What the Importance of Scalar and dot Product in Mathematics ?

SCALAR OR DOT PRODUCT :

Having defined vectors, we now proceed to combine them. The laws for combining vectors
must be mathematically consistent. From the possibilities that are consistent we select two
that are both mathematically and physically interesting. A third possibility is introduced in
Chapter 2, in which we form tensors.
The projection of a vector A onto a coordinate axis, which gives its Cartesian components
in Eq. (1.4), defines a special geometrical case of the scalar product of A and the
coordinate unit vectors:
Ax = Acosα ≡ A · ˆx, Ay = Acosβ ≡ A · ˆy, Az = Acos γ ≡ A · ˆz. (1.22)
7The n-dimensional vector space of real n-tuples is often labeled Rn and the n-dimensional vector space of complex n-tuples is
labeled Cn.
1.3 Scalar or Dot Product 13
This special case of a scalar product in conjunction with general properties the scalar product
is sufficient to derive the general case of the scalar product.
Just as the projection is linear in A, we want the scalar product of two vectors to be
linear in A and B, that is, obey the distributive and associative laws
A · (B+C) = A · B+A · C (1.23a)
A · (yB) = (yA) · B = yA · B, (1.23b)
where y is a number. Now we can use the decomposition of B into its Cartesian components
according to Eq. (1.5), B = Bx ˆx+By ˆy+Bz ˆz, to construct the general scalar or dot product
of the vectors A and B as
A · B = A · (Bx ˆx+ By ˆy +Bz ˆz)
= BxA · ˆx+ByA · ˆy+BzA · ˆz upon applying Eqs. (1.23a) and (1.23b)
= BxAx +ByAy +BzAz upon substituting Eq. (1.22).
Hence
A · B ≡

i
BiAi =

i
AiBi = B · A. (1.24)
If A = B in Eq. (1.24), we recover the magnitude A = (

A2
i )1/2 of A in Eq. (1.6) from
Eq. (1.24).
It is obvious from Eq. (1.24) that the scalar product treats A and B alike, or is symmetric
in A and B, and is commutative. Thus, alternatively and equivalently, we can first
generalize Eqs. (1.22) to the projection AB of A onto the direction of a vector B
= 0
as AB = Acos θ ≡ A · Bˆ , where Bˆ = B/B is the unit vector in the direction of B and θ
is the angle between A and B, as shown in Fig. 1.7. Similarly, we project B onto A as
BA = B cos θ ≡ B · Aˆ . Second, we make these projections symmetric in A and B, which
leads to the definition
A · B ≡ ABB = ABA = AB cos θ. (1.25)
FIGURE 1.7 Scalar product A · B = AB cos θ .
14 Chapter 1 Vector Analysis
FIGURE 1.8 The distributive law
A · (B +C) = ABA +ACA = A(B +C)A, Eq. (1.23a).
The distributive law in Eq. (1.23a) is illustrated in Fig. 1.8, which shows that the sum of
the projections of B and C onto A, BA + CA is equal to the projection of B + C onto A,
(B+C)A.
It follows from Eqs. (1.22), (1.24), and (1.25) that the coordinate unit vectors satisfy the
relations
ˆx · ˆx = ˆy · ˆy = ˆz · ˆz = 1, (1.26a)
whereas
ˆx · ˆy = ˆx · ˆz = ˆy · ˆz = 0. (1.26b)
If the component definition, Eq. (1.24), is labeled an algebraic definition, then Eq. (1.25)
is a geometric definition. One of the most common applications of the scalar product in
physics is in the calculation of work = force·displacement·cos θ , which is interpreted as
displacement times the projection of the force along the displacement direction, i.e., the
scalar product of force and displacement, W = F · S.
If A · B = 0 and we know that A
= 0 and B
= 0, then, from Eq. (1.25), cos θ = 0, or
θ = 90◦, 270◦, and so on. The vectors A and B must be perpendicular. Alternately, we
may say A and B are orthogonal. The unit vectors ˆx, ˆy, and ˆz are mutually orthogonal. To
develop this notion of orthogonality one more step, suppose that n is a unit vector and r is
a nonzero vector in the xy-plane; that is, r = ˆxx + ˆyy (Fig. 1.9). If
n · r = 0
for all choices of r, then n must be perpendicular (orthogonal) to the xy-plane.
Often it is convenient to replace ˆx, ˆy, and ˆz by subscripted unit vectors em,m = 1, 2, 3,
with ˆx = e1, and so on. Then Eqs. (1.26a) and (1.26b) become
em · en = δmn. (1.26c)
For m
= n the unit vectors em and en are orthogonal. For m = n each vector is normalized
to unity, that is, has unit magnitude. The set em is said to be orthonormal. A major
advantage of Eq. (1.26c) over Eqs. (1.26a) and (1.26b) is that Eq. (1.26c) may readily be
generalized to N-dimensional space: m,n = 1, 2, . . . , N. Finally, we are picking sets of
unit vectors em that are orthonormal for convenience – a very great convenience.
1.3 Scalar or Dot Product 15
FIGURE 1.9 A normal vector.
Invariance of the Scalar Product Under Rotations
We have not yet shown that the word scalar is justified or that the scalar product is indeed
a scalar quantity. To do this, we investigate the behavior of A · B under a rotation of the
coordinate system. By use of Eq. (1.15),
A′xB′x +A′yB′y +A′zB′z =

i
axiAi

j
axjBj +

i
ayiAi

j
ayjBj
+

i
aziAi

j
azjBj . (1.27)
Using the indices k and l to sum over x,y, and z, we obtain

k
A′kB′k =

l

i

j
aliAialjBj , (1.28)
and, by rearranging the terms on the right-hand side, we have

k
A′kB′k =

l

i

j
(alialj)AiBj =

i

j
δijAiBj =

i
AiBi . (1.29)
The last two steps follow by using Eq. (1.18), the orthogonality condition of the direction
cosines, and Eqs. (1.20), which define the Kronecker delta. The effect of the Kronecker
delta is to cancel all terms in a summation over either index except the term for which the
indices are equal. In Eq. (1.29) its effect is to set j = i and to eliminate the summation
over j . Of course, we could equally well set i = j and eliminate the summation over i.
16 Chapter 1 Vector Analysis
Equation (1.29) gives us

k
A′kB′k =

i
AiBi , (1.30)
which is just our definition of a scalar quantity, one that remains invariant under the rotation
of the coordinate system.
In a similar approach that exploits this concept of invariance, we take C = A + B and
dot it into itself:
C · C = (A +B) · (A+ B)
= A · A+B · B+ 2A · B. (1.31)
Since
C · C = C2, (1.32)
the square of the magnitude of vector C and thus an invariant quantity, we see that
A · B =
1
2

C2 − A2 − B2
, invariant. (1.33)
Since the right-hand side of Eq. (1.33) is invariant—that is, a scalar quantity—the lefthand
side, A · B, must also be invariant under rotation of the coordinate system. Hence
A · B is a scalar.
Equation (1.31) is really another form of the law of cosines, which is
C2 = A2 +B2 +2AB cos θ. (1.34)
Comparing Eqs. (1.31) and (1.34), we have another verification of Eq. (1.25), or, if preferred,
a vector derivation of the law of cosines (Fig. 1.10).
The dot product, given by Eq. (1.24), may be generalized in two ways. The space need
not be restricted to three dimensions. In n-dimensional space, Eq. (1.24) applies with the
sum running from 1 to n. Moreover, n may be infinity, with the sum then a convergent infinite
series (Section 5.2). The other generalization extends the concept of vector to embrace
functions. The function analog of a dot, or inner, product appears in Section 10.4.
FIGURE 1.10 The law of cosines.
1.3 Scalar or Dot Product 17
Exercises
1.3.1 Two unit magnitude vectors ei and ej are required to be either parallel or perpendicular
to each other. Show that ei · ej provides an interpretation of Eq. (1.18), the direction
cosine orthogonality relation.
1.3.2 Given that (1) the dot product of a unit vector with itself is unity and (2) this relation is
valid in all (rotated) coordinate systems, show that ˆx′ · ˆx′ = 1 (with the primed system
rotated 45◦ about the z-axis relative to the unprimed) implies that ˆx · ˆy = 0.
1.3.3 The vector r, starting at the origin, terminates at and specifies the point in space (x, y, z).
Find the surface swept out by the tip of r if
(a) (r −a) · a = 0. Characterize a geometrically.
(b) (r −a) · r = 0. Describe the geometric role of a.
The vector a is constant (in magnitude and direction).
1.3.4 The interaction energy between two dipoles of moments μ1 and μ2 may be written in
the vector form
V =−
μ1 · μ2
r3 +
3(μ1 · r)(μ2 · r)
r5
and in the scalar form
V =
μ1μ2
r3 (2 cos θ1 cos θ2 − sin θ1 sin θ2 cos ϕ).
Here θ1 and θ2 are the angles of μ1 and μ2 relative to r, while ϕ is the azimuth of μ2
relative to the μ1–r plane (Fig. 1.11). Show that these two forms are equivalent.
Hint: Equation (12.178) will be helpful.
1.3.5 A pipe comes diagonally down the south wall of a building, making an angle of 45◦
with the horizontal. Coming into a corner, the pipe turns and continues diagonally down
a west-facing wall, still making an angle of 45◦ with the horizontal. What is the angle
between the south-wall and west-wall sections of the pipe?
ANS. 120◦.
1.3.6 Find the shortest distance of an observer at the point (2, 1, 3) from a rocket in free
flight with velocity (1, 2, 3) m/s. The rocket was launched at time t = 0 from (1, 1, 1).
Lengths are in kilometers.
1.3.7 Prove the law of cosines from the triangle with corners at the point of C and A in
Fig. 1.10 and the projection of vector B onto vector A.
FIGURE 1.11 Two dipole moments.
18 Chapter 1 Vector Analysis