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Chapter 5
Vector Geometry
In this chapter we will look more closely at certain ge- vector in Rn as a position vector as described in section
ometric aspects of vectors in Rn. We will first develop an 1.3 of Lay’s textbook. A position vector is just a pointer
intuitive understanding of some basic concepts by looking to a certain location in Rn. When using position vectors
at vectors in R2 and R3 where visualization is easy, then it is not necessary to make a firm distinction between a
wewill extend these geometric intuitions to Rn for any n. vector and its endpoint. For example, when we say that a
The basic geometric concepts that we will look at involve line is a set of vectors we mean that the endpoints of the
measurable quantities such as length, angle, area and vol- vectors lie on the line. If we want to stress the direction
ume. We also take a closer look at the two main types of of the vector we will usually represent it as an arrow. If
equations covered in this course: parametric-vector equa- we want to stress the particular location that the vector
tions and linear equations. is pointing to we will usually represent it by a point.
Webegin with a reminder. We defined a vector in Rn
as an n-tuple, i.e., as an n×1 matrix. This is an algebraic EXAMPLE 5.1. If A = (x ,x ,...,xn) and B =
definition of a vector where a vector is just a list of num- 1 2
(y ,y ,...,yn) are two points then the vector from A
bers. The geometric objects we will look at in this chapter 1 2 −→
should be seen as geometric interpretations of this alge- to B (represented by AB ) is defined as follows
braic definition. One difficulty that students encounter at 2y −x 3
1 1
this stage is that there are many different geometric in- 6y −x 7
−→ 6 2 27
terpretations that can be given to a vector. For example, AB=6 . 7
n 4 . 5
a vector in R can be interpreted geometrically as .
y −x
• an arrow starting at the origin. n n
Youcanthink of this as letting A be the origin of a new
• an arrow with a certain length and direction but no −→
fixed location. coordinate system and then the entries in AB give the lo-
cation of B relative to A. Or you can imagine translating
• a point (or more exactly, the coordinates of a point both A and B by subtracting A from both points so that
relative to some reference point). A is translated to the origin. Finally, you can think of
−→
• a directed line segment between two points. AB as an arrow from A to B.
So, for example, if we have P(1,5,2) and Q(7,7,0)
• a displacement (i.e., a translation). −−→ 27−13 263
This multiplicity of interpretations is a strength of the then PQ = 47−55 = 4 2 5. The entries in this vector
vector concept not a weakness. Vectors have many appli- 0−2 −2
cations and depending on the application one geometric indicate that when you travel from P to Q you move 6
units in the x direction, 2 units in the x direction and 2
interpretation may be more relevant than another but no 1 2
matter what geometric interpretation is chosen the under- units in the negative x3 direction. These entries express
−−→
lying vector algebra remains the same. We will interpret a the location of Q relative to P. If PQ is drawn with the
1
2 Chapter 5. Vector Geometry
1
intial point at the origin then the terminal point would be
(6, 2, -2).
±1 1 2 3 4 5 6
0
We will usually represent a vector as an n × 1 matrix
but there is another standard way of representing vectors
that is frequently used. In R2 we define ±1
» – » –
i = 1 j = 0
0 1 ±2
2
It then follows that any vector in R can be written as
» – » – » – ±3
a = a + 0 =ai+bj
b 0 b (5,±3)
Similarly in R3 we define ±4
2 3 2 3 2 3
1 0 0 Figure 5.1.
4 5 4 5 4 5
i = 0 j = 1 k= 0
0 0 1
and then any vector in R3 can be written In R3 a similar argument based on the Pythagorean
2a3 Theorem gives q
4b5=ai+bj+ck kuk = u2 +u2 +u2
c 1 2 3
YoushouldrealizethatinR2 thevectorsiandjarejust 2u13
the vectors which we have called e1 and e2, the standard for any vector u = 4u25.
basis of R2. Similarly in R3 the vectors i, j and k are the u3
Wecan extend the above formulas to Rn by defining
standard basis of R3.
q 2 2 2
kuk = u1 +u2 +···+un
5.1 Distance and Length 2u13
6u27
Notice that if u = 6 7 is any vector in Rn then
6 . 7
The first geometric concept we want to look at is the 4 . 5
.
the length of a vector. We define this to be the usual un
Euclidean distance from the intial point (the origin) to 2 3
the end point of the vector. The length any vector v u1
in Rn will be represented by kvk. This quantity is also ˆ ˜6u27
T u u · · · u 6 7 2 2 2
referred to as the magnitude or norm of v. u u= 1 2 n 6 . 7 = u1 +u2 +···+un
» – 4 . 5
Let u = u1 be a vector in R2. The length of this .
u2 un
vector would be the distance from the origin (0,0) to the We then have the following concise formula which is
point (u1,u2) and this is given by the Pythagorean The- valid for vectors in Rn for all n
orem as q
2 2 2 T
kuk = u1 +u2 kuk =u u
» – EXAMPLE5.3. Let u be any vector in Rn and k be a
EXAMPLE 5.2. Let u = 5 . Figure 5.1 shows u scalar then
−3 2 “ T”
and by the Pythagorean Theorem we can find the norm kkuk = ku (ku)
of u as q
√ 2 T
kuk = 52 +(−3)2 = 34 =k u u
=k2kuk2
5.1. Distance and Length 3
Taking square roots then gives
kkuk = |k|kuk
This shows that multiplying any vector in Rn by a scalar
k scales the length of the vector by |k|. We will sometimes
make a distinction between the sense of a vector and the
direction of a vector. When a vector is multiplied by a
negative scalar the reversal of the arrow is described by
saying the sense has been reversed but the direction has
stayed the same.
Definition 5.1. The distance between two vectors u
and v in Rn is defined as ku−vk.
EXAMPLE 5.4. The distance between u = i+k and
v=j−kis q √
2 2 2
ku−vk=ki−j+2kk= 1 +(−1) +2 = 6
Unit Vectors
Aunit vector is a vector whose length is 1.
If u is any non-zero vector in Rn then 1 u is a unit
kuk 2
vector. This can be seen by applying the formula kvk =
vTv to the vector 1 u. This gives:
kuk
„ 1 «T „ 1 « 1 T
kuku kuku = kuk2u u
1 2
= kuk2kuk
=1
Theprocess of multiplying a vector by the reciprocal of
its length to obtain a unit vector is called normalization.
Notice that this procedure doesn’t alter the direction or
sense of the vector.
2 2 3
6 2 7
EXAMPLE5.5. Normalize the vector v = 6 7.
4 0 5
√ √ −1
Wehave kvk= 4+4+0+1= 9=3so
2 2 3 2 2/3 3
1 6 2 7 6 2/3 7
6 7=6 7
3 4 0 5 4 0 5
−1 −1/3
is a unit vector parallel to v. Note: Just to avoid any pos-
sible confusion, when we say that two non-zero vectors,
u and v, are parallel we mean that they have the same
direction. Each one is a scalar multiple of the other.
4 Chapter 5. Vector Geometry
» –
Problems a. 3 d. 3i −5j+2k
» – 4 » –
−→ 3 233 e. 1
1. If A = (4,−2) and AB = −1 what is B? b. 445 »t –
−→ 2−63 25 3 f. cost+sint
2. If B = (5,−4,7) and AB = 4 2 5 what is A? 1 cost−sint
2 6 1 7
c. 6 7
3. Find the length of the following vectors: 4−15
» 3 – »cosθ– −1
a. −2 d. sinθ −→ −−→
9. If kABk = 5 and kBCk = 3 what are the possible
2 3 2 3 −→
1 cos(s)sin(t) values for kACk?
b. 4 4 5 e. 4cos(s)cos(t)5 » – » –
10. Let u = cos(s) and v = cos(t) . These are two
−1 sin(s) sin(s) sin(t)
2 3 f. i + j + k unit vectors in R2. Show that the distance from u
4 p
637 to v is 2−2cos(s−t) 2 3
c. 6 7 g. 4i − j − 3k
425 √ 2 3 4v15
1 h. 1−2t i+tj+tk 11. Prove that in R the length of v2 is given by
2 3 q v3
1 v2 +v2 +v2.
617 1 2 3
4. Let v = 6 7 be the vector in Rn all of whose 12. True or False:
6.7
4.5
. 2 T
1 a. kuk =uu
entries are 1. What is kvk? 2 T
b. kuk =u u
5. Find the lengths of the sides of triangle ABC where 2 T
the vertices are given by c. k2uk = 4u u
2 T T
d. ku+vk =u u+v v
a. A(0,0),B(3,3),C(5,−1) C(0,0,1) e. If kuk = kvk then ku+vk = kuk+kvk.
b. A(−1,2),B(1,5),C(3,1) 2 T T
d. A(3,1,2), f. kAuk =u A Au
c. A(1,0,0), B(4,−1,−2), 13. Under what conditions will ku+vk = kuk+kvk?
B(0,1,0), C(−2,0,1) T
14. Suppose A is an n×n matrix such that A A = I.
−→ » – −−→ » – Let v be any vector in Rn. Show kAvk = kvk
6. a. If AB = 2 , BC = −3 , and A is the » –
−1 5 15. Let A = cosθ −sinθ . Show that if v is any
point (3,7) what is C? Draw a diagram illus- sinθ cosθ
tating this problem. vector in R2 then kAvk = kvk.
2 3 2 3
−−→ 3 −→ 2
4 5 4 5
b. If PQ = 1 , QR = −1 , and R is the
0 1
point (−3,5,2) what is P?
213 2 k 3
4 5 4 5
7. Let u = 2 and v = k+1 . Use calculus to
4 k+2
find the value of k for which the distance from u to
v is a minimum.
8. Find a unit vector parallel to each of the following
vectors:
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