We have discussed vectors in physics in various earlier topics. Essentially a vector in physics is an arrow with direction,
magnitude and origin. Position vectors, velocity vectors and acceleration vectors are considered in Chapter 1, and force
vectors discussed in Topics 2-01, 2-02, and 2-03. Mathematical vectors are different from physics vectors because in mathematics
the origin of a vector is irrelevant. Two arrows with the same direction and the same magnitude, but originates from different
points, are considered to be the same vector in mathematics. The necessity in physics to distinguish vectors with the same
magnitude and direction but different origins should be obvious. For example, consider a force vector f applying on
a rock, and a force vector f' that has the same direction and magnitude as f but applies on a nearby
egg; the physical effects of those two force vectors are obviously very different. Nevertheless the understanding about the
properties of mathematical vectors is important. After all we can always add the notion of the origin later on for the usage in
physics. That is why we review vector analysis from the view of mathematics in this topic.
Throughout this topic we will discuss three-dimensional vectors. The reduction to two-dimensional case is straightforward except the case of vector product that has no meaning in two-dimensional space.
1. Definition of Vectors
We can draw an arrow in a three-dimensional space; the arrow has an origin, a magnitude or length, and a direction. The arrow can be moved without changing the magnitude and the direction such that it will originate from any arbitrary point in the three-dimensional space. Thus we have a family of countless arrows that all have the same magnitude and direction but originate from different points in the three-dimensional space. We call such a family of arrows a mathematical vector or just a vector. A vector is denoted by a boldface letter, like a. Each specific arrow in the family of a vector is called a representation of the vector. The magnitude of each arrow of the family of vector a is denoted as |a|.
Let c be a positive real number, and a a vector. The vector ca is defined as a family of arrows that are parallel to the arrows in the family of vector a, and
|ca| = c|a| . (1)
The vector -ca has a direction opposite to the direction of a and a magnitude
|-ca| = c|a| . (2)
If c = 0, then ca is a zero vector that is represented by any point in the three-dimensional space.
2. Summation and Subtraction of Vectors
Let vector a be represented by an arrow that originates from point O and ends at point P as shown in Fig. 1. Vector b can be represented by an arrow that originates from point P. Let the end point of the arrow that represents vector b be denoted as point Q. The arrow OQ is denoted as vector d, as shown in Fig. 1. We define that d is the summation of a and b, that is,
d = a + b. (3)
In Fig. 1 points O, P and Q lie in a plane. The arrow PQ can be moved within the plane, without changing the magnitude and the direction, until its origin coincides with point O. The tip of this new arrow is denoted as point R as shown in Fig. 2. PQ and OR are parallel and are equal in length. Angle PQO equals angle ROQ, and OQ is shared by triangle OPQ and triangle ROQ. Thus triangle OPQ and triangle ROQ are congruent. This means that angle POQ equals angle RQO. This also means that OP and RQ are parallel and equal in length. Therefore, OPQR is a parallelogram. Since the arrow OR is also a representation of vector b, we get the summation of Eq. (3) through the rule of parallelogram. These two methods of vector summation, that is, the method of Fig. 1 and the method of parallelogram of Fig. 2 are identical in the mathematical approach, though they have different meanings in physics as discussed in Topic 1-7.
To define the subtraction of vector b from vector a, we draw vector a and vector b as in Fig. 3(a). The arrow drawn from the tip of b to the tip of a, denoted as f, is defined as the subtraction a - b. In Fig. 3(b), the subtraction a - b is defined by the rule of parallelogram. Thus the vector subtraction defined as in Fig. 3(a) or in Fig. 3(b) by the rule of parallelogram, can be written as
f = a - b. (4)
3. Scalar Product of Vectors
In Fig. 4 two arrows representing two vectors, a and b are drawn to originate from a single point. The two vectors span an angle θ. The scalar product of a and b, denoted as a·b, is defined as
a·b = |a|· |b|·cosθ. (5)
>From the definition of scalar product in the above equation, it is clear that
a·b = b·a . (6)
The next step that we want to prove is the distributive relation of scalar products,
a·(b + c) = (b + c)· a = a·b + a·c . (7)
In Fig. 5 vectors a, b and c are drawn from point O, with the angle spanned between a and b denoted as θ and the angle between c and a denoted as φ. It should be noted that three vectors a, b and c do not need to lie in a single plane. Also drawn is a unit vector u with |u| = 1 that originates from point O and has the same direction as a. Vector b can be decomposed as
b =(|b|cosθ) u + b⊥, (7A)
where b⊥ is perpendicular to a. Similarly c can be decomposed as
c =(|c|cosφ) u + c⊥, (7B)
where c⊥ is perpendicular to a. Thus vector d=b+c can be written as
d =(|b|cosθ + |c|cosφ )u + (b⊥ + c⊥). (7C)
The angle spanned between d and a is denoted as Δ. Vector d can also be written as
d =(|d|cosΔ) u + d⊥, (7D)
where d⊥ is perpendicular to a. Comparing Eq. (7C) and Eq. (7D), we get
|d|cosΔ = |b|cosθ + |c|cosφ.
(|d|cosΔ) |a| = |b|cosθ|a| + |c|cosφ|a|. (7E)
The left-hand side of Eq. (7E) equals to a·d, and two terms at the right-hand side of the equation equal to a· b and a· c respectively. Thus the validity of Eq. (7) is proven.
4. Vector Product
Let vector a and vector b originate from point O as shown in Fig. 6. The plane that contains these two vectors is denoted as L. We define a unit vector u⊥ that is perpendicular to the plane L and points at the direction of the advance of a right-hand screw as we turn from a to b. The angle spanned between a and b is denoted as θ. The vector (|a|·|b|sinθ)u⊥ is defined as the vector product between a and b, and is denoted as a x b. Thus
a x b = (|a|·|b|sinθ)u⊥. (8)
>From the definition of vector product, it is obvious that
a x b = -b x a. (9)
Again we need to prove the distributive property of vector product, that is,
a x (b + c) = a x b + a x c. (10)
In Fig. 7 vectors a, b and c are drawn to originate from point O. We consider a plane that passes through point O and is perpendicular to a. Let the projection of vector b on this plane be denoted as b⊥, and the angle between a and b be denoted as θ. Then
|b⊥| = |b|sinθ. (10A)
Similarly the projection of vector c on the plane is c⊥, and the angle between a and c is denoted as φ. Then we have
|c⊥| = |c|sinφ. (10B)
Let us define
d = b + c. (10C)
Again the angle between d and a is denoted as Δ, and the projection of d on the plane is denoted as d⊥. Then we have
|d⊥| = |d|sinΔ. (10D)
Eq. (10C) implies
d⊥ = b⊥ + c ⊥. (10E)
Now rotate counter-clockwise b, c and d by 90 degrees, and call the resulting vectors b'⊥, c'⊥ and d'⊥ respectively. Apparently we have
d'⊥ = b'⊥ + c'⊥. (10F)
|a|b'⊥ = a x b,
|a|c'⊥ = a x c,
|a|d'⊥ = a x d,
we have from Eq. (10F)
a x d = a x b + a x c.
Then from Eq. (10C) we get
a x (b + c) = a x b + a x c.
Thus Eq. (10) is proven.
5. Rectangular Coordinate Representation
Let us choose a rectangular coordinate system with x-, y- and z-axes as shown in Fig. 8. The origin of the coordinate system is point O. We define a unit vector i that originates from O and lies on the x-axis, pointing at the + x direction. Similarly a unit vector j lies along the y-axis and a unit vector k lies along the z-axis. The rectangular coordinate representation of those unit vectors are obviously,
i = (1, 0, 0) ,
j = (0, 1, 0) , (11)
k = (0, 0, 1).
A vector a is drawn in Fig. 8, originating from point O. The lengths of the projections of the vector on x-, y- and z-axes are denoted as ax, ay and az respectively. Thus we write
a = (ax, ay, az)
= axi + ayj + azk. (12)
According to the definition of scalar product and vector product,
i·j = j·k = k·i = 0,
i x j = k,
j x k = i, (13)
i x j = k.
Repeatedly applying distributive rules for scalar and vector products and using Eqs. (13), we get
a·b = ax·bx + ay·by + az·bz, (14)
a x b = (aybz - azby)i + (azbx - axbz)j + (axby - aybx/)k. (15)
Exercise 1: Prove Eq. (14).
>From Eq. (12) we can write
a = axi + ayj + azk,
b = bxi + byj + bzk.
Then we get
a·b = axbxi + aybyj + azbzk + (axby+aybx)i·j + (aybz+azby)j·k + (azbx+axbz)k·i.
>From the first one of Eqs. (13) and the rule i·i = j·j = k·k = 1, the above equation becomes Eq. (14).
Exercise 2: Prove Eqs. (15).
>From Eq. (12) and Eq. (9) we get
a x b = (axbx)i x i + (ayby)j x j + (azbz)k x k + (aybz-azby)j x k + (azbx-axbz)k x i + (axby-aybx)i x j.
>From the definition of vector product, we get a x a = 0. Then with Eqs. (13), we get Eq. (15).