Recalling that the standard unit vectors in Eq. (A1 –
19a-c) are i = (1 0 0), j = (0 1 0), and k = (0 0 1), from Eq. (A1
– 21)
I.3 The Dot Product and the Cross
Product
i  i 1
(A1 – 23)
Figure A1 –17
ij0
ik  0
j i  0 j j 1 jk  0
k  i  0 k  j  0 k  k  1.
Equation (A1 – 23) states that the dot products of the unit
vectors with themselves are equal to one and that the dot
products of perpendicular unit vectors are equal to zero.
These relationships are called the unit vector dot products.
The way in which a vector is multiplied by a number
was defined in the last section and this section defines two ways
a vector is multiplied by a vector. Consider the vectors A and
B. One way that vector A can be multiplied by vector B is
written as A  B with a dot between the vectors. This is called
the dot product. The second way that A can be multiplied by B
is written as A B with a cross between them. This is called the
cross product. The dot product produces a number and the
cross product produces a vector. For this reason, the dot product
is also called the scalar product (The word “scalar” means a
number that is not a vector.) and the cross product is also called
the vector product.
The dot product of any two vectors is calculated using
the rules of vector algebra and the unit vector dot products.
You’ll need to memorize the unit vector dot products that are
given in Eq. (A1 – 23) but not necessarily the original
definition of the dot product given in Eq. (A1 – 21).
Finding a Vector’s Components
The dot product becomes very useful when the interest
lies in finding a component of a vector in a given direction.
Consider two vectors A and B, as shown in Fig. A1 – 18. Let’s
find the component of A in the direction of B, denoted by AB.
The Dot Product
The dot product of the two vectors A = Axi + Ayj +
Azk and B = Bxi + Byj + Bzk produces the number
(A1 – 21)
A  B = Ax Bx  Ay By  Az Bz
Figure A1 – 18: Component of A in the
direction of B
Much like multiplying a vector by a number, you can verify
that the dot product satisfies the following three rules of vector
algebra.
(A1 – 22a,b)
A  B  B  A,
A  (B  C)  A  B  A  C
Equations (A1 – 22a,b) are the commutative rule of the dot
product, and the distributive rule of the dot product.
Toward this end, notice that the dot product of A with
the
standard
unit
vector
i
is
( Ax i  Ay j  Az k )  i  Ax (i  i)  Ay ( j  i)  Az (k  i)  Ax . Sim
Although Eq. (A1 – 21) is the definition of a dot
product, you’ll see momentarily that it isn’t needed when the
dot product is actually being calculated. Instead, the dot product
of any two vectors uses the dot products of the standard unit
vectors.
ilarly, A  j  Ay and A  k  Az . The components of A in the x,
y, and z directions are equal to the dot products of A with the
corresponding unit vectors. Of course, the interest lies in
finding the component of the vector A in the direction of B, not
in the direction of one of the axes. To find the component of A
in the direction of B simply recognize that the direction of a
1
unit vector is really arbitrary, so the component of a vector A in
the direction of B is
(A1 – 24)
AB  A  u B ,
uB 
standard unit vectors are i = (1 0 0), j = (0 1 0), and k = (0 0 1),
so from Eq. (A1 – 24)
B
B
(A1 – 27)
ii  0
i j k
i  k  j
j  i  k
ki  j
j j  0
j k  i
k  j  i k  k  0.
where uB is the unit vector associated with B. Thus, to find AB,
first calculate the unit vector associated with B, and then take
its dot product with A.
where 0 is the zero vector. These relationships are called the
unit vector cross products. Notice in Eq. (A1 – 27) that the
zero vector is produced by unit vectors that are multiplied by
themselves. On the other hand, the unit vectors that are
multiplied by different unit vectors produce the remaining unit
vector, positive or negative. The signs that are produced can be
committed to memory by drawing the unit vector circle shown
in Fig. A1 – 19. The unit vector circle is a circle inside which
the unit vectors i, then j, and then k are drawn counter-clockwise. When rotating around the circle counter-clock-wise,
positive unit vectors are produced and when rotating in the
opposite direction, the negative of the unit vectors are
produced. Thus, Equation (A1 – 27) states that the cross
products of identical unit vectors are equal to the zero
vector, and that the cross products of two different unit
vectors are equal to the third unit vector or the negative of
the third unit vector, as determined from the unit vector
circle.
The Cross Product
The cross product of the two vectors A = Axi + Ayj +
Azk and B = Bxi + Byj + Bzk is defined as the vector
(A1 – 25)
A  B  ( Ay Bz  Az By )i  ( Az Bx  Ax By ) j  ( Ax By  Ay Bx )k
You can verify that the cross product satisfies the following
three rules of vector algebra.
A  B  B  A,
(A1 – 26a-c)
A  (B  C)  B  (C  A)
A  (B  C)  A  B  A  C
Figure A1 – 19: Unit vector circle.
Equations (A1 – 26a-c) are the commutative rule of the cross
product, the associative rule of the dot and cross products,
and the distributive rule of the cross product.
By comparing Eq. (A1 – 22a,b) with Eq. (A1 – 26a-c)
you’ll see that there are several notable differences between the
rules that the dot product obeys and the rules that the cross
product obeys. First, notice the minus sign that appears in Eq.
(A1 – 26a). Reversing the order of the cross product changes
the sign of the result. When performing vector calculations, you
must keep track of the order of the cross products.
Otherwise you can make a sign error. Secondly, notice that the
associative rule in Eq. (A1 – 26b) involves both the dot product
and the cross product. The vectors are “cyclically shifted” in
Eq. (A1 – 26b). For example, it follows from Eq. (A1 – 26b) by
renaming the vectors that B  (C  A)  C  ( A  B) . Also,
notice that no associative rule exists for a dot product of a dot
product, nor of a cross product of a cross product. Finally,
notice that Eq. (A1 – 26c) is analogous to Eq. (A1 – 22b).
Instead of drawing the unit vector circle, the signs of the cross
products in Eq. (A1 – 27) can also be found using handedness.
Handedness
As Fig. A1 – 20 shows, coordinate systems are either
right-handed or left-handed depending on how they’re
oriented.
Figure A1 – 20: Right-handed and lefthanded coordinate systems
Like for the dot product, the original definition of a
cross product that was given in Eq. (A1 – 25) is not needed
when the cross product of two vectors is actually calculated.
Instead, the cross product of any two vectors is calculated using
the cross products of the standard unit vectors. Recall that the
2
To explain this, first consider a planar coordinate system with
its x axis positive to the right and its y axis positive up. This x-y
coordinate system can be rotated to produce any other x’-y’
coordinate system that exists in three-dimensional space. In
other words, any planar coordinate system can be rotated in
three-dimensional space to produce any other planar coordinate
system. On the other hand, this is not the case for threedimensional coordinate systems. For three-dimensional
coordinate systems there are two physically different types of
coordinate systems, called right-handed coordinate systems and
left-handed coordinate systems. Their difference is physically
significant. Any two right handed coordinate systems can be
rotated from one to the other and any two left-handed
coordinate systems can be rotated from one to the other. But
there is no way to physically rotate a right-handed coordinate
system to align it with a left-handed coordinate system.
Finding a Vector Perpendicular to a Plane
The general question arises how to find a vector C that
is perpendicular to a plane of vectors. To find C, you’ll first
need two vectors in the plane; we’ll call them A and B. The
answer is simply that the vector C = A B is perpendicular to
A and B, as depicted in Fig. A1 – 23.
Figure A1 – 23: The Perpendicular to a plane
Two coordinate systems S1 and S2 are shown in Fig.
A1 – 21. S1 is defined as a right-handed coordinate system.
Let’s determine whether S2 is right-handed or left-handed.
Toward this end, S1 will be rotated, axis-by-axis until it either
aligns with S2, in which case S2 is right-handed, or it doesn’t
align with S2, in which case S2 is left-handed. First, rotate S1
counter-clockwise 90◦ about the z axis to align the x axis of S1
with the x axis of S2. Next, rotate clockwise 90◦ about the x’
axis of S1 to align the z axis of S1 with the z axis of S2. Notice
that the y’’ axis of S1 is in the opposite direction of the y axis of
S2. So, S2 is a left-handed coordinate system. As a convention,
engineers always use right-handed coordinate systems.
You can see why this is true by making several
observations. First, notice in Eq. (A1 – 25) letting B = A
that A A = 0. Next, notice that saying that A B is
perpendicular to A, is the same as saying that the component of
A B in the direction of A is zero. Likewise saying that A B
is perpendicular to B amounts to saying that the component of
A B in the direction of B is zero. Let’s show that the
components of A B are zero in the directions of A and B, and
hence perpendicular to A and B. From Eqs. (A1 – 24) and (A1
– 26b),
A  C  A  (A  B)  B  (A  A)  0 so C is
perpendicular to A. Also, B  C  B  (A  B)  A  (B  B)  0
so C is also perpendicular to B. Thus, A B is perpendicular
to A and B.
Figure A1 – 21: Coordinate Systems S1 can
not be rotated to produce coordinate system
S2.
Finding the Angle between Two Vectors
The angle between two vectors A and B can be found
using the dot product or the cross product. First let’s use the dot
product. From Fig. A1 – 24, the cosine of the angle between A
A
1
B
and B is cos  B  ( A  ) so
A
A
B
The handedness of a coordinate system can also be
determined by positioning the axes of your right hand, as shown
in Fig A1 – 22. The thumb, index finger, and the middle finger
are lined up in their natural order along the x, y, and z axes,
respectively.
(A1 – 28)
Figure A1 – 22: Right-hand rule
cos 
1
AB
AB
Figure A1 – 24: The angle between two
vectors can be found using the dot product.
3
Next consider using the cross product to find the angle Key Terms
between the vectors A and B. Let’s set up two perpendicular
unit vectors i and j that are in the plane of the vectors A and B
so that i is in the direction of A. The unit vector k is Associative Rule of the Dot and Cross Products,
perpendicular to the plane. Then, A = Ai and B = Bxi + Byj. The Commutative Rule of the Cross Product, Commutative Rule
cross product of A and B is A  B  Ai  ( Bxi  By j)  ABy k. of the Dot Product, Cross Product, Distributive Rule of the
Cross Product, Distributive Rule of the Dot Product, Dot
The
length
of
is
then
equal
to Product, Left-Handed, Right-Handed, Scalar Product,
A B
Vector Product, Unit Vector Circle, Unit Vector Cross
A  B  AB y k  AB y  AB sin  , so
Products, Unit Vector Dot Products.
(A1 – 29)
sin  
1
AB
AB
Review Questions
1. What is another name for the dot product? What is another
The validity of Eq. (A1 – 29) is not restricted by the fact that i name for the Cross Product?
and j were originally set up to be in the plane of A and B. 2. Which of the rules of vector algebra involve the dot product?
Indeed, i and j were set up that way only for the purposes of 3. Describe how a component of a vector A in the direction of the
vector B is found.
generating Eq. (A1 – 29).
4. What is handedness?
5. How can you tell whether a coordinate system is right-handed
or left-handed?
6. Describe two ways to find an angle between two vectors.
4