Multiplication of vectors

7.6 MULTIPLICATION OF VECTORS
b
O
θ
a
b cos θ
Figure 7.8 The projection of b onto the direction of a is b cos θ. The scalar
product of a and b is ab cos θ.
in the previous example, the speed of the second particle relative to the first is
given by
√
u = |u| = (−3)2 + (−8)2 = 73.
A vector whose magnitude equals unity is called a unit vector. The unit vector
in the direction a is usually notated â and may be evaluated as
â =
a
.
|a|
(7.14)
The unit vector is a useful concept because a vector written as λâ then has magnitude λ and direction â. Thus magnitude and direction are explicitly separated.
7.6 Multiplication of vectors
We have already considered multiplying a vector by a scalar. Now we consider
the concept of multiplying one vector by another vector. It is not immediately
obvious what the product of two vectors represents and in fact two products
are commonly defined, the scalar product and the vector product. As their names
imply, the scalar product of two vectors is just a number, whereas the vector
product is itself a vector. Although neither the scalar nor the vector product
is what we might normally think of as a product, their use is widespread and
numerous examples will be described elsewhere in this book.
7.6.1 Scalar product
The scalar product (or dot product) of two vectors a and b is denoted by a · b
and is given by
a · b ≡ |a||b| cos θ,
0 ≤ θ ≤ π,
(7.15)
where θ is the angle between the two vectors, placed ‘tail to tail’ or ‘head to head’.
Thus, the value of the scalar product a · b equals the magnitude of a multiplied
by the projection of b onto a (see figure 7.8).
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VECTOR ALGEBRA
From (7.15) we see that the scalar product has the particularly useful property
that
a·b =0
(7.16)
is a necessary and sufficient condition for a to be perpendicular to b (unless either
of them is zero). It should be noted in particular that the Cartesian basis vectors
i, j and k, being mutually orthogonal unit vectors, satisfy the equations
i · i = j · j = k · k = 1,
(7.17)
i · j = j · k = k · i = 0.
(7.18)
Examples of scalar products arise naturally throughout physics and in particular in connection with energy. Perhaps the simplest is the work done F · r in
moving the point of application of a constant force F through a displacement r;
notice that, as expected, if the displacement is perpendicular to the direction of
the force then F · r = 0 and no work is done. A second simple example is afforded
by the potential energy −m · B of a magnetic dipole, represented in strength and
orientation by a vector m, placed in an external magnetic field B.
As the name implies, the scalar product has a magnitude but no direction. The
scalar product is commutative and distributive over addition:
a·b=b·a
a · (b + c) = a · b + a · c.
(7.19)
(7.20)
Four non-coplanar points A, B, C, D are positioned such that the line AD is perpendicular
to BC and BD is perpendicular to AC. Show that CD is perpendicular to AB.
Denote the four position vectors by a, b, c, d. As none of the three pairs of lines actually
intersect, it is difficult to indicate their orthogonality in the diagram we would normally
draw. However, the orthogonality can be expressed in vector form and we start by noting
that, since AD ⊥ BC, it follows from (7.16) that
(d − a) · (c − b) = 0.
Similarly, since BD ⊥ AC,
(d − b) · (c − a) = 0.
Combining these two equations we find
(d − a) · (c − b) = (d − b) · (c − a),
which, on mutliplying out the parentheses, gives
d · c − a · c − d · b + a · b = d · c − b · c − d · a + b · a.
Cancelling terms that appear on both sides and rearranging yields
d · b − d · a − c · b + c · a = 0,
which simplifies to give
(d − c) · (b − a) = 0.
From (7.16), we see that this implies that CD is perpendicular to AB. 220
7.6 MULTIPLICATION OF VECTORS
If we introduce a set of basis vectors that are mutually orthogonal, such as i, j,
k, we can write the components of a vector a, with respect to that basis, in terms
of the scalar product of a with each of the basis vectors, i.e. ax = a·i, ay = a·j and
az = a · k. In terms of the components ax , ay and az the scalar product is given by
a · b = (ax i + ay j + az k) · (bx i + by j + bz k) = ax bx + ay by + az bz ,
(7.21)
where the cross terms such as ax i · by j are zero because the basis vectors are
mutually perpendicular; see equation (7.18). It should be clear from (7.15) that
the value of a · b has a geometrical definition and that this value is independent
of the actual basis vectors used.
Find the angle between the vectors a = i + 2j + 3k and b = 2i + 3j + 4k.
From (7.15) the cosine of the angle θ between a and b is given by
cos θ =
a·b
.
|a||b|
From (7.21) the scalar product a · b has the value
a · b = 1 × 2 + 2 × 3 + 3 × 4 = 20,
and from (7.13) the lengths of the vectors are
√
and
|a| = 12 + 22 + 32 = 14
|b| =
22 + 32 + 42 =
√
29.
Thus,
cos θ = √
20
√ ≈ 0.9926
14 29
⇒
θ = 0.12 rad. We can see from the expressions (7.15) and (7.21) for the scalar product that if
θ is the angle between a and b then
cos θ =
ay by
az bz
ax bx
+
+
a b
a b
a b
where ax /a, ay /a and az /a are called the direction cosines of a, since they give the
cosine of the angle made by a with each of the basis vectors. Similarly bx /b, by /b
and bz /b are the direction cosines of b.
If we take the scalar product of any vector a with itself then clearly θ = 0 and
from (7.15) we have
a · a = |a|2 .
Thus the magnitude of a can be written in a coordinate-independent form as
√
|a| = a · a.
Finally, we note that the scalar product may be extended to vectors with
complex components if it is redefined as
a · b = a∗x bx + a∗y by + a∗z bz ,
where the asterisk represents the operation of complex conjugation. To accom221
VECTOR ALGEBRA
a×b
b
θ
a
Figure 7.9 The vector product. The vectors a, b and a×b form a right-handed
set.
modate this extension the commutation property (7.19) must be modified to
read
a · b = (b · a)∗ .
(7.22)
In particular it should be noted that (λa) · b = λ∗ a · b, whereas a · (λb) = λa · b.
√
However, the magnitude of a complex vector is still given by |a| = a · a, since
a · a is always real.
7.6.2 Vector product
The vector product (or cross product) of two vectors a and b is denoted by a × b
and is defined to be a vector of magnitude |a||b| sin θ in a direction perpendicular
to both a and b;
|a × b| = |a||b| sin θ.
The direction is found by ‘rotating’ a into b through the smallest possible angle.
The sense of rotation is that of a right-handed screw that moves forward in the
direction a × b (see figure 7.9). Again, θ is the angle between the two vectors
placed ‘tail to tail’ or ‘head to head’. With this definition a, b and a × b form a
right-handed set. A more directly usable description of the relative directions in
a vector product is provided by a right hand whose first two fingers and thumb
are held to be as nearly mutually perpendicular as possible. If the first finger is
pointed in the direction of the first vector and the second finger in the direction
of the second vector, then the thumb gives the direction of the vector product.
The vector product is distributive over addition, but anticommutative and nonassociative:
(a + b) × c = (a × c) + (b × c),
b × a = −(a × b),
(a × b) × c = a × (b × c).
222
(7.23)
(7.24)
(7.25)
7.6 MULTIPLICATION OF VECTORS
P
F
θ
R
r
O
Figure 7.10 The moment of the force F about O is r×F. The cross represents
the direction of r × F, which is perpendicularly into the plane of the paper.
From its definition, we see that the vector product has the very useful property
that if a × b = 0 then a is parallel or antiparallel to b (unless either of them is
zero). We also note that
a × a = 0.
(7.26)
Show that if a = b + λc, for some scalar λ, then a × c = b × c.
From (7.23) we have
a × c = (b + λc) × c = b × c + λc × c.
However, from (7.26), c × c = 0 and so
a × c = b × c.
(7.27)
We note in passing that the fact that (7.27) is satisfied does not imply that a = b. An example of the use of the vector product is that of finding the area, A, of
a parallelogram with sides a and b, using the formula
A = |a × b|.
(7.28)
Another example is afforded by considering a force F acting through a point R,
whose vector position relative to the origin O is r (see figure 7.10). Its moment
or torque about O is the strength of the force times the perpendicular distance
OP , which numerically is just Fr sin θ, i.e. the magnitude of r × F. Furthermore,
the sense of the moment is clockwise about an axis through O that points
perpendicularly into the plane of the paper (the axis is represented by a cross
in the figure). Thus the moment is completely represented by the vector r × F,
in both magnitude and spatial sense. It should be noted that the same vector
product is obtained wherever the point R is chosen, so long as it lies on the line
of action of F.
Similarly, if a solid body is rotating about some axis that passes through the
origin, with an angular velocity ω then we can describe this rotation by a vector
ω that has magnitude ω and points along the axis of rotation. The direction of ω
223
VECTOR ALGEBRA
is the forward direction of a right-handed screw rotating in the same sense as the
body. The velocity of any point in the body with position vector r is then given
by v = ω × r.
Since the basis vectors i, j, k are mutually perpendicular unit vectors, forming
a right-handed set, their vector products are easily seen to be
i × i = j × j = k × k = 0,
(7.29)
i × j = −j × i = k,
(7.30)
j × k = −k × j = i,
(7.31)
k × i = −i × k = j.
(7.32)
Using these relations, it is straightforward to show that the vector product of two
general vectors a and b is given in terms of their components with respect to the
basis set i, j, k, by
a × b = (ay bz − az by )i + (az bx − ax bz )j + (ax by − ay bx )k.
For the reader who is familiar with determinants
this can also be written as
i
j k
a × b = ax ay az
b b b
x
y
z
(7.33)
(see chapter 8), we record that
.
That the cross product a × b is perpendicular to both a and b can be verified
in component form by forming its dot products with each of the two vectors and
showing that it is zero in both cases.
Find the area A of the parallelogram with sides a = i + 2j + 3k and b = 4i + 5j + 6k.
The vector product a × b is given in component form by
a × b = (2 × 6 − 3 × 5)i + (3 × 4 − 1 × 6)j + (1 × 5 − 2 × 4)k
= −3i + 6j − 3k.
Thus the area of the parallelogram is
√
A = |a × b| = (−3)2 + 62 + (−3)2 = 54. 7.6.3 Scalar triple product
Now that we have defined the scalar and vector products, we can extend our
discussion to define products of three vectors. Again, there are two possibilities,
the scalar triple product and the vector triple product.
224
7.6 MULTIPLICATION OF VECTORS
v
P
c
φ
O
θ
b
a
Figure 7.11 The scalar triple product gives the volume of a parallelepiped.
The scalar triple product is denoted by
[a, b, c] ≡ a · (b × c)
and, as its name suggests, it is just a number. It is most simply interpreted as the
volume of a parallelepiped whose edges are given by a, b and c (see figure 7.11).
The vector v = a × b is perpendicular to the base of the solid and has magnitude
v = ab sin θ, i.e. the area of the base. Further, v · c = vc cos φ. Thus, since c cos φ
= OP is the vertical height of the parallelepiped, it is clear that (a × b) · c = area
of the base × perpendicular height = volume. It follows that, if the vectors a, b
and c are coplanar, a · (b × c) = 0.
Expressed in terms of the components of each vector with respect to the
Cartesian basis set i, j, k the scalar triple product is
a · (b × c) = ax (by cz − bz cy ) + ay (bz cx − bx cz ) + az (bx cy − by cx ),
(7.34)
which can also be written as a determinant:
ax ay
a · (b × c) = bx by
c c
x
y
az
bz
cz
.
By writing the vectors in component form, it can be shown that
a · (b × c) = (a × b) · c,
so that the dot and cross symbols can be interchanged without changing the result.
More generally, the scalar triple product is unchanged under cyclic permutation
of the vectors a, b, c. Other permutations simply give the negative of the original
scalar triple product. These results can be summarised by
[a, b, c] = [b, c, a] = [c, a, b] = −[a, c, b] = −[b, a, c] = −[c, b, a].
225
(7.35)