# Multiplying Vectors by David Halliday, Robert Resnick

There are three ways in which vectors can be multiplied, but none is exactly like the usual algebraic multiplication. As you read this material, keep in mind that a vector-capable calculator will help you multiply vectors only if you understand the basic rules of that multiplication.

Multiplying a Vector by a Scalar

If we multiply a vector $\stackrel{\to }{a}$ by a scalar s, we get a new vector. Its magnitude is the product of the magnitude of $\stackrel{\to }{a}$ and the absolute value of s. Its direction is the direction of $\stackrel{\to }{a}$ if s is positive but the opposite direction if s is negative. To divide $\stackrel{\to }{a}$ by s, we multiply $\stackrel{\to }{a}$ by 1/s.

Multiplying a Vector by a Vector

There are two ways to multiply a vector by a vector: one way produces a scalar (called the scalar product), and the other produces a new vector (called the vector product). (Students commonly confuse the two ways.)

The Scalar Product

The scalar product of the vectors and in Fig. 3-18a is written as $\stackrel{\to }{a}\cdot \stackrel{\to }{b}$ and defined to be

$\stackrel{\to }{a}\cdot \stackrel{\to }{b}=\mathrm{ab}cos\theta$ (3-20)

where a is the magnitude of $\stackrel{\to }{a}$, b is the magnitude of ath>b, and θ is the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ (or, more properly, between the directions of $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$).There are actually two such angles: θ and 360o - θ. Either can be used in Eq. 3-20, because their cosines are the same.

Note that there are only scalars on the right side of Eq. 3-20 (including the value of cos θ). Thus $\stackrel{\to }{a}$$\stackrel{\to }{b}$ on the left side represents a scalar quantity. Because of the notation, $\stackrel{\to }{a}$$\stackrel{\to }{b}$ is also known as the dot product and is spoken as "a dot b."

A dot product can be regarded as the product of two quantities: (1) the magnitude of one of the vectors and (2) the scalar component of the second vector along the direction of the first vector. For example, in Fig. 3-18b, $\stackrel{\to }{a}$ has a scalar component a cos θ along the direction of $\stackrel{\to }{b}$; note that a perpendicular dropped from the head of $\stackrel{\to }{a}$ onto $\stackrel{\to }{b}$ determines that component. Similarly, math>b has a scalar component b cos θ along the direction of $\stackrel{\to }{a}$.

If the angle between two vectors is 0o, the component of one vector along the other is maximum, and so also is the dot product of the vectors. If, instead, θ is 90°, the component of one vector along the other is zero, and so is the dot product.

Equation 3-20 can be rewritten as follows to emphasize the components:

$\stackrel{\to }{a}\cdot \stackrel{\to }{b}=\left(a\mathrm{cos}\theta \right)\left(b\right)=\left(a\right)\left(b\mathrm{cos}\theta \right)$(3-21)

The commutative law applies to a scalar product, so we can write
$\stackrel{\to }{a}\cdot \stackrel{\to }{b}=\stackrel{\to }{b}\cdot \stackrel{\to }{a}$

When two vectors are in unit-vector notation, we write their dot product as
$\stackrel{\to }{a}\cdot \stackrel{\to }{b}=\left({a}_{x}\stackrel{^}{i}+{a}_{y}\stackrel{^}{j}+{a}_{z}\stackrel{^}{k}\right)\cdot \left({b}_{x}\stackrel{^}{i}+{b}_{y}\stackrel{^}{j}+{b}_{z}\stackrel{^}{k}\right)$(3-22)

which we can expand according to the distributive law: Each vector component of the first vector is to be dotted with each vector component of the second vector. By doing so, we can show that
$\stackrel{\to }{a}\cdot \stackrel{\to }{b}={a}_{x}{b}_{x}+{a}_{y}{b}_{y}+{a}_{z}{b}_{z}$(3-23)

Figure 3-18 (a) Two vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$, with an angle θ between them. (b) Each vector has a component along the direction of the other vector.

The Vector Product

The vector product of $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$, written

$\stackrel{\to }{a}×\stackrel{\to }{b}$, produces a third vector c whose magnitude is

C = ab sin θ (3-24)

where θ is the smaller of the two angles between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$. (You must use the smaller of the two angles between the vectors because sin θ and sin(360o sin - θ) differ in algebraic sign.) Because of the notation, $\stackrel{\to }{a}×\stackrel{\to }{b}$ is also known as the cross product, and in speech it is "a cross b."

If $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are parallel or antiparallel, $\stackrel{\to }{a}×\stackrel{\to }{b}=0$. The magnitude of $\stackrel{\to }{a}×\stackrel{\to }{b}$, which can be written as |$\stackrel{\to }{a}×\stackrel{\to }{b}$|, is maximum when $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are perpendicular to each other.

The direction of $\stackrel{\to }{c}$ is perpendicular to the plane that contains $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$. Figure 3-19a shows how to determine the direction of $\stackrel{\to }{c}=\stackrel{\to }{a}×\stackrel{\to }{b}$ with what is known as a right-hand rule. Place the vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ tail to tail without altering their orientations, and imagine a line that is perpendicular to their plane where they meet. Pretend to place your right hand around that line in such a way that your fingers would sweep $\stackrel{\to }{a}$ into $\stackrel{\to }{b}$ through the smaller angle between them. Your outstretched thumb points in the direction of $\stackrel{\to }{c}$.

The order of the vector multiplication is important. In Fig. 3-19b, we are determining the direction of $\stackrel{\to }{{c}^{\prime }}=$ $\stackrel{\to }{b}×\stackrel{\to }{a}$, so the fingers are placed to sweep $\stackrel{\to }{b}$ into $\stackrel{\to }{a}$ through the smaller angle. The thumb ends up in the opposite direction from previously, and so it must be that $\stackrel{\to }{{c}^{\prime }}=\stackrel{\to }{\mathrm{-c}}$; that is,

$\stackrel{\to }{b}×\stackrel{\to }{a}=-\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)$.(3-25)

In other words, the commutative law does not apply to a vector product. In unit-vector notation, we write

$\stackrel{\to }{a}×\stackrel{\to }{b}=\left({a}_{x}\stackrel{^}{i}+{a}_{y}\stackrel{^}{j}+{a}_{z}\stackrel{^}{k}\right)×\left({b}_{x}\stackrel{^}{i}+{b}_{y}\stackrel{^}{j}+{b}_{z}\stackrel{^}{k}\right)$,(3-26)

which can be expanded according to the distributive law; that is, each component of the first vector is to be crossed with each component of the second vector. For example, in the expansion of Eq. 3-26, we have

${a}_{x}\stackrel{^}{i}×{b}_{y}\stackrel{^}{j}={a}_{x}{b}_{y}\left(\stackrel{^}{i}×\stackrel{^}{j}\right)={a}_{x}{b}_{y}\stackrel{^}{k}$.

In the last step we used Eq. 3-24 to evaluate the magnitude of n the last step we used Eq. 3-24 to evaluate the magnitude of ' as unity. (These vectors and each have a magnitude of unity, and the angle between $\stackrel{^}{i}×\stackrel{^}{j}$ as unity. (These vectors $\stackrel{^}{i}$ and $\stackrel{^}{j}$ each have a magnitude of unity, and the angle between them is 90o.) Also, we used the right-hand rule to get the direction of $\stackrel{^}{i}×\stackrel{^}{j}$ as being in the positive direction of the z axis (thus in the direction of $\stackrel{^}{k}$).

Continuing to expand Eq. 3-26, you can show that

$\stackrel{\to }{a}×\stackrel{\to }{b}=\left({a}_{y}{b}_{z}-{b}_{y}{a}_{z}\right)\stackrel{^}{i}+\left({a}_{z}{b}_{x}-{b}_{z}{a}_{x}\right)\stackrel{^}{j}+\left({a}_{x}{b}_{y}-{b}_{x}{a}_{y}\right)\stackrel{^}{k}$ (3-27)

A determinant or a vector-capable calculator can also be used. To check whether any xyz coordinate system is a right-handed coordinate system, use the right-hand rule for the cross product $\stackrel{^}{i}×\stackrel{^}{j}=\stackrel{^}{k}$ with that system. If your fingers sweep $\stackrel{^}{i}$ (positive direction of x) into $\stackrel{^}{j}$ (positive direction of y) with the outstretched thumb pointing in the positive direction of z (not the negative direction), then the system is right-handed.

Figure 3-19 Illustration of the right-hand rule for vector products. (a) Sweep vector $\stackrel{\to }{a}$ into vector $\stackrel{\to }{b}$ with the fingers of your right hand. Your outstretched thumb shows the direction of vector $\stackrel{\to }{c}=\stackrel{\to }{a}×\stackrel{\to }{b}$. (b) Showing that $\stackrel{\to }{a}×\stackrel{\to }{b}$ is the reverse of $\stackrel{\to }{b}×\stackrel{\to }{a}$.