 γ − 2 cosα cosβ cos γ

.

Here w2 = 1, since this volume is symmetric in α,β, γ . That is, w =±1 and is independent

of Aˆ , Bˆ , Cˆ . Using again the special case xˆ × (xˆ × yˆ) = −yˆ in Eq. (1.55) finally

gives w = 1. (An alternate derivation using the Levi-Civita symbol εij k of Chapter 2 is the

topic of Exercise 2.9.8.)

It might be noted here that just as vectors are independent of the coordinates, so a vector

equation is independent of the particular coordinate system. The coordinate system only

determines the components. If the vector equation can be established in Cartesian coordinates,

it is established and valid in any of the coordinate systems to be introduced in

Chapter 2. Thus, Eq. (1.55) may be verified by a direct though not very elegant method of

expanding into Cartesian components (see Exercise 1.5.2).

Exercises

1.5.1 One vertex of a glass parallelepiped is at the origin (Fig. 1.18). The three adjacent

vertices are at (3, 0, 0), (0, 0, 2), and (0, 3, 1). All lengths are in centimeters. Calculate

the number of cubic centimeters of glass in the parallelepiped using the triple scalar

product.

1.5.2 Verify the expansion of the triple vector product

A×(B ×C) = B(A · C)−C(A · B)

FIGURE 1.18 Parallelepiped: triple scalar product.

30 Chapter 1 Vector Analysis

by direct expansion in Cartesian coordinates.

1.5.3 Show that the first step in Eq. (1.43), which is

(A×B) · (A×B) = A2B2 −(A · B)2,

is consistent with the BAC–CAB rule for a triple vector product.

1.5.4 You are given the three vectors A, B, and C,

A = ˆx+ ˆy,

B = ˆy+ ˆz,

C = ˆx− ˆz.

(a) Compute the triple scalar product, A · B × C. Noting that A = B + C, give a geometric

interpretation of your result for the triple scalar product.

(b) Compute A×(B×C).

1.5.5 The orbital angular momentum L of a particle is given by L = r × p = mr × v, where

p is the linear momentum. With linear and angular velocity related by v = ω ×r, show

that

L = mr2

ω − ˆr(ˆr · ω)

.

Here ˆr is a unit vector in the r-direction. For r · ω = 0 this reduces to L = Iω, with the

moment of inertia I given by mr2. In Section 3.5 this result is generalized to form an

inertia tensor.

1.5.6 The kinetic energy of a single particle is given by T = 1

2mv2. For rotational motion this

becomes 1

2m(ω ×r)2. Show that

T =

1

2

m

r2ω2 −(r · ω)2

.

For r · ω = 0 this reduces to T = 1

2Iω2, with the moment of inertia I given by mr2.

1.5.7 Show that13

a×(b×c)+b×(c×a)+c×(a×b) = 0.

1.5.8 A vector A is decomposed into a radial vector Ar and a tangential vector At. If ˆr is a

unit vector in the radial direction, show that

(a) Ar = ˆr(A · ˆr) and

(b) At =−ˆr×(ˆr×A).

1.5.9 Prove that a necessary and sufficient condition for the three (nonvanishing) vectors A,

B, and C to be coplanar is the vanishing of the triple scalar product

A · B×C = 0.

13This is Jacobi’s identity for vector products; for commutators it is important in the context of Lie algebras (see Eq. (4.16) in

Section 4.2).

1.5 Triple Scalar Product, Triple Vector Product 31

1.5.10 Three vectors A, B, and C are given by

A = 3ˆx− 2ˆy+2ˆz,

B = 6ˆx+ 4ˆy−2ˆz,

C=−3ˆx−2ˆy −4ˆz.

Compute the values of A · B×C and A×(B×C),C×(A×B) and B×(C×A).

1.5.11 Vector D is a linear combination of three noncoplanar (and nonorthogonal) vectors:

D = aA+bB+cC.

Show that the coefficients are given by a ratio of triple scalar products,

a =

D · B×C

A · B×C

, and so on.

1.5.12 Show that

(A×B) · (C×D) = (A · C)(B · D)− (A · D)(B · C).

1.5.13 Show that

(A×B)×(C×D) = (A · B×D)C− (A · B×C)D.

1.5.14 For a spherical triangle such as pictured in Fig. 1.14 show that

sinA

sinBC =

sinB

sinCA =

sinC

sinAB

.

Here sinA is the sine of the included angle at A, while BC is the side opposite (in

radians).

1.5.15 Given

a′ =

b×c

a · b×c

, b′ =

c×a

a · b×c

, c′ =

a×b

a · b×c

,

and a · b×c 

= 0, show that

(a) x · y′ = δxy, (x, y = a,b, c),

(b) a′ · b′ ×c′ = (a · b×c)−1,

(c) a =

b′ ×c′

a′ · b′ ×c′

.

1.5.16 If x · y′ = δxy, (x, y = a,b, c), prove that

a′ =

b×c

a · b×c

.

(This is the converse of Problem 1.5.15.)

1.5.17 Show that any vector V may be expressed in terms of the reciprocal vectors a′, b′, c′ (of

Problem 1.5.15) by

V = (V · a)a′ +(V · b)b′ +(V · c)c′.

32 Chapter 1 Vector Analysis

1.5.18 An electric charge q1 moving with velocity v1 produces a magnetic induction B given

by

B =

μ0

4π

q1

v1 × ˆr

r2 (mks units),

where ˆr points from q1 to the point at which B is measured (Biot and Savart law).

(a) Show that the magnetic force on a second charge q2, velocity v2, is given by the

triple vector product

F2 =

μ0

4π

q1q2

r2 v2 ×(v1 × ˆr).

(b) Write out the corresponding magnetic force F1 that q2 exerts on q1. Define your

unit radial vector. How do F1 and F2 compare?

(c) Calculate F1 and F2 for the case of q1 and q2 moving along parallel trajectories

side by side.

ANS.

(b) F1 =−

μ0

4π

q1q2

r2 v1 ×(v2 × ˆr).

In general, there is no simple relation between

F1 and F2. Specifically, Newton’s third law, F1 =−F2,

does not hold.

(c) F1 =

μ0

4π

q1q2

r2 v2 ˆr=−F2.

Mutual attraction.