There are three ways in which vectors can be multiplied, but none is exactly like the usual algebraic multiplication. As you read this section, 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 
by a scalar s, we get a new vector. Its magnitude is the product of the magnitude of and the absolute value of s. Its direction is the direction of if s is positive but the opposite direction if s is negative. To divide by s, we multiply 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, and so starting now, you should carefully distinguish between them.
The Scalar Product
The scalar product of the vectors and 
in Fig. 3-19a is written as · and
Fig. 3-19 (a) Two vectors and , with an angle 
between them. (b) Each vector has a component along the direction of the other vector.
where a is the magnitude of , b is the magnitude of , and is the angle between and (or, more properly, between the directions of and ). There are actually two such angles: and 360° − . 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 ). Thus · on the left side represents a scalar quantity. Because of the notation, · 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. 3-19b, has a scalar component a cos along the direction of ; note that a perpendicular dropped from the head of onto determines that component. Similarly, has a scalar component b cos along the direction of .
If the angle between two vectors is 0°, the component of one vector along the other is maximum, and so also is the dot product of the vectors. If, instead, 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:
The commutative law applies to a scalar product, so we can write
When two vectors are in unit-vector notation, we write their dot product as
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
CHECKPOINT 4 Vectors 
and 
have magnitudes of 3 units and 4 units, respectively. What is the angle between the directions of and if · equals (a) zero, (b) 12 units, and (c) −12 units?
What is the angle between = 3.0
− 4.0
and = −2.0 + 3.0
Solution: First, a caution: Although many of the following steps can be bypassed with a vector-capable calculator, you will learn more about scalar products if, at least here, you use these steps.
One Key Idea here is that the angle between the directions of two vectors is included in the definition of their scalar product (Eq. 3-20):
In this equation, a is the magnitude of , or
and b is the magnitude of , or
A second Key Idea is that we can separately evaluate the left side of Eq. 3-24 by writing the vectors in unit-vector notation and using the distributive law:
We next apply Eq. 3-20 to each term in this last expression. The angle between the unit vectors in the first term ( and ) is 0°, and in the other terms it is 90°. We then have
Substituting this result and the results of Eqs. 3-25 and 3-26 into Eq. 3-24 yields
The Vector Product
The vector product of and , written × , produces a third vector 
whose magnitude is
where is the smaller of the two angles between and . (You must use the smaller of the two angles between the vectors because sin and sin(360° − ) differ in algebraic sign.) Because of the notation, × is also known as the cross product, and in speech it is “a cross b.”
If and are parallel or antiparallel, × = 0. The magnitude of × , which can be written as | × |, is maximum when and are perpendicular to each other.
The direction of is perpendicular to the plane that contains and . Figure 3-20a shows how to determine the direction of = × with what is known as a right-hand rule. Place the vectors and 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 into through the smaller angle between them. Your outstretched thumb points in the direction of .
Fig. 3-20 Illustration of the right-hand rule for vector products. (a) Sweep vector into vector with the fingers of your right hand. Your outstretched thumb shows the direction of vector = × . (b) Showing that × is the reverse of × .
The order of the vector multiplication is important. In Fig. 3-20b, we are determining the direction of ′= × , so the fingers are placed to sweep into through the smaller angle. The thumb ends up in the opposite direction from previously, and so it must be that ′ = −; that is,
In other words, the commutative law does not apply to a vector product.
In unit-vector notation, we write
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. The cross products of unit vectors are given in Appendix E (see “Products of Vectors”). For example, in the expansion of Eq. 3-29, we have
because the two unit vectors and are parallel and thus have a zero cross product. Similarly, we have
In the last step we used Eq. 3-27 to evaluate the magnitude of × as unity. (These vectors and each have a magnitude of unity, and the angle between them is 90°.) Also, we used the right-hand rule to get the direction of × as being in the positive direction of the z axis (thus in the direction of ).
Continuing to expand Eq. 3-29, you can show that
You can also evaluate a cross product by setting up and evaluating a determinant (as shown in Appendix E) or by using a vector-capable calculator.
To check whether any xyz coordinate system is a right-handed coordinate system, use the right-hand rule for the cross product × = with that system. If your fingers sweep (positive direction of x) into (positive direction of y) with the outstretched thumb pointing in the positive direction of z, then the system is right-handed.
CHECKPOINT 5 Vectors and have magnitudes of 3 units and 4 units, respectively. What is the angle between the directions of and if the magnitude of the vector product × is (a) zero and (b) 12 units?
In Fig. 3-21, vector lies in the xy plane, has a magnitude of 18 units, and points in a direction 250° from the positive direction of the x axis. Also, vector has a magnitude of 12 units and points along the positive direction of the z axis. What is the vector product = × ?
Solution: One Key Idea is that when we have two vectors in magnitude-angle notation, we find the magnitude of their cross product with Eq. 3-27:
A second Key Idea is that with two vectors in magnitude-angle notation, we find the direction of their cross product with the right-hand rule of Fig. 3-20. In Fig. 3-21, imagine placing the fingers of your right hand around a line perpendicular to the plane of and (the line on which is shown) such that your fingers sweep into. Your outstretched thumb then gives the direction of . Thus, as shown in the figure, lies in the xy plane. Because its direction is perpendicular to the direction of , it is at an angle of
Fig. 3-21 Vector (in the xy plane) is the vector (or cross) product of vectors and .
from the positive direction of the x axis.
Solution: The Key Idea is that when two vectors are in unit-vector notation, we can find their cross product by using the distributive law. Here that means we can write
We next evaluate each term with Eq. 3-27, finding the direction with the right-hand rule. For the first term here, the angle between the two vectors being crossed is 0. For the other terms, is 90°. We find
This vector is perpendicular to both and , a fact you can check by showing that · = 0 and · = 0; that is, there is no component of along the direction of either or .
PROBLEM-SOLVING TACTICS
TACTIC 5 : Common Errors with Cross Products
Several errors are common in finding a cross product. (1) Failure to arrange vectors tail to tail is tempting when an illustration presents them head to tail; you must mentally shift (or better, redraw) one vector to the proper arrangement without changing its orientation. (2) Failing to use the right hand in applying the right-hand rule is easy when the right hand is occupied with a calculator or pencil. (3) Failure to sweep the first vector of the product into the second vector can occur when the orientations of the vectors require an awkward twisting of your hand to apply the right-hand rule. Sometimes that happens when you try to make the sweep mentally rather than actually using your hand. (4) Failure to work with a right-handed coordinate system results when you forget how to draw such a system (see Fig. 3-14).
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