CHAPTER 3. VECTOR ALGEBRA
Part 1: Addition and Scalar Multiplication for Vectors.
§1. Basics. Geometric or physical quantities such as length, area, volume, temperature, pressure, speed, energy, capacity etc. are given by specifying a single numbers. Such
quantities are called scalars, because many of them can be measured by tools with scales.
Simply put, a scalar is just a number. Quantities such as force, velocity, acceleration,
momentum, angular velocity, electric or magnetic field at a point etc are vector quantities,
which are represented by an arrow. If the ‘base’ and the ‘head’ of this arrow are B and
−−→
H repectively, then we denote this vector by BH:
Figure 1.
Often we use a single block letter in lower case, such as u, v, w, p, q, r etc. to denote
−−→
−−→
a vector. Thus, if we also use v to denote the above vector BH, then v = BH. A
vector v has two ingradients: magnitude and direction. The magnitude is the length
−−→
of the arrow representing v, and is denoted by |v|. In case v = BH, certainly we
−−→
have |v| = |BH| for the magnitude of v. The meaning of the direction of a vector is
self–evident. Two vectors are considered to be equal if they have the same magnitude
and direction. You recognize two equal vectors in drawing, if their representing arrows
are parallel to each other, pointing in the same way, and have the same length
1
Figure 2.
For example, if A, B, C, D are vertices of a parallelogram, followed in that order, then
−→ −−→
−−→ −−→
AB = DC and AD = BC:
Figure 3.
Vectors are added according to the parallelogram law or the triangular law:
Figure 4.
Figure 5.
2
In figure 4, vectors u and v lying on the neighboring sides of a parallelogram are given
−→
−−→
by u = AB and v = AD. Then their sum u + v lies on the diagonal, given by
−→
u + v = AC. In Figure 5, vectors u and v lying on two sides of a triangle △ABC
−→
−−→
are given by u = AB and v = BC. Then their sum u + v lies on the remaining side,
−→
given by u + v = AC. Thus we have
−→ −−→ −→
AB + BC = AC.
The last identity can be generalized. Given a number of points, say A1 , A2 , . . . , An , we
−−−→ −−−→
−−−−−→ −−−→
have A1 A2 + A2 A3 + · · · + An−1 An = A1 An .
Figure 6.
−−−→ −−−→ −−−→ −−−→ −−−→
We only check this for five points, that is, for n = 5: A1 A2 + A2 A3 + A3 A4 + A4 A5 = A1 A5 .
−−−→ −−−→ −−−→
−−−−→ −−−→ −−−→
The quick way begins by observing that A1 A2 + A2 A3 = A1 A3 and A3 , A4 + A4 A5 = A3 A5
−−−→ −−−→ −−−−→ −−−→ −−−→ −−−→ −−−→
and finishes with A1 A2 + A2 A3 + A3 , A4 + A4 A5 = A1 A3 + A3 A5 = A1 A5 :
Figure 7.
3
The ‘step by step’ way described next is slower; however, it can be easily adapted to the
general case with arbitrary number of points:
−−−→ −−−→ −−−→
A1 A2 + A2 A3 = A1 A3
−−−→ −−−→ −−−→ −−−→ −−−→ −−−→
A1 A2 + A2 A3 + A3 A4 = A1 A3 + A3 A4 = A1 A4
−−−→ −−−→ −−−→ −−−→ −−−→ −−−→ −−−→
A1 A2 + A2 A3 + A3 A4 + A4 A5 = A1 A4 + A4 A5 = A1 A5
The picture below gives the idea of this ‘step by step’ approach:
Figure 8.
The next example is crucial because it describes a general situation that frequently occurs.
−→
−−→
−→
Example. Suppose that OA = u and OB = v. We are asked to determine AB in
terms of u and v.
Figure 9.
4
−→ −→ −−→
First, we write down OA + AB = OB. Hence
−→ −−→ −→
AB = OB − OA = v − u,
which is the answer. The reader should consult Figure 9 and reading this answer and
then keep the whole picture in mind.
We denote by 0 the zero vector. It has the property that 0 + v = v and v + 0 = v.
−→
Geometrically it is represented by a point. For any point A, we have AA = 0. Clearly
the magnitude of the zero vector is zero: |0| = 0.
Exercise 1. Let A, B, C, D be the vertices of a parallelogram; see Figure XXX above.
−−→
−−→
−−→
−−→
−→
−−→
(a) Write BD in terms of AB and AD. (b) Write BD in terms of CB and CD.
−−→
−→
−→
(c) Write BC in terms of AB and AC.
Exercise 2. Let A, B, C, D, E, F be the vertices of a regular hexagon and let O be
−→
−−→
−→
−→
its center; see Figure 10 below. Let a = OA and b = OB. Write OC and AE in
terms of a and b.
Figure 10.
−→
Exercise 3. Let A, B, C, D, E, F, G, H be the vertices of a cube and let u = AB
−−→
−→
−→ −→
−→
v = AD and w = AE; see Figure 11 below. Write F C AC and AG in terms of u,
v and w.
5
Figure 11.
Besides addition, another operations in vector algebra is scalar multiplication, that
is, multiplying vectors by scalars. Recall that a scalar is just a number, such as 2. A
vector v multiplied by 2, denoted by 2v is in the same direction as v but its length
is doubled. (You can tell that this is the most natural way to define 2v.) What about
(−2)v? Well, you can rewrite it as 2(−v), that is, −v multiplied by 2. Now −v is
obtained by turning v in the opposite direction. Now it is clear that the magnitude of
(−2)v is doubled and its direction is opposite to v.
Figure 12.
In general, given a scalar a and a vector v, the scalar multiplication av of v by
a can be defined. The magnitude of av is determined by |av| = |a||v|; (for example,
|2v| = |2|v| = 2v| and |(−2)v| = | − 2|v| = 2v|). The direction of av depends on the
sign of a: if a is positive, then av has the same direction as v, and if a is negative,
then av has the oppsite direction. Certainly, if a = 0 or v = 0, then av is the zero
vector.
The two operations in vector algebra, namely, addition and scalar multiplication, give
the so–called linear structure for a space of vectors (or simply called a vector space). The
6
theory based on the conceptual frame work of linear structure is called linear algebra,
which is a basic mathematical course for almost all diciplines which need mathematics.
−→
Exercise 4. Let ∆ABC be a equilateral triangle with O as its center, and let b = AB
−→
−→
and c = AC; see figure 13 below. Write AO in terms of b and c. (Hint: notice that
−→ −−→ −→
OA + OB + OC = 0.
Figure 13.
Problem 5. (a) Draw a picture to check the identity −(u + v) = (−u) + (−v). (b)
Draw a picture to convince yourself the validity of the identity a(bf u + v) = au + av for
a positive scalar a.
(c) Explain why the last identity also holds for negative a.
Problem 6. (a) Convince yourself that if vectors v and w are parallel and if v
is nonzero, then there is a scalar a such that w = av. Why do we need to assume
that v is nonzero here? (b) Let e1 and e2 be two nonzero vectors in the Euclidean
plane. Assume that these two vectors are not parallel to each other. Convince yourself
by drawing that any planar vector v can be written in the form a1 e1 + a2 e2 for some
scalars a1 , a2 . Why the scalars are uniquely determined by v?
Next we use Cartesian coordinates to study vectors. The reader is assumed to be
familiar with the Cartesian coordinate system. Let us recall some of its salient features.
Normally we draw the x–axis horizontally and the y–axis vertically. The intersection of
these two axis is denoted by O, called the origin of the coordinates system. Both axis
are scaled so that each point of the plane corresponds to a unique (ordered) pair of real
numbers, called the coordinates of this point. In the Figure 14 below we depict the point
P with coordinates (3, 2) so that we may write P = (3, 2).
7
Figure 14.
−→
When we have u = OP with P = (3, 2), we also call (3, 2) the coordinated of u and
we write u = (3, 2); see Figure 14 above.
But now we have a poblem: when we write (3, 2), does it mean the point P or
−→
the vector OP ? Isn’t there some conflict with our notation? Our answer is, whether
−→
(3, 2) should be P or should be OP is only our subjective opinion, that is, we can
interpret it in either way. All we need is to put a word in front we indicate our choice of
interpretation, such as “point (3, 2)” or “vector (3, 2)”.
The importance of using coordiantes for vectors lies in the fact that the two basic vector
operations, addition and scalar multiplication, are carried out in the “coordinatewise
manner”:
If v1 = (x1 , y1 ) and v2 = (x2 , y2 ), then v1 + v2 = (x1 + x2 , y1 + y2 ) and av1 = (ax1 , ay1 ).
(Here a is any scalar.) Instead of giving a routine but pretty boring proof os this basic
statement, we let you check its validity for some special cases to convince you the validity
of it in the following easy exercises.
−→
Exercise 7. In each of the following cases, use graphic paper to draw the vector AB
−→
−−→ −→
and check the identity AB = OB − OA: (a) A = (1, 2), B = (3, 5); (b) A = (−2, 5),
B = (3, 2); (c) A = (−1, −2), B = (1, 2); (d) A = (−3, −2), B = (−1, −4).
Exercise 8. (a) Find the midpoint M between P (−5, 4) and Q(9, 6).
(b) If M (2, 3)
is the midpoint of P (1, 1), what is Q? Use graphic paper to check your answer in both
parts.
8
Exercise 9.. Let A B and C be arbitrary points and let M be the midpoint of AB
−−→
and N be the midpoint of BC. Express the vector M N as a linear combination of
−→
−−→
u = AB and v = BC.
Exercise 10. Use graphic paper to draw the square in the Cartesian plane with its
vertices given as follows:
A(1, 1), B(−1, 1), C(−1, −1) and D(1, −1).
Locate the midpoints of four sides in your picture, give as follows:
P (1, 0), Q(0, 1), R(0, −1), S(−1, 0).
−→
−−→
Let u = OP and v = OB. Find each of the vectors
−→ −→ −−→ −→ −→ −→
OA, OC, OD, OQ, OR, OS
−→
as a linear combination of u and v; for example OQ = u + v. Note: the accuracy of your
drawing helps you to figure out the answer.
Exercise 11.. Use graphic paper to draw the regular hexagon in the Cartesian plane
with its vertices given as follows:
A(2, 0), B(1,
√
3), C(−1,
√
√
√
3), D(0, −2), E(−1, − 3), F (1, − 3).
−→
−→
−−→ −−→ −−→ −→
Let u = OA and v = OC. Find each of the vectors OB, OD, OE, OF as a linear
combination of u and v.
Problem 12. It is easy to check that the sum of vectors u, v, w is zero, if we can find
vectors x, y and z such that
u = y − x, v = z − y, w = x − z.
Prove that the converse is also true: if u + v + w = 0, then we can find vectors x, y and
z such that u = y − x, v = z − y, w = x − z.
Exercise 13. In each of the following parts, use graphic paper to plot the points A and
−→
−−→
B, draw the vectors a = OA and b = OB. Then draw the vectors
−−→
vt ≡ OPt = (1 − t)a + tb
−→
−−→
for given values of t. Finally, check the identities APt = tAB for the given values of t.
9
(a) A = (3, 7), B = (11, 3); given values of t: t = −1/4, 0, 1/4, 1/2, 3/4, 1, 5/4.
(b) A = (−1, −2), B = (8, 1); given values of t: t = −1/3, 0, 1/3, 2/3, 1, 4/3.
Part 2: The dot product
.
−−→
−−→
Let us begin with two vectors OA1 = v1 = (x1 , y1 ) and OA2 = v2 = (x2 , y2 ).
Assume that these two vectors are perpendicular to each other, that is, ∠A1 OA2 = 90o .
What can we say about them?
By assumption, ∆AOB is a right triangle. A right triangle suggests the Pythagoras
theorem, which is applied here without hesitation to get |A1 A2 |2 = |OA1 |2 + |OA2 |2 .
Now |OA1 |2 = |v1 |2 = x21 + y12 and |OA2 |2 = |v2 |2 = x22 + y22 . On the other hand, we
−−−→ −−→ −−→
have A1 A2 = OA2 − OA1 = v2 − v1 = (x2 − x1 , y2 − y1 ) and hence
¯
¯
¯−−−→¯2
2
|A1 A2 | = ¯A1 A2 ¯ = (x2 − x1 )2 + (y2 − y1 )2 = x21 − 2x1 x2 + x22 + y12 − 2y1 y2 + y22 .
Thus |A1 A2 |2 = |OA1 |2 + |OA2 |2 gives
x21 − 2x1 x2 + x22 + y12 − 2y1 y2 + y22 = x21 + y12 + x22 + y22 .
Upon cancelling of certain terms from both sides, we arrive at −2x1 x2 − 2y1 y2 = 0, or
x1 x2 + y1 y + 2 = 0. We conclude: that vectors v1 = (x1 , y1 ) and v2 = (x2 , y2 ) are
perpendicular if and only if x1 x2 + y1 y + 2 = 0. The expression x1 x2 + y1 y2 is called
the dot product of vectors v1 and v2 and is denoted by v1 • v2 . The above discussion
can be summarize as follows:
v1 ⊥ v2
if and only if
10
v1 • v2 = 0.
We give the formal definition of the dot product as follows:
Definition. The dot product, (or the scalar product, or the inner product)
v1 • v2 of two vectors v1 = (x1 , y1 ) and v2 = (x2 , y2 ) is defined to be
v1 • v2 = x1 x2 + y1 y2 ,
which is a scalar.
For example, if a = (3, 4) and b = (−6, 5), then a • b = 3 × (−6) + 4 × 5 = 2.
Exercise 14. In each of the following cases, find a • b. (a) a = (1, 2), b = (3, 4);
(b)
a = (5, 3), b = (3, −5); (c) a = (1 − s, s), b = (s, s − 1), where s is any number.
Exercise 15. (a) Check the identities u • v = v • u and
u • (av + bw) = au • v + bu • w,
where u, v, w are arbitrary vectors and a, b are arbitrary scalars. (b) Use the
results in the previous part to deduce (u + v) • (u + v) = u • u + 2u • v + v • v and
(u − v) • (u − v) = u • u − 2u • v + v • v.
We state two basic facts about dot products. The first one is the following identity
which holds for any vector v:
|v|2 = v • v
This is checked as follows. Let v = (x, y). Letting v1 = v2 = v = (x, y) in v1 • v2 =
x1 x2 + y1 y2 , we get v • v = x2 + y 2 , which is clearly |v|2 . The second fact (which is
already mentioned above) is about the perpendicular relation between two vectors v1 and
v2 :
v1 ⊥ v2 ⇔ v1 • v2 = 0.
(Here the symbols “⇔” stands for “means the same as”, or “is equivalent to”, or “if and
only if”.) Expressed in words, the first fact says that the square of the magnitude of a
vector is equal to the dot product of this vector with itself, and the second fact says that
two vectors are perpendicular to each other if and only if their dot product is zero.
Exercise 16.. In each of the following parts, draw the given points A, B, C, and
−→
−→
the vectors u = AB and v = AC. Compute u • v to find if these two vectors are
11
perpendicular. Check if your conclusion agrees with your drawing. (a) A = (4, 2), B =
(1, 3), C = (6, 8).
(b) A = (2, −1), B = (−1, −3), C = (−2, 5).
Let us give an “in depth” study about the wonderful little identity |v|2 = v • v. The left
hand side contains the term |v|, the length of a vector, which is a geometric quantity.
The right hand side is the dot product v • v is the result of an algebraic manipulation.
So this little identity actually plays a big role: it bridges a gap between two different
subjects, namely, geometry and algebra. This following example gives a good illustration
of this role it plays.
Example. In Euclidean geometry there is a highly nontrivial fact: the sum of the
squares of four sides of parallelogram is the sum of the squares of its diagonals. that is,
if A, B, C, D are vertices of a parallelogram, then
|AB|2 + |BC|2 + |CD|2 + |CA|2 = |AC|2 + |DB|2 ,
(∗)
that is, the sum of the squares of four sides is equal to the sum of the squares of two
diagonals; (|AB| stands for the length of the line segment AB, which is the same as the
distance between the points A and B). Here we use vector algebra to give a computational
−→
−−→
proof. To start the proof by introducing some notation: let u = AB and v = AD; see
the following figure.
Then |AB| = |CD| = |u|, |AD| = |BC| = |v| and hence |AB|2 + |BC|2 + |CD|2 +
−→
−→ −−→
|CA|2 = 2u|2 + 2|v|2 . On the other hand, we have AC = AB + BC = u + v and
−−→ −→ −−→
DB = AB − AD = u − v. Hence the right hand side of (∗) is equal to
−→
−−→
|AC|2 + |BD|2 = |AC|2 + |BD|2 = |u + v|2 + |u − v|2
= (u + v) • (u + v) + (u − v) • (u − v)
= u • u + 2u • v + v • v + u • u2u • v + v • v
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= 2u • u + 2v • v = 2|u|2 + |v|2 ,
which matches the left hand side.
The following exercise requires the efficient usage of the two basic facts stated above. If
you do it in the right way, it only needs one line to finish!
Exercise 17. Prove that if |u| = |v|, then the vectors u+v and u−v are perpendicular.
The identity |v|2 = v • v tells you that “| |” can be expressed in terms of “•”. The
following identity, often called the polarization formula, tells us a way to turn the table
around: express “•” in terms of “| |”:
1
u • v = (|u + v|2 + |u − v|2 ).
4
(2.1)
This identity is useful in some circumstances, especially in some high level subjects, but
you are not asked to memorize it!
Exercise 18. Verify (2.1) above.
What is the big deal about (2.1)? Well, for one thing, it shows that the inner product
u • v is unchanged under rotation. To see this, rotate the plane (with the origin (0, 0)
as the center of rotation) and assume that u and v turn into vectors u′ and v′ . Then the
parallelogram spanned by u, v turn into the parallelogram spanned by u′ and v′ . Hence
u + v and u − v after the rotation will become u′ + v′ and u′ − v′ respectively. Since
a rotation does not alter the magnitude of a vector, we have |u + v| = |u′ + v′ | and
|u − v| = |u′ − v′ |. The polarization formula tells us that u • v = u′ • v′ . This shows that
a rotation does not change the dot product u • v.
Now, let us assume that u and v are unit vectors, that is |u| = 1 and v| = 1. Let θ
(the Greek letter read as theta) be the angle between them. Rotate the plane in such a
way that u becomes u′ ≡ (1, 0). The angle between the vectors remains unchanged after
rotation. So the angle between u′ and v′ is still θ. A good picture of the situation tells
us that v′ = (cos θ, sin θ). So we have
u • v = u′ • v′ = (1, 0) • (cos θ, sin θ) = cos θ.
The above argument uses the assumption that u and v are unit vectors. Now we drop this
assumption. We will run into some complication if we follow the same line of argument.
13
To avoid this, we introduce new vectors
U=
u
,
|u|
V=
v
.
|v|
Now U is the unit vector in the direction of u and V is the unit vector in the direction
of v. So the angle between U and V is still θ. We can repeat the above argument for the
unit vectors U and V to arrive at
U • V = cos θ,
or
u
v
•
= cos θ.
|u| |v|
Beautifying the last identity, we arrive at the all important and lovely identity:
♥♥♥♥♥♥♥♥
u • v = |u||v| cos θ.
♥♥♥♥♥♥♥♥
Remember that θ here is the angle between u and v.
Exercise 19. In each of the following parts, find the angle between the given vectors
√
√
u and v. (a) u = (1, 3), |v = ( 3, 1) (b) u = (1, 2), v = (3, 1) (c) u = (3, 7),
√
√
v = (−2, 5) (d) u = (2, 1), v = (2 3 + 1, 3 − 2).
(As you can see, I always design questions in such a way that the final answers turn out
to be neat and pleasing, and sometimes the answer give you a pleasant surprise. This is
not the case with many questions from our textbook.)
14
Answers to Exercises
−→
−→
−→
−→
1.1. (a) AB = (2, 3), (b) AB = (5, −3), (c) AB = (2, 4), (a) AB = (2, −2).
−−→
−−→
−−→
1.2. (a) M = (2, 1), (b) First method: Note that P M = M Q, with P M =
−→ −−→ −−→ −−→ −−→
(2, 3) − (1, 1) = (1, 2) and hence OQ = OM + M Q = OM + P M = (2, 3) + (1, 2) = (3, 5),
which gives Q = (3, 5). Second method: Write Q = (u, v). Then 2 = (1 + u)/2 and
3 = (1 + v)/2. Solving this linear system of equations in u, v, we get u = 3 and v = 5.
Hence Q = (3, 5).
−−→
−−→ −−→
1.3. Notice that M N = M B + BN . Since M is the midpoint of AB, we have
−−→ 1 −→
−−→
−−→
M B = 2 AB. Similarly, we have BN = 21 BC. Hence
µ
¶
1 −→
−−→ 1 −→ 1 −−→
= AC
.
M N = AB + BC
2
2
2
−→
−→
−→
−−→
−→ −→ −→
1.4. OQ = u + v, OR = −OP = −u, OD = −v, OA = OP + OQ = 2u + v,
−→
−→
−→
−→
OC = −OA = −2u − v, OS = −OQ = −u − v.
−→
−−→
−−→
−−→
−−→
1.5. OB = u + v, OD = −u, OE = −OB = −u − v, OF = −v.
−→
1.6. That u+v +w = 0 means that we can draw a triangle ∆ABC such that u = AB,
−−→
−→
−→
−−→
v = BC and w = CA. Fix any point O in the plane. Let OA = x, OB = y and
−→
OC = z. We can check that this works!
−→
1.7. (a) AB = (11, 3) − (3, 7) = (8, −4).
−−−−→
P−1/4 = (1, 8), AP−1/4 = (−2, 1),
−−→
P0 = (3, 7),
AP0 = (0, 0),
−−−→
AP1/4 = (2, −1),
P1/4 = (5, 6),
−−−→
P1/2 = (7, 5),
AP1/2 = (4, −2),
−−−→
P3/4 = (9, 4),
AP3/4 = (6, −3),
−−→
AP1 = (8, −4),
P1 = (11, 3),
−−−→
P5/4 = (13, 2),
AP5/4 = (10, −5).
−−−→
We only give the detail for finding OP1/4 as follows:
µ
¶
3
1
3
1
1
3
1
−−−→ 3
OP1/4 = a + b = (3, 7) + (11, 3) =
× 3 + × 11, × 7 + × 3 = (5, 6).
4
4
4
4
4
4
4
4
−→
(b) AB = (8, 1) − (−1, −2) = (9, 3).
P−1/3 = (−4, −3),
−−−−→
AP−1/3 = (−3, −1)
15
−−→
AP0 = (0, 0),
−−−→
P1/3 = (2, −1), AP1/3 = (3, 1),
−−−→
P1/2 = (5, 0), AP1/2 = (6, 2),
−−→
P1 = (8, 1), AP1 = (9, 3),
−−−→
P5/4 = (11, 2), AP4/3 = (12, 4).
P0 = (−1, −2),
2.1. (a) 11,
(b) 0
(c) 0.
2.2. (a) u = (−3, 1), v = (2, 6), u • v = 0. (b) u = (−3, −2), v = (−4, 6), u • v = 0.
2.3. Here is the one line proof:
|u| = |v| ⇒ (u + v) • (u − v) = u • u − v • v = |u|2 − |v|2 = 0 ⇒ u + v ⊥ u − v.
As usual “⇒” stands for “implies” and “⊥” stands for “perpendicular to ”.
2.4. We start form the right hand side, without the factor 1/4 for tidiness: Now (2.1)
is clear.
√
√
2.5 (a) u • v = 2 3, |u| = 2, |v| = 2, cos θ = 3/2, θ = 60o (or π/3);
√
√
√
(b) u • v = 5, |u| = 5, |v| = 10, cos θ = 2/2, θ = 45o (or π/4).
√
√
√
(c) u • v = 29, |u| = 58, |v| = 29, cos θ = 2/2, θ = 45o (or π/4).
√
√
√
√
(d) u • v = 5 3, |u| = 5, |v| = 20, cos θ = 3/2, θ = 60o (or π/3).
16