# 3.1 Basic Laws of Vector Algebra

A vector is a mathematical object that resembles an arrow. Vector A in Fig. 3-1 has magnitude (or length) A=|A| and unit vector a ˆ:

A=a ˆ|A|=a ˆ A. (3.1)

# Figure 3-1 Vector A=a ˆ A has magnitude A=|A| and points in the direction of unit vector a ˆ=A/A.

The unit vector a ˆ has a magnitude of one (|a ˆ|=1) and points from A’s tail or anchor to its head or tip. From Eq. (3.1),

a ˆ=A|A|=A A. (3.2)

In the Cartesian (or rectangular) coordinate system shown in Fig. 3-2(a), the x, y, and z coordinate axes extend along directions of the three mutually perpendicular unit vectors x ˆ, y ˆ, and z ˆ, which are also called base vectors. The vector A in Fig. 3-2(b) may be decomposed as

A=x ˆ A x+y ˆ A y+z ˆ A z, (3.3)

where A x, A y, and A z are A’s scalar components along the x-, y-, and z axes, respectively. The component A z is equal to the perpendicular projection of A onto the z axis, and similar definitions apply to A x and A y. Application of the Pythagorean theorem—first to the right triangle in the x–y plane to express the hypotenuse A r in terms of A x and A y and then again to the vertical right triangle with sides A r and A z and hypotenuse A—yields the following expression for the magnitude of A:

A=|A|=A x 2+A y 2+A z 2+. (3.4)

Since A is a nonnegative scalar, only the positive root applies. From Eq. (3.2), the unit vector a ˆ is

a ˆ=A A=x ˆ A x+y ˆ A y+z ˆ A z A x 2+A y 2+A z 2+. (3.5)

# Figure 3-2 Cartesian coordinate system: (a) base vectors x ˆ, y ˆ, and z ˆ and (b) components of vector A.

(a) Base vectors

(b) Components of A

Figure 3 2 Full Alternative Text

Occasionally, we use the shorthand notation A=(A x,A y,A z) to denote a vector with components A x, A y, and A z in a Cartesian coordinate system.

# 3.1.1 Equality of Two Vectors

Two vectors A and B are equal if they have equal magnitudes and identical unit vectors. Thus, if

A=a ˆ A=x ˆ A x+y ˆ A y+z ˆ A z, (3.6a)
B=b ˆ B=x ˆ B x+y ˆ B y+z ˆ B z, (3.6b)

then A=B if and only if A=B and a ˆ=b ˆ, which requires that A x=B x, A y=B y, and A z=B z.

# 3.1.2 Vector Addition and Subtraction

The sum of two vectors A and B is a vector

C=x ˆ C x+y ˆ C y+z ˆ C z,

given by

C=A+B=(x ˆ A x+y ˆ A y+z ˆ A z)+(x ˆ B x+y ˆ B y+z ˆ B z)=x ˆ(A x+B x)+y ˆ(A y+B y)+z ˆ(A z+B z)=x ˆ C x+y ˆ C y+z ˆ C z, (3.7)

with C x=A x+B x, etc.

Graphically, vector addition can be accomplished by either the parallelogram or the head-to-tail rule (Fig. 3-3). Vector C is the diagonal of the parallelogram with sides A and B. With the head-to-tail rule, we may either add A to B or B to A. When A is added to B, it is repositioned so that its tail starts at the tip of B while keeping its length and direction unchanged. The sum vector C starts at the tail of B and ends at the tip of A.

# Figure 3-3 Vector addition by (a) the parallelogram rule and (b) the head-to-tail rule.

(a) Parallelogram rule

Figure 3 3-a Full Alternative Text

(b) Head-to-tail rule

Subtraction of vector B from vector A is equivalent to the addition of A to negative B. Thus,

D=A−B=A+(−B)=x ˆ(A x−B x)+y ˆ(A y−B y)+z ˆ(A z−B z). (3.9)

Graphically, the same rules used for vector addition are also applicable to vector subtraction; the only difference is that the arrowhead of (−B) is drawn on the opposite end of the line segment representing the vector B (i.e., the tail and head are interchanged).

# 3.1.3 Position and Distance Vectors

The position vector of a point P in space is the vector from the origin to P. Assuming points P 1 and P 2 are at (x 1,y 1,z 1) and (x 2,y 2,z 2) in Fig. 3-4, their position vectors are

R 1=O P 1→=x ˆ x 1+y ˆ y 1+z ˆ z 1, (3.10a)
R 2=O P 2→=x ˆ x 2+y ˆ y 2+z ˆ z 2, (3.10b)

where point O is the origin.

The distance vector from P 1 to P 2 is defined as

R 12=P 1 P 2→=R 2−R 1=x ˆ(x 2−x 1)+y ˆ(y 2−y 1)+z ˆ(z 2−z 1), (3.11)

and the distance d between P 1 and P 2 equals the magnitude of R 12:

d=|R 12|=[(x 2−x 1)2+(y 2−y 1)2+(z 2−z 1)2]1/2. (3.12)

Note that the first and second subscripts of R 12 denote the locations of its tail and head, respectively (Fig. 3-4).

# Figure 3-4 Distance vector R 12=P 1 P 2→=R 2−R 1, where R 1 and R 2 are the position vectors of points P 1 and P 2, respectively.

Figure 3 4 Full Alternative Text

# 3.1.4 Vector Multiplication

There exist three types of products in vector calculus: the simple product, the scalar (or dot) product, and the vector (or cross) product.

# Simple Product

The multiplication of a vector by a scalar is called a simple product. The product of the vector A=a ˆ A by a scalar k results in a vector B with magnitude B=k A and direction the same as A. That is, b ˆ=a ˆ. In Cartesian coordinates,

B=k A=a ˆ k A=x ˆ(k A x)+y ˆ(k A y)+z ˆ(k A z)=x ˆ B x+y ˆ B y+z ˆ B z. (3.13)

# Scalar or Dot Product

The scalar (or dot) product of two co-anchored vectors A and B, denoted A⋅B and pronounced “A dot B,” is defined geometrically as the product of the magnitude of A and the scalar component of B along A, or vice versa. Thus,

where θ A B is the angle between A and B (Fig. 3-5) measured from the tail of A to the tail of B. Angle θ A B is assumed to be in the range 0≤θ A B≤180°. The scalar product of A and B yields a scalar whose magnitude is less than or equal to the products of their magnitudes (equality holds when θ A B=0) and whose sign is positive if 0<θ A B<90° and negative if 90°<θ A B<180°. When θ A B=90°, A and B are orthogonal, and their dot product is zero. The quantity A cos θ A B is the scalar component of A along B. Similarly, B cos θ B A is the scalar component of B along A.

# Figure 3-5 The angle θ A B is the angle between A and B, measured from A to B between vector tails. The dot product is positive if 0≤θ A B<90°, as in (a), and it is negative if 90°<θ A B≤180°, as in (b).

The commutative property follows from Eq. (3.14) and the fact that θ A B=θ B A. The distributive property expresses the fact that the scalar component of the sum of two vectors along a third one equals the sum of their respective scalar components.

The dot product of a vector with itself gives

A⋅A=|A|2=A 2, (3.16)

which implies that

A=|A|=A⋅A+. (3.17)

Also, θ A B can be determined from

θ A B=cos−1[A⋅B A⋅A+B⋅B+]. (3.18)

Since the base vectors x ˆ, y ˆ, and z ˆ are each orthogonal to the other two, it follows that

If A=(A x,A y,A z) and B=(B x,B y,B z), then

A⋅B=(x ˆ A x+y ˆ A y+z ˆ A z)⋅(x ˆ B x+y ˆ B y+z ˆ B z). (3.20)

Use of Eqs. (3.19a) and (3.19b) in Eq. (3.20) leads to

A⋅B=A x B x+A y B y+A z B z. (3.21)

# Vector or Cross Product

The vector (or cross) product of two vectors A and B, denoted A×B and pronounced “A cross B,” yields a vector defined as

where n ˆ is a unit vector normal to the plane containing A and B (Fig. 3-6(a)). The magnitude of the cross product, A B|sin θ A B|, equals the area of the parallelogram defined by the two vectors. The direction of n ˆ is governed by the right-hand rule (Fig. 3-6(b)): n ˆ points in the direction of the right thumb when the fingers rotate from A to B through the angle θ A B. Note that, since n ˆ is perpendicular to the plane containing A and B, A×B is perpendicular to both vectors A and B.

# Figure 3-6 Cross product A×B points in the direction n ˆ, which is perpendicular to the plane containing A and B and defined by the right-hand rule.

(a) Cross product

Figure 3 6-a Full Alternative Text

(b) Right-hand rule

Figure 3 6-b Full Alternative Text

The anticommutative property follows from the application of the right-hand rule to determine n ˆ. The distributive property follows from the fact that the area of the parallelogram formed by A and (B+C) equals the sum of those formed by (A and B) and (A and C):

The cross product of a vector with itself vanishes. That is,

A×A=0. (3.24)

From the definition of the cross product given by Eq. (3.22), it is easy to verify that the base vectors x ˆ, y ˆ, and z ˆ of the Cartesian coordinate system obey the right-hand cyclic relations:

Note the cyclic order (x y z x y z…). Also,

If A=(A x,A y,A z) and B=(B x,B y,B z), then use of Eqs. (3.25) and (3.26) leads to

A×B=(x ˆ A x+y ˆ A y+z ˆ A z)×(x ˆ B x+y ˆ B y+z ˆ B z)=x ˆ(A y B z−A z B y)+y ˆ(A z B x−A x B z)+z ˆ(A x B y−A y B x). (3.27)

The cyclical form of the result given by Eq. (3.27) allows us to express the cross product in the form of a determinant:

# Example 3-1: Vectors and Angles

In Cartesian coordinates, vector A points from the origin to point P 1=(2,3,3), and vector B is directed from P 1 to point P 2=(1,−2,2). Find:

1. vector A, its magnitude A, and unit vector a ˆ,

2. the angle between A and the y axis,

3. vector B,

4. the angle θ A B between A and B, and

5. perpendicular distance from the origin to vector B.

Solution: (a) Vector A is given by the position vector of P 1=(2,3,3) (Fig. 3-7). Thus,

A=x ˆ 2+y ˆ 3+z ˆ 3,A=|A|=2 2+3 2+3 2=22,a ˆ=A A=(x ˆ 2+y ˆ 3+z ˆ 3)/22.

# Figure 3-7 Geometry of Example 3-1.

Figure 3 7 Full Alternative Text

(b) The angle β between A and the y axis is obtained from

A⋅y ˆ=|A||y ˆ|cos β=A cos β,

or

β=cos−1 A⋅y ˆ A=cos−1 3 22=50.2°.

(c)

B=x ˆ(1−2)+y ˆ(−2−3)+z ˆ(2−3)=−x ˆ−y ˆ 5−z ˆ.

(d)

θ A B=cos−1[A⋅B|A||B|]=cos−1[(−2−15−3)22 27]=145.1∘.

(e) The perpendicular distance between the origin and vector B is the distance |O P 3→| shown in Fig. 3-7. From right triangle O P 1 P 3,

|O P 3→|=|A|sin(180°−θ A B)=22 sin(180°−145.1°)=2.68.

# Example 3-2: Cross Product

Given vectors A=x ˆ 2−y ˆ+z ˆ 3 and B=y ˆ 2−z ˆ 3, compute (a) A×B, (b) y ˆ×B, and (c) (y ˆ×B)⋅A.

Solution: (a) Application of Eq. (3.28) gives

A×B=x ˆ y ˆ z ˆ 2−1 3 0 2−3=x ˆ((−1)×(−3)−3×2)−y ˆ(2×(−3)−3×0)+z ˆ(2×2−(−1×0))=−x ˆ 3+y ˆ 6+z ˆ 4.

(b) y ˆ×B=y ˆ×(y ˆ 2−z ˆ 3)=−x ˆ 3.

(c) (y ˆ×B)⋅A=−x ˆ 3⋅(x ˆ 2−y ˆ+z ˆ 3)=−6.

# 3.1.5 Scalar and Vector Triple Products

When three vectors are multiplied, not all combinations of dot and cross products are meaningful. For example, the product

A×(B⋅C)

does not make sense because B⋅C is a scalar, and the cross product of the vector A with a scalar is not defined under the rules of vector algebra. Other than the product of the form A(B⋅C), the only two meaningful products of three vectors are the scalar triple product and the vector triple product.

# Scalar Triple Product

The dot product of a vector with the cross product of two other vectors is called a scalar triple product, so named because the result is a scalar. A scalar triple product obeys the cyclic order:

The equalities hold as long as the cyclic order (A B C A B C…) is preserved. The scalar triple product of vectors A=(A x,A y,A z), B=(B x,B y,B z), and C=(C x,C y,C z) can be expressed in the form of a 3×3 determinant:

A⋅(B×C)=A x A y A z B x B y B z C x C y C z. (3.30)

The validity of Eqs. (3.29) and (3.30) can be verified by expanding A, B, and C in component form and carrying out the multiplications.

# Vector Triple Product

The vector triple product involves the cross product of a vector with the cross product of two others, such as

A×(B×C). (3.31)

Since each cross product yields a vector, the result of a vector triple product is also a vector. The vector triple product does not obey the associative law. That is,

A×(B×C)≠(A×B)×C, (3.32)

which means that it is important to specify which cross multiplication is to be performed first. By expanding the vectors A, B, and C in component form, it can be shown that

which is known as the “bac-cab” rule.

# Example 3-3: Vector Triple Product

Given A=x ˆ−y ˆ+z ˆ 2, B=y ˆ+z ˆ, and C=−x ˆ 2+z ˆ 3, find (A×B)×C and compare it with A×(B×C).

Solution:

A×B=x ˆ y ˆ z ˆ 1−1 2 0 1 1=−x ˆ 3−y ˆ+z ˆ

and

(A×B)×C=x ˆ y ˆ z ˆ−3−1 1−2 0 3=−x ˆ 3+y ˆ 7−z ˆ 2.

A similar procedure gives A×(B×C)=x ˆ 2+y ˆ 4+z ˆ. The fact that the results of two vector triple products are different demonstrates the inequality stated in Eq. (3.32).

Concept Question 3-1: When are two vectors equal and when are they identical?

Concept Question 3-2: When is the position vector of a point identical to the distance vector between two points?

Concept Question 3-3: If A⋅B=0, what is θ A B?

Concept Question 3-4: If A×B=0, what is θ A B?

Concept Question 3-5: Is A(B⋅C) a vector triple product?

Concept Question 3-6: If A⋅B=A⋅C, does it follow that B=C?