chapter 1 introduction to vectors
TRANSCRIPT
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MATH1131 Algebra, 2014
Group 4 Mon 9 am, Thu 12 noon
Dr. Chi Mak
School of Mathematics and StatisticsUniversity of New South Wales
Chi Mak (UNSW) MATH1131 Algebra 1 / 50
Contact Details
Office: Red Centre Room 4073Email: [email protected]
Consultation time: TBA
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Chapter 1 Introduction to Vectors
A scalar quantityis a quantity which can be specified by a number.
A vector quantity is a quantity which is specified by a magnitude (orlength) and a direction.
Naturally, two vectors are equal if and only if they have the samemagnitude and direction.
In notation, a vector is usually denoted by a boldface lower case letter, for
example v (v orvwhen written by hand). The magnitude ofv is denotedby|v|.An arrow from a point Ato another point B is evidently a vector. Themagnitude is the distance between Aand Band the direction is Ato B.
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This vector is denoted byAB, and we call Athe initial point(the tail) andB the terminal point(the head). With a choice of a reference point, the
origin O, any point Acan be specified byOAwhich is called the position
vectorofA.
Example
Draw the vector with initial point Cwhich is equal to the vectorAB.
Draw two vectors which equal a.
Solution
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Definition (Addition of Vectors.)
Translate the vectors a and b so that the initial point ofb lies at theterminal point ofa. The arrow which goes from the initial point ofa tothe terminal point ofb represents the vector a+b.
Remark
If we draw the vectors a, b, and a+b from the same initial point,
this initial point and the three terminal points form a parallelogram.AB+
BC=
AC
Vector addition satisfies the commutative law and associative law:a + b = b + a, (a + b) + c = a + (b + c)
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The Zero Vector and the Negative of a Vector
The sum of two vectors of equal magnitude and opposite direction is anarrow of zero magnitude. Logically, the direction will have no meaning.For the closure of vector addition (i.e the sum of 2 vectors is always avector), this quantity is considered as a vector.
Definition (Zero Vector.)The zero vector is the vector 0 of magnitude zero (direction undefined).
For any vector a, we have a+0= 0+a= a.
Given a vector b, there is a unique vector of the same magnitude butopposite direction. We call this the negative ofb.
Definition
The negativeofb, writtenb is a vector such thatb+ (b) =b+b= 0.Ifa and b are vectors, we define the subtraction by
a b= a+ (b).Chi Mak (UNSW) Vectors 6 / 50
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Geometrically, we can do a b bya+ (b). On the other hand, we cantranslate a and b so that they have the same initial point, then a b isthe vector from the terminal point ofb to the terminal point ofa.
a b is also a diagonal of the parallelogram mentioned in the Remark onp.5.
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Example
Draw the vectorsAB,AB+ CD, andABCD.
Solution
Example
Algebraically, AB= , PQ+ QR=PQ+
QR+
RP= , and
ABAC =
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Definition (Multiplication of a Vector by a Scalar.)
Let a and b be vectors and let R.1 If >0, then a is the vector whose magnitude is |a| and whose
direction is the same as that ofa.
2 If= 0, then a= 0.
3 If
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Application to Geometry
Example
Let O, A, Bbe points andOA= a,
OB = b. Suppose that P divides
ABin the ratio 3 : 1. FindOP in terms ofa and b.
Solution
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Example
Prove that the midpoints of the sides of a quadrilateral form aparallelogram.
Solution
Attempt Problems 1.1Chi Mak (UNSW) Vectors 12 / 50
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Two-dimensional Analytic Geometry and R2.
With a chosen coordinatesystem, it is clear that we
can specify the vectorAB
by two numbers 1, 2 (in
order) and the vectorCD
by3, 1.
We represent the vectors in column form as
12
and
31
, called
column vectors. Note that write the vectors as columns or rows is a
matter of choice. In this course and the subsequent ones, we use columnvectors to represent vectors in R2 (2-dimensional space), R3
(3-dimensional space), and in general, Rn. Beware that some authorsprefer row vectors and be careful about what you read from other booksand the Internet.
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We denote the vectors with unit length in the direction ofx-axis andy-axis by i and j. We call these two vectors the standard basisvectors forR
2. With these basis vectors we can write all vectors in R2 in terms of i, j.
For the vectors on the previous pages:
AB= andCD= .
It is clear that ifa=
a1a2
, b=
b1b2
R2 and R, we have
a+b=a
1+b
1a2+b2
and a=a
1a2
.
By Pythagoras theorem,|a|=
a21+a22.
We call the set of all column vectors R2 because each vector is composedof 2 real coordinates. Formally,
R2 =
x1x2
:x1, x2 R
.
We can use the vector
a1a2
to represent the position vector of the point
(a1, a2).Chi Mak (UNSW) Rn and Analytic Geometry 14 / 50
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Example
Town A is in the direction N 25 E and 3 km from O; B is N58 W and 4km from A. What are the distance and the bearing ofB from O?
Solution
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Three-dimensional Analytic Geometry and R3.
A three-dimensional coordinate system consists of an origin and threemutually perpendicular axes, the x, y and z-axes using the same unitlength. We shall use the right-handed system using your right hand, ifyou point your index finger in the x-direction, your middle finger in they-direction, then your extended thumb will point in the z-direction.
The point A, as a 3-tuple, is(x1, y1, z1). The position vectorofA is given by
OA=
x1y1
z1
.
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We also choose the unit vectors i, j, and k in the x, y and z directions,
respectively. Any vector v=ABcan be specified by the directed distances
v1, v2, v3 from A to Balong the x, y and z directions. The vector can bewritten in terms of the standard basis vectors i, j, k or as a column vector.
v= v1i+v2j+v3k=v1
v2v3
.
It is also clear that ifa=
a1a2
a3
, b=
b1b2
b3
R3 and R, we have
a+b=a1+b1a2+b2
a3+b3
and a= a1
a2a3
.
By Pythagoras theorem, a is |a|=
a21+a22+a
23.
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Example
Let u=
13
2
and v=
211
.
a) Find|2u v|.b) Find u+ 3v in terms of the standard basis vectors i, j, k.
Solution
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Solution (Continued.)
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Vectors in Rn
Naturally, we can generalise vectors in 2 or 3 dimensional space to vectorsin the n-dimensional space as column vectors with n real coordinates.
Definition
Let n be a positive integer. The set Rn is defined by
Rn =
a1...
an
:a1, . . . , an R
.
An element
a1...
an
in Rn is called an n-vector or simply a vector; and
a1, . . . , an are the componentsof the vector.Two vectors in are said to be equalif and only if all the correspondingcomponents are equal.
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In 2 and 3 dimensional spaces, we already have the notion of length, whileaddition and scalar multiplication are results from the arrow arithmetics.In Rn forn>3, we need to define these notions.
Definition
Let a=
a1
...an
, b= b1
...bn
be vectors in Rn and be a real number.
Vector addition is defined by a+b=
a1+b1...
an+ bn
.
The scalar multiplication is defined by a=
a1
...an
.
The length of a vector a is defined as|a|=
a21+ +a2n.
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Definition
The standard basis vectors for Rn are
Example
320
2
+
23
31
= , 3
e26
= ,
10.42
+
2e
10
Example
The vector a= R4 has the property that for all v R4,
a+v = v+a = v? This vector is called the zero vectorin R4.
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Example
Find the vector v such that v+
13
12
=
0000
.
Remarks
Let a, b, c Rn and , R.1 a+b= b+a.
2 (a+b) +c= a+ (b+c).
3 The vector, 0, with all entries 0 satisfies a+0 = 0+a = a.
4 For any a
Rn, there exists
a
Rn such that
a+ (a) = (a) +a= 0.5 (a) = () a.
6 ( + ) a= a+ a.
7 (a+b) =a+ b.
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Similar to arrows, for any a, b Rn, we define a b= a+ (b).Example
23
14
22
01
=
Geometrically, a vector a=
a1...
an
Rn defines the point A(a1, . . . , an) in
the n-dimensional space. That is position vector ofA isOA= a.
In general, suppose that a and b are position vectors of the points A andB, respectively. The vector with initial point Aand terminal point B is
given by AB=
OBOA= b a.
The distance between two points A and B is defined to beAB.
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Direction is a bit subtle.
Definition
Two non-zero vectors a and b in Rn are said to be parallel ifb= a forsome non-zero real number .They are said to be in the same direction if >0.
Example
Let A= (1, 3, 2, 4), B= (4, 2, 4, 8), C = (5,2,1, 8),D= (2,1,3, 4) be four points.
a) FindAB and
AB.b) Are A, B, C collinear?
c) Is ABCDis a parallelogram?
Solution
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Solution (Continued.)
Attempt Problems 1.2, 1.3
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Parametric Vector Form of a Line.
In the xy-plane, a line can be written as a linear equation y=mx+borax+by+c= 0, the Cartesian equations of the line. The unknownsrepresent the Cartesian coordinates of a variable point on the line. A point
is on the line if and only if the coordinates of the point satisfy theequation. However, we shall see that in higher dimensions, we cannotwrite a line as a single Cartesian equation.
On the contrary, we can alwaysexpress a line as a single vectorequation. The unknown vector of theequation represents the positionvector of a variable point on the line.
The position vector x Rn of anypoint Pon a line through the originparallel to a vector v Rn is of theform v for some scalar R.
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Definition
The expression x= v for R is called a parametric vector form of theline through O parallel to v.
Remarks
A point A with position vector a=OA lies on the line x= v iff(abbreviation for if and only if) we can find a real value k for theparameter such that a= kv.
We called this equation a parametric vector form because it is avector equation involving a parameter.
Parametric vector form of a line is not unique.
The set{x Rn :x= v, for some R} is called the span ofvand is denoted by span(v). Geometrically, the points whose positionvectors are in this set form a line the line in Rn spanned by v.
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Example
Find two parametric vector forms of a line in R4 spanned by
1
32
2
.
Solution
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A line through A and parallel to v
Definition
Line in Rn. A line in Rn is any set of vectors of the form
S={x Rn
:x = a+ v, for some R},where a and v=0 are fixed vectors in Rn. The expression,x= a+ v, R, is called the parametric vector form of the line.
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Example
Find a vector parallel to the line
x=
21
1
+
123
, R.
Write three points on the line.
Solution
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Example
Find a parametric vector form of a line through (1, 2, 5) parallel to
31
2
.
Solution
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Example
Find a parametric vector form of the line through the point (1, 1, 3, 1) andparallel to the line through (0, 1, 1, 2), (1, 2, 3, 1).
Solution
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Cartesian Form of a Line.
From a parametric vector form of a line in Rn x= a+ v, we can getthe Cartesian equation by setting variables for the coordinates theneliminating the parameter.
Example
Convert x= 123
+ 243
, R to Cartesian form.
Solution
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If a parametric vector form of a line is x= a+ v, where
x=
x1...
xn
, a=
a1...
an
, v=
v1...
vn
,
and none of the coordinates ofv is 0, the Cartesian form of the line is
x1 a1v1
= = xn anvn
.
Conversely, if the Cartesian form of a line is given as above (none of thecoordinate variables is missing and the coefficient of each variable is 1),
then (a1, . . . , an) is a point on the line and
v1...
vn
is a vector parallel tothe line.
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Cartesian form to a parametric vector form.
Example
Convert the following lines to parametric vector forms.a) 2x+ 3y= 4 b) x= 2
Solution
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Example
Convert the following line in R4 to parametric vector form
x1 12
= x2
3 =x4+ 2 and x3= 1.
Solution
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Example
Convert the following line to parametric vector form
3x1 12
= x2
3 = 2x3+ 2.
Solution
Attempt Problems 1.4
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Parametric Vector Form of a Plane.
Span of a non-zero vector is a line through the origin. How about twonon-parallel vectors, neither of which is 0?
Definition (Linear Combination and Span of Two Vectors.)
A linear combination of two vectors v1 and v2 is a sum of scalar multiplesofv1 and v2,
1v1+ 2v2,
where 1 and 2 are real numbers.
The span of two vectors v1 and v2, written span(v1, v2), is the set of alllinear combinations ofv1 and v2, that is
{x: x = 1v1+ 2v2, for some 1, 2 R}.
A plane through the origin is the span of any two (non-zero) non-parallelvectors.
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Two non-zero, non-parallel vectors will span a plane.
Hence the parametric vector equationof the through A(with positionvector a) and parallel to v1, v2 is
x= a+ 1v1+ 2v2,
for1, 2 R.
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Example
Describe geometrically the following sets.
a) span
13
2
,
264
.
b) x= 1
1302
+ 22
31
4
; 1, 2 R.
c) x=
12
3
+ 1
13
2
+ 2
26
4
; 1, 2 R.
Solution
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Solution (Continued.)
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Example
Find a parametric vector form of the plane through (1, 2, 3, 1) and
parallel to
2130
and the line 3x1 1
2 =
x2
3 = 2x3+ 2 =x4.
Solution
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Cartesian form of a plane in R3.
In the special case ofR3, the equation of a plane can be written as a linearequation in the three coordinate variables,
ax+by+cz=d.
Example
Find a parametric vector form of the plane
x+ 2y 3z= 4.
Solution
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Solution (Continued.)
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Example
Find the Cartesian form of the plane
x=
12
3
+ 1
10
2
+ 2
21
3
; 1, 2 R.
Solution
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Solution (Continued.)
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Example
Find the intersection of the line
x=
11
2
+
11
2
, R
and the plane 2x y+z= 8.Solution
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Solution (Continued.)
Attempt Problems 1.5
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End of Chapter 1