What is VECTOR AND CROSS PRODUCT ?

 VECTOR OR CROSS PRODUCT

A second form of vector multiplication employs the sine of the included angle instead

of the cosine. For instance, the angular momentum of a body shown at the point of the

distance vector in Fig. 1.12 is defined as

angular momentum = radius arm×linear momentum

= distance×linear momentum×sin θ.

For convenience in treating problems relating to quantities such as angular momentum,

torque, and angular velocity, we define the vector product, or cross product, as

C = A×B, with C = AB sin θ. (1.35)

Unlike the preceding case of the scalar product, C is now a vector, and we assign it a

direction perpendicular to the plane of A and B such that A,B, and C form a right-handed

system. With this choice of direction we have

A×B=−B×A, anticommutation. (1.36a)

From this definition of cross product we have

ˆx× ˆx = ˆy× ˆy = ˆz× ˆz = 0, (1.36b)

whereas

ˆx× ˆy = ˆz, ˆy× ˆz = ˆx, ˆz× ˆx = ˆy,

ˆy× ˆx=−ˆz, ˆz× ˆy=−ˆx, ˆx× ˆz=−ˆy.

(1.36c)

Among the examples of the cross product in mathematical physics are the relation between

linear momentum p and angular momentum L, with L defined as

L = r×p,

FIGURE 1.12 Angular momentum.

1.4 Vector or Cross Product 19

FIGURE 1.13 Parallelogram representation of the vector product.

and the relation between linear velocity v and angular velocity ω,

v = ω ×r.

Vectors v and p describe properties of the particle or physical system. However, the position

vector r is determined by the choice of the origin of the coordinates. This means that

ω and L depend on the choice of the origin.

The familiar magnetic induction B is usually defined by the vector product force equation8

FM = qv×B (mks units).

Here v is the velocity of the electric charge q and FM is the resulting force on the moving

charge.

The cross product has an important geometrical interpretation, which we shall use in

subsequent sections. In the parallelogram defined by A and B (Fig. 1.13), B sin θ is the

height if A is taken as the length of the base. Then |A × B| = AB sin θ is the area of the

parallelogram. As a vector, A×B is the area of the parallelogram defined by A and B, with

the area vector normal to the plane of the parallelogram. This suggests that area (with its

orientation in space) may be treated as a vector quantity.

An alternate definition of the vector product can be derived from the special case of the

coordinate unit vectors in Eqs. (1.36c) in conjunction with the linearity of the cross product

in both vector arguments, in analogy with Eqs. (1.23) for the dot product,

A×(B +C) = A×B+A×C, (1.37a)

(A+ B)×C = A×C+ B×C, (1.37b)

A×(yB) = yA×B = (yA)×B, (1.37c)

8The electric field E is assumed here to be zero.

20 Chapter 1 Vector Analysis

where y is a number again. Using the decomposition of A and B into their Cartesian components

according to Eq. (1.5), we find

A×B ≡ C = (Cx,Cy,Cz) = (Ax ˆx +Ay ˆy +Az ˆz)×(Bx ˆx +By ˆy+Bz ˆz)

= (AxBy −AyBx )ˆx× ˆy+(AxBz −AzBx )ˆx× ˆz

+(AyBz −AzBy )ˆy× ˆz

upon applying Eqs. (1.37a) and (1.37b) and substituting Eqs. (1.36a), (1.36b), and (1.36c)

so that the Cartesian components of A×B become

Cx = AyBz −AzBy, Cy = AzBx −AxBz, Cz = AxBy −AyBx , (1.38)

or

Ci = AjBk − AkBj , i, j, k all different, (1.39)

and with cyclic permutation of the indices i, j , and k corresponding to x,y, and z, respectively.

The vector product C may be mnemonically represented by a determinant,9

C =

ˆx ˆy ˆz

Ax Ay Az

Bx By Bz

≡ ˆx

Ay Az

By Bz

− ˆy

Ax Az

Bx Bz

+ ˆz

Ax Ay

Bx By

, (1.40)

which is meant to be expanded across the top row to reproduce the three components of C

listed in Eqs. (1.38).

Equation (1.35) might be called a geometric definition of the vector product. Then

Eqs. (1.38) would be an algebraic definition.

To show the equivalence of Eq. (1.35) and the component definition, Eqs. (1.38), let us

form A · C and B · C, using Eqs. (1.38). We have

A · C = A · (A×B)

= Ax(AyBz − AzBy ) +Ay(AzBx −AxBz)+Az(AxBy − AyBx )

= 0. (1.41)

Similarly,

B · C = B · (A×B) = 0. (1.42)

Equations (1.41) and (1.42) show that C is perpendicular to both A and B (cos θ = 0, θ =

±90◦) and therefore perpendicular to the plane they determine. The positive direction is

determined by considering special cases, such as the unit vectors ˆx× ˆy = ˆz (Cz =+AxBy ).

The magnitude is obtained from

(A×B) · (A×B) = A2B2 − (A · B)2

= A2B2 − A2B2 cos2 θ

= A2B2 sin2 θ. (1.43)

9See Section 3.1 for a brief summary of determinants.

1.4 Vector or Cross Product 21

Hence

C = AB sin θ. (1.44)

The first step in Eq. (1.43) may be verified by expanding out in component form, using

Eqs. (1.38) for A × B and Eq. (1.24) for the dot product. From Eqs. (1.41), (1.42), and

(1.44) we see the equivalence of Eqs. (1.35) and (1.38), the two definitions of vector product.

There still remains the problem of verifying that C = A × B is indeed a vector, that

is, that it obeys Eq. (1.15), the vector transformation law. Starting in a rotated (primed

system),

C′i = A′jB′k −A′kB′j , i,j,and k in cyclic order,

=

l

ajlAl

m

akmBm −

l

aklAl

m

ajmBm

=

l,m

(ajlakm −aklajm)AlBm. (1.45)

The combination of direction cosines in parentheses vanishes for m = l.We therefore have

j and k taking on fixed values, dependent on the choice of i, and six combinations of

l and m. If i = 3, then j = 1, k = 2 (cyclic order), and we have the following direction

cosine combinations:10

a11a22 − a21a12 = a33,

a13a21 − a23a11 = a32,

a12a23 − a22a13 = a31

(1.46)

and their negatives. Equations (1.46) are identities satisfied by the direction cosines. They

may be verified with the use of determinants and matrices (see Exercise 3.3.3). Substituting

back into Eq. (1.45),

C′3 = a33A1B2 + a32A3B1 +a31A2B3 −a33A2B1 −a32A1B3 −a31A3B2

= a31C1 +a32C2 +a33C3

=

n

a3nCn. (1.47)

By permuting indices to pick up C′1 and C′2, we see that Eq. (1.15) is satisfied and C is

indeed a vector. It should be mentioned here that this vector nature of the cross product

is an accident associated with the three-dimensional nature of ordinary space.11 It will be

seen in Chapter 2 that the cross product may also be treated as a second-rank antisymmetric

tensor.

10Equations (1.46) hold for rotations because they preserve volumes. For a more general orthogonal transformation, the r.h.s. of

Eqs. (1.46) is multiplied by the determinant of the transformation matrix (see Chapter 3 for matrices and determinants).

11Specifically Eqs. (1.46) hold only for three-dimensional space. See D. Hestenes and G. Sobczyk, Clifford Algebra to Geometric

Calculus (Dordrecht: Reidel, 1984) for a far-reaching generalization of the cross product.

22 Chapter 1 Vector Analysis

If we define a vector as an ordered triplet of numbers (or functions), as in the latter part

of Section 1.2, then there is no problem identifying the cross product as a vector. The crossproduct

operation maps the two triples A and B into a third triple, C, which by definition

is a vector.

We now have two ways of multiplying vectors; a third form appears in Chapter 2. But

what about division by a vector? It turns out that the ratio B/A is not uniquely specified

(Exercise 3.2.21) unless A and B are also required to be parallel. Hence division of one

vector by another is not defined.

Exercises

1.4.1 Show that the medians of a triangle intersect in the center, which is 2/3 of the median’s

length from each corner. Construct a numerical example and plot it.

1.4.2 Prove the law of cosines starting from A2 = (B−C)2.

1.4.3 Starting with C = A+ B, show that C×C = 0 leads to

A×B=−B×A.

1.4.4 Show that

(a) (A −B) · (A +B) = A2 −B2,

(b) (A −B) ×(A +B) = 2A×B.

The distributive laws needed here,

A · (B+ C) = A · B+A · C,

and

A×(B+ C) = A×B+A×C,

may easily be verified (if desired) by expansion in Cartesian components.

1.4.5 Given the three vectors,

P = 3ˆx+2ˆy− ˆz,

Q=−6ˆx−4ˆy+2ˆz,

R = ˆx− 2ˆy− ˆz,

find two that are perpendicular and two that are parallel or antiparallel.

1.4.6 If P = ˆxPx + ˆyPy and Q = ˆxQx + ˆyQy are any two nonparallel (also nonantiparallel)

vectors in the xy-plane, show that P×Q is in the z-direction.

1.4.7 Prove that (A×B) · (A×B) = (AB)2 −(A · B)2.

1.4 Vector or Cross Product 23

1.4.8 Using the vectors

P = ˆx cos θ + ˆy sin θ,

Q = ˆx cosϕ − ˆy sin ϕ,

R = ˆx cosϕ + ˆy sin ϕ,

prove the familiar trigonometric identities

sin(θ + ϕ) = sin θ cosϕ +cos θ sin ϕ,

cos(θ + ϕ) = cos θ cosϕ −sin θ sin ϕ.

1.4.9 (a) Find a vector A that is perpendicular to

U = 2ˆx+ ˆy − ˆz,

V = ˆx− ˆy+ ˆz.

(b) What is A if, in addition to this requirement, we demand that it have unit magnitude?

1.4.10 If four vectors a,b, c, and d all lie in the same plane, show that

(a×b)×(c×d) = 0.

Hint. Consider the directions of the cross-product vectors.

1.4.11 The coordinates of the three vertices of a triangle are (2, 1, 5), (5, 2, 8), and (4, 8, 2).

Compute its area by vector methods, its center and medians. Lengths are in centimeters.

Hint. See Exercise 1.4.1.

1.4.12 The vertices of parallelogram ABCD are (1, 0, 0), (2,−1, 0), (0,−1, 1), and (−1, 0, 1)

in order. Calculate the vector areas of triangle ABD and of triangle BCD. Are the two

vector areas equal?

ANS. AreaABD =−1

2 (ˆx + ˆy+2ˆz).

1.4.13 The origin and the three vectors A, B, and C (all of which start at the origin) define a

tetrahedron. Taking the outward direction as positive, calculate the total vector area of

the four tetrahedral surfaces.

Note. In Section 1.11 this result is generalized to any closed surface.

1.4.14 Find the sides and angles of the spherical triangle ABC defined by the three vectors

A = (1, 0, 0),

B =

1

√2

, 0,

1

√2

,

C =

0,

1

√2

,

1

√2

.

Each vector starts from the origin (Fig. 1.14).

24 Chapter 1 Vector Analysis

FIGURE 1.14 Spherical triangle.

1.4.15 Derive the law of sines (Fig. 1.15):

sinα

|A| =

sinβ

|B| =

sin γ

|C|

.

1.4.16 The magnetic induction B is defined by the Lorentz force equation,

F = q(v×B).

Carrying out three experiments, we find that if

v = ˆx,

F

q = 2ˆz−4ˆy,

v = ˆy,

F

q = 4ˆx− ˆz,

v = ˆz,

F

q = ˆy−2ˆx.

From the results of these three separate experiments calculate the magnetic induction B.

1.4.17 Define a cross product of two vectors in two-dimensional space and give a geometrical

interpretation of your construction.

1.4.18 Find the shortest distance between the paths of two rockets in free flight. Take the first

rocket path to be r = r1 + t1v1 with launch at r1 = (1, 1, 1) and velocity v1 = (1, 2, 3)

1.5 Triple Scalar Product, Triple Vector Product 25

FIGURE 1.15 Law of sines.

and the second rocket path as r = r2 + t2v2 with r2 = (5, 2, 1) and v2 = (−1,−1, 1).

Lengths are in kilometers, velocities in kilometers per hour.