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THE UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1141 Algebra
Section 2: - Introduction to Vectors.
You may have already met the notion of a vector in physics. There you will have thoughtof it as an arrow, that was used to represent a force. Adding forces corresponded to ‘addingarrows’. In this topic we are going to look at vectors from a geometric view point, althoughwe will include some examples based on simple ideas from physics.
One of the most powerful developments in Mathematics came from the simple idea of theco-ordinate plane. Indeed 2-dimensional co-ordinate geometry was crucial in the develop-ment of the Calculus. The obvious question arises as to how we can generalise this to higherdimensions. Vectors give us a way of generalising co-ordinate geometry into higher dimen-sions in a very straight forward manner.
Definition: A vector is a directed line segment which represents a displacement fromone point P to another point Q.
The word vector comes from the Latin veho (cf. vehicle), meaning to carry.
We represent a vector either using the notation−→
PQ or by using v. In the algebra notes(and in these notes), vectors are represented using bold letters, v. You should representvectors by underlining the letter, viz v. This is important, because you will need to carefullydistinguish between vectors, scalars (and later matrices).
v
P
R
S
−→
PQ
−→
SR
Q
w
A vector has both direction and length (or magnitude). Two vectors are equal ifthey have the same direction and the same magnitude. Hence in the diagram v = w.
We will denote the length of the vector v by |v|.
1
Two vectors are parallel if they have the same direction.
Position Vectors: We choose a fixed point O, in whatever dimensional space we hap-pen to be and call this the origin. The position vector of a point in any number ofdimensions will be represented by a vector from the origin to that point.
O
P
Hence the vector−→
OP in the diagram is called the position vector of the point P . A positionvector gives the position of a point in space, whereas a direction vector is simply a vectorhaving direction and magnitude (length).
Addition of vectors: To geometrically add two vectors there are two different methods(each important). If we think of a force vector, then, the obvious way to add two vectorsis to put them tip to tail and join the tail of the first to the tip of the second, as in thediagram.
vw
v + w
w
To add the vectors v and w, we move w and then complete the triangle. This method ofaddition is known as the triangle law of addition.
You can see from this that one could obtain the same vector by forming a parallelogramfrom the two vectors and taking the diagonal (often called the resultant) as the sum of thetwo vectors.
2
v
w
v + w
This method is known as the parallelogram law.
Subtraction of vectors is performed in a similar way:
w
v
w − v
To check this makes sense, add the vectors that are tip to tail, v+(w − v) = w as expected.Observe that the vector labelled w − v is not a position vector.
Thus if P and Q have position vectors v and w respectively, then−→
PQ= w − v. In gen-eral,
−→
PQ=−→
OQ −−→
OP .
w
v
w − v
O
P
Q
3
Example: Suppose ABCDEF is a regular hexagon with the vector p on the side AB andvector q on the side BC. Express the vectors on the sides: CD, DE, EF, FA and the diag-onals AC, AD, AE in terms of p and q.
CP: In a paralleogram ABCD,−→
AB= a,−→
AD= b, and M is the intersection of the diagonals.
Express, in terms of a and b the vectors,−→
MA,−→
MB,−→
MC ,−→
MD.
The Triangle Inequality:
Let us restrict ourselves, for the moment, to the plane.
Since the sum of any two sides of a triangle must exceed the third side, we can write
|u + v| ≤ |u| + |v|
for any vectors u and v.
u
v
u + v
4
Q: When do we have equality?
CP: By writing |u| = |u + v − v|, prove that |u + v| ≥ ||u| − |v||.
Scalar Multiplication:
We can multiply a vector by a scalar λ (generally just a real number).
This has the geometric effect of stretching the vector if λ > 1, stretching and reversingits direction if λ < −1.
u
λu
CP: a. Suppose that a and b are non-collinear (non-zero) vectors and that λa + µb = 0.
Deduce that λ = µ = 0.
b. Generalise this result to three non-coplanar (non-zero) vectors.
Commutative and Associative Laws:
The commutative law of vector addition states that a + b = b + a.
Geometrically this is obvious:
a
b
a + b = b + a
b
a
The associative law of vector addition states that a + (b + c) = (a + b) + c. Again thefollowing geometric proof (?) will suffice.
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a
b
b + c
a + b
c(a + b) + c =
a + (b + c)
Dependence:
Suppose we have a collection of vectors a1, a2, ...an. Suppose we start from the origin andmove along the vector, λ1a1, then along λ2a2, then along λ3a3 and so on until we move alongλnan. Can we choose the values of the λi’s NOT ALL ZERO so that we arrive back at theorigin?
In general- NO. However, if we can, then we say that the vectors a1, a2, ..., an are lin-
early dependent. In other words:
The vectors a1, a2, ..., an are linearly dependent if there exist real numbers λ1, λ2, ..., λn,NOT ALL ZERO such that
λ1a1 + λ2a2 + ... + λnan = 0.
In the diagram, the vectors a,b, c are linearly dependent:
2a
3b−c
0
Example: Suppose a,b are non-zero vectors. Show that p = a+b,q = b−c, r = a−b+c
and s = b + 1
2c are linearly dependent.
6
Simple Applications to Physics:
Ex: The center of mass of a system of particles is a specific point at which, for manypurposes, the system’s mass behaves as if it were concentrated.
Suppose masses m1, m2 and m3 are placed at the points A, B and C respectively, withposition vectors a,b and c.
Let M , with position vector m be the centre of mass.
b
A
B
C
M
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Geometric Proofs:
Ex: Prove (using vectors) that the line joining the midpoint of two sides of a triangle isparallel to the third side and half its length.
Ex: Prove that the medians of a triangle meet at a point G which divides each median inthe ratio 2:1.
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CP: Suppose P, Q, R are the midpoints of AB, BC, CA respectivley in ∆ABC. Let O beany point inside the triangle. Use vectors to prove that OP + OQ + OR = OA + OB + OC.Is the result still true if O is outside the triangle?
Co-ordinates:
Thus far, much of what we have done works in any number of dimensions. We are nowgoing to define n dimensional space and introduce a co-ordinate system in which to placeour vectors.
We take an n-tuple
a1
a2
.
.
an
of real numbers and think of each ai as lying on an axis xi. In
2 and 3 dimensions, we identify these axes as the XY and XY Z axes respectively, whichare mutually orthogonal.
The set of all such n-tuples will be called Rn.
We say that the vector−→
PQ in Rn has co-ordinates
a1
a2
.
.
.
an
, if we must move a1 units
along the x1 axis, a2 units along the x2 axis, and so on, when moving from P to Q.
Hence an n-tuple
a1
a2
.
.
an
in Rn can be interpreted as the position vector of a point P
in Rn.
For example, in R3, the point P in the diagram has position vector
231
.
9
P
231
0 =
000
We can then define the addition of two vectors (algebraically) in Rn by
a1
a2
.
.
an
+
b1
b2
.
.
bn
=
a1 + b1
a2 + b2
.
.
an + bn
and multiplication by a scalar λ to be
λ
a1
a2
.
.
an
=
λa1
λa2
.
.
λan
Multiplying a vector by a scalar λ merely stretches the vector (if λ > 1 ) or shrinks it if0 ≤ λ < 1. If λ is negative then the vector reverses direction. (Note: The algebraic definitionof addition agrees with the geometric definition.)
v
2v
0
−v
b
10
Note that we can now prove such rules as the commutative law algebraically, viz:
a + b =
a1
a2
.
.
an
+
b1
b2
.
.
bn
=
a1 + b1
a2 + b2
.
.
an + bn
=
b1 + a1
b2 + a2
.
.
bn + an
=
b1
b2
.
.
bn
+
a1
a2
.
.
an
= b + a.
Parallel Vectors.
Two vectors are defined to be parallel if one is a non-zero multiple of the other. Thatis, v is parallel to w if v = λw for some scalar λ 6= 0.
For example,
123
is parallel to
−2−4−6
.
Ex: Find the vectors−→
PQ, and−→
QP if P =
7−13
and Q =
21−3
.
Ex: Suppose that A =
(
00
)
B =
(
14
)
, C =
(
35
)
, D =
(
21
)
are the position vectors
for four points A, B, C, D. Prove that the quadrilateral ABCD is a parallelogram.
11
CP: ABCD is a parallelogram with vertices A, B, C, D which have the following position
vectors: A =
(
12
)
, B =
(
25
)
, C =
(
36
)
. Find the three possible position vectors of
D. (Hint: Keep it simple!)
Basis Vectors:
The standard basis vectors in R2 are the vectors
(
10
)
and
(
01
)
which are often
denoted by i and j. Observe that every vector in R2 can be written in terms of these basis
vectors. For example
(
1−2
)
can be written as i − 2j.
In 3-dimensions, the basis vectors, i, j,k are
100
,
010
,
001
.
Note that every vector in R3 can be expressed in terms of these basis vectors, viz:
a1
a2
a3
can be expressed as a1i + a2j + a3k.
In higher dimensions, we label the basis vectors as e1, e2,... and so on.Thus, in R
4, we have
e1 =
1000
, e2 =
0100
, e3 =
0010
, e4 =
0001
.
Once again, we can represent any vector in Rn in terms of the standard basis vectors in R
n.
Distances and Lengths:
Given a vector x =
(
a
b
)
in R2, we can use Pythagoras’ Theorem to compute the length
of this vector as√
a2 + b2. We use the notation |x| =√
a2 + b2. In R3, given a vector
x =
a
b
c
, we can see from the diagram that |OP | =√
a2 + b2 and then in ∆OAP we
have |OA| = |x| =√
a2 + b2 + c2.
12
X
O
b
a
c
A
P
In higher dimensions, we can define the length of a vector by generalising this formula, i.e.
Definition: A vector x =
a1
a2
...an
in Rn has length |x| given by |x| =
√
a2
1+ a2
2+ · · · + a2
n.
Ex: Find the lengths of a =
−132
and b =
1234
.
13
The distance between two points A and B in Rn will be defined as the length of the vector
−→
AB,
in other words, if A has position vector a =
a1
a2
...an
and B has position vector b =
b1
b2
...bn
,
then the length of−→
AB is
|−→
AB | = |b− a| =√
(b1 − a1)2 + · · · + (bn − an)2.
Ex: Find the distance between
(
1−2
)
and
(
3−4
)
and between
−125
and
36−1
.
The ‘length function’ | · |, (sometimes called a norm) has the following properties:
1. |a| ≥ 0.
2. |a| = 0 if and only if a = 0.
3. |λa| = |λ||a|, for λ ∈ R.
A vector which has unit length is called a unit vector. Any vector can be made into aunit vector by dividing by its length.
Ex: Find a unit vector parallel to the vector
2−31
.
14
CP: The point P =
x
y
z
makes angles α, β, γ respectively with the X, Y and Z axes.
Prove that cos2 α + cos2 β + cos2 γ = 1.
Ex: Suppose A and B are points with position vectors a and b.Find a vector (in terms of a and b) which bisects the angle AOB, where O is the origin.
15
Equations of Lines:
We seek to find the equation of a line in vector form. The vector equation of a line is aformula which gives the position vector x of every point on that line. This equation issometimes referred to at the parametric vector form of the line. I will generally just say‘vector equation’ of the line.
Suppose we have a line passing through the origin which contains a vector u in R2. Ev-
ery point on that line will have a position vector which is a multiple of u. Conversely,every multiple of u will correspond to the position vector of a point on that line. Hence theequation of the line can be written as x = λu where λ is any real number.
u
x = λu
0
For example, if u were the vector
(
23
)
then the equation of the line through u passing
through the origin would be x = λ
(
23
)
, λ ∈ R.
Another way of denoting the set of all real multiples of a given vector is to call it thespan of the vector. Thus we could write {λu : λ ∈ R} as span(u). This idea of span isextremely important.
16
Ex: In R2 what is the span of
(
10
)
? What is span
(
11
)
?
The advantage of this definition of the equation of a line is that it easily generalises to anynumber of dimensions. For example, the equation of the line in R
3 which passes through the
origin and is parallel to the vector
−13−6
is simply x = λ
−13−6
.
If the line does not pass through the origin, then we proceed as follows: To find the vectorequation of a line we need to know two things:
1. The position vector a of a point A on the line
2. The direction of the line, i.e. a vector−→
AB= b parallel to the line.
a
x = a + λb
0
b
λb
A
B
X
x
Thus, the span of b will give a line through the origin parallel to b and adding a will shiftthe line to its proper position. Thus we can find the position vector x of any point X on theline by going from the origin along the vector a and then moving along the line, by adding
some multiple of b until we reach X. Thus, the vector−→
OX is given by−→
OX=−→
OA +−→
AX and−→
AX is some multiple of b. Hence the equation of the line is x = a + λb, λ ∈ R.
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Ex: Find the vector equation of the line passing through the point P with position vec-
tor
2−35
and parallel to the vector
16−4
.
Ex: Find the vector equation of the line passing through the two points P, Q with position
vectors P =
−126
and Q =
4−23
.
CP: Let A, B, C be a triangle and X be the midpoint of BC. Let Y be the point on AC
which divides AC in the ratio 3 : 1. Let T be the point on AX that divides AX in the ratio6 : 1. Prove that T lies on the line through Y B.
Ex: Find the vector equation of the line in 2-dimensions with cartesian equation y = 2x+1.
18
Ex: Does the point
−123
lie on the line x =
3411
+ λ
214
?
Configurations:
In R3 two lines can
• meet at a point
• be parallel
• neither meet nor be be parallel.
CP: Two non-parallel lines in space that never meet are called skew lines. Prove that
the lines x =
2−14
+ λ
123
and x =
316
+ µ
2−31
are skew lines.
Line Segments:
The set
S =
x ∈ R3 : x =
123
+ λ
−23−1
,−1 ≤ λ ≤ 3
represents that line segment from
3−14
to
−5110
.
19
CP: Decide whether or not the line segment ℓ1: joining
0311
and
245
meets the
line segment ℓ2: joining
111
and
45−1
.
Cartesian Equations of the Line: Given the vector form of a line in 3-dimensions, we canwrite down the Cartesian equations (note the plural) as follows. For example, suppose the
vector equation is x =
−123
+ λ
3−21
. Recall that x is simply an abbreviation for
x1
x2
x3
. Hence, equating co-ordinates, we can write x1 = −1 + 3λ, x2 = 2− 2λ, x3 = 3 + λ.
Eliminating λ from these equations we have
x1 + 1
3=
x2 − 2
−2=
x3 − 3
1.
These are called the cartesian equations of the line. Clearly this can be done for any suchline and so the general form is
x1 − a
α=
x2 − b
β=
x3 − c
γ,
where (a, b, c) is a point on the line and (α, β, γ) is a direction vector of the line, providedthat none of the numbers α, β, γ is zero.
Hence in vector form this would be x =
a
b
c
+ λ
α
β
γ
.
The following example tells us what to do when one of the components of the directionvector is zero.
Ex: Convert x =
21−5
+ µ
20−1
into cartesian form.
20
Observe that two lines will be parallel if their direction vectors are parallel. Two directionvectors are parallel if and only if one is a nonzero multiple of the other. For example thelines
x =
156
+λ
32−1
and x =
−247
+λ
64−2
are parallel since their direction vec-
tors are
32−1
and
64−2
respectively and the second vector is simply a multiple of the
first. Observe also that the equations x =
156
+λ
32−1
and x =
156
+λ
64−2
represent the same lines, since they are parallel and pass through the same point.
Ex: Find the equation of the line passing through
1−25
and parallel to
x + 1
3=
y − 1
−2=
z + 6
−4.
Ex: Find the intersection (if possible) of the lines x =
−103
+ λ
12−1
and
x =
35−2
+ µ
21−3
.
21
CP: Prove that the angle bisectors of a triangle are concurrent.
Equations of Planes:
In 3-dimensions and higher, we can construct planes. Suppose we seek the vector equa-tion of the plane passing through the origin parallel to two given non-parallel vectors a andb.
a λa0
b
µb
X
P
Q
x
To reach any point X with position vector x on the plane we need to ‘stretch’ the vector a
to P and ‘stretch’ the vector b to Q in such a way that−→
OX=−→
OP +−→
OQ. Thus, x = λa+µb.Conversely, if we take the vector which results from adding a multiple of a and a multipleof b then this will be the position vector of a point on the plane.
Ex: Find the vector equation of the plane passing through the origin parallel to the vectors
15−3
and
−247
.
The plane generated by two such vectors is called the span of the two vectors. So for example,
22
the span of
15−3
and
−247
is simply the set
λ
15−3
+ µ
−247
: λ, µ ∈ R
.
This set is also referred to as the set of all linear combinations of the two vectors. Thus,given any two vectors a,b, span{a,b} = {λa+ µb : λ, µ ∈ R} and we say that x is a linear
combination of a and b if x = λa + µb for some particular λ and µ.
Ex: Describe the span of
123
,
−135
. Repeat for span
123
,
246
.
As with the equation of a line, to get the vector equation of a plane not through the origin,we simply shift the plane by adding any position vector of a point which lies on the plane.Thus to obtain the vector equation of a plane we need:
1. The position vector of a point on the plane.
2. Two non-parallel vectors which are parallel to (or lie on) the plane.
Ex: Find the vector equation of the plane passing through the point P with position vector
2−35
and parallel to the vectors
16−4
and
2−51
.
23
Ex: Find the vector equation of the plane passing through the three points P, Q, R with
position vectors P =
−1262
, Q =
4−23−1
and R =
17−25
.
Configurations:
In R3, two (distinct) planes can be
• parallel
• meet in a line
In R3, three (distinct) planes can be arranged so that
• the three planes are parallel
• two planes are parallel and the third plane is parallel to neither the first two.
• they meet at a single point
• they meet in a line
• none are parallel, but no point lies on all three planes.
In the next chapter we will learn how to analyse these scenarios algebraically.
24
Regions:
Ex: The set
S =
x = λ
120
+ µ
−124
, 0 ≤ λ ≤ 1, 0 ≤ µ ≤ 1
represents the parallelogram with vertices O,
−124
,
120
,
044
in R3.
Ex: The set
S =
x = λ
120
+ µ
−124
, 0 ≤ λ ≤ 1, 0 ≤ µ ≤ λ
represents the triangle with vertices O,
120
,
044
in R3
CP: Suppose that a,b, c are the position vectors of 3 non-collinear points in R3. Let
S = {x ∈ R3 : x = αa + βb + γc, where α, β, γ ≥ 0 and α + β + γ = 1}.
Describe S in geometric terms and give a proof that your claim is correct.
Cartesian Equation of a Plane:
As with lines, we can find the cartesian equation of a plane by eliminating the two pa-rameters λ and µ. This is generally quite fiddly to do algebraically. Later in this course, youwill see a much better method, but for the moment, we will do it by algebra.
Ex: Find the cartesian equation of the plane x =
123
+ λ
240
+ µ
−103
.
25
The procedure can be reversed to find the vector equation of a plane from the cartesianequation.
Ex: Find the vector equation of the plane 3x − 6y + 2z = 12.
Further Examples:
Ex: Find the intersection of the planes 2x + y − z = 10 and 3x + 4y + 2z = 29.
26
Ex: Show that the line x = λ
11−5−3
is parallel to the plane
x =
256
+ µ
426
+ ν
−135
.
Ex: Find the intersection of x =
123
+ λ
−14−2
and 2x + 3y − z = 29.
27