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Mathematics 1229A/B

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Math 1229A/B
Unit 4:
Vectors in ℜ m
(text reference: Section 2.1)
▯V. Olds 2010 Unit 4 51
m
4 Vectors in ℜ
We’ve learnt about ℜ and ℜ , which you’ve seen before. Now, we’re going to extend some of
m
the same ideas to spaces with more than 3 dimensions. We refer to m-dimensional space as ℜ or
m-space. “But”, you’re saying to yourself, “how can you have more than 3 dimensions?”. Well, for
starters, time is often referred to as the fourth dimension of physical space. The world we live in
has depth, breadth and height, and also time. Where you’re sitting right this minute is a particular
location that could be described by x-, y- and z-coordinates. But you’re in that location at this
particular instant. At other times you weren’t. However the location still existed. So time is another
axis along which space can be measured.
And after that? Well, for higher dimensions, it’s just much easier if you don’t try to relate it
5
to physical space. Because it tends to make your brain hurt if you try to actually picture ℜ or
ℜ or ℜ , and so forth. But to mathematicians, that’s no reason not to talk about them. The-
oretically, there could be more dimensions. And even if there aren’t, there are still situations in
which it is useful to use constructs which correspond to things we’ve already seen (points, vectors,
etc.) but with more parts to them, as if they were from a higher-dimensional space. For instance, a
company which produces 10 diﬀerent products might have a use for considering a “point” with 10
coordinates, where each coordinate corresponds to a diﬀerent one of the 10 products and indicates,
for instance, how many of that product are in stock at the moment. There are many uses of the
kind of mathematical constructs we’ve been working with, other than literally as points in space
or directed line segments, in which the kind of arithmetic we’ve been doing still makes sense. And
if these constructs aren’t being used to represent physical space, then there’s no reason that they
would need to be limited to 3 dimensions.
m
We start our study of ℜ with a number of deﬁnitions, which will all look familiar to you from
ℜ and ℜ . These deﬁnitions just extend the construct we’ve been using, vectors, to higher dimen-
sional space, and then extend (most of) the things we were doing with them, i.e. tell us how to
do arithmetic etc. with these higher-dimensional vectors. And along the way we’ll also see some
2
theoretical results. But the arithmetic works just the same in higher dimensions as it did in ℜ and
ℜ , and the theorems will look familiar too, so it will be really easy for you to learn this. There’s
m
not much that’s really new in this unit. It’s just a matter of getting used to the idea ofor m > 3.
Deﬁnition: For any positive integer m ≥ 2, we use ℜ , also called m-space, to denote
the set of all ordered m-tuplesu = (u1,u 2...,um), where each value u iay be any real
number (i.e. for alliu ∈ ℜ). For anyu ∈ ℜ , we refer to u as a vector, or an m-vector.
The numbers u ,1u ,2..., um are called the components of the m-vector ▯ u.
Note: As we have already learnt, the terminology is sometimes a bit diﬀerent for ℜ and ℜ . For
2
instance, we generally refer to an ordered pair or (rarely) ordered duple in ℜ and an ordered triple
in ℜ , rather than saying ordered 2-tuple or ordered 3-tuple. Also note that, as we have been doing
2 3
with vectors in ℜ and ℜ , when a vector is given a certain name, such as ▯u, the components (when
not speciﬁc numbers) are assumed to be named with the same letter, with subscripts. For instance
referring to the vector u we would understand that the ﬁrst component is called u ▯1, the second
component is called u , and so forth. The only exception to this is, again as we’ve already seen,
2 2 3
that often in ℜ we use ▯x = (x,y) and likewise in ℜ we use ▯ x = (x,y,z).
Examples: (1,2,3,4) and (−0.2,61.3,0.04,0) are both vectors in ℜ . (5,3,6,3,−1,−10) is a 6-vector.
5
Ifv ∈ ℜ , then ▯v = (v1,v2,v3,v4,v5). 52 Unit 4
m
Deﬁnition: The zero vector in ℜ is the m-vector whose components are all 0. Also,
two m-vectors are equal if and only if their corresponding components are identical.
m
That is, for anu,▯v ∈ ℜ , u = v if and only i1 u =1v 2 u =2v , ..., amd= vm.
Note: Only vectors which have the same number of components can be equal. That isu can only
be equal tov if they are both m-vectors, for the same value of m. (And of course will only actually
be equal if their corresponding components are identical.)
4 12
Examples: (0,0,0,0) is the zero vector in ℜ , and for 0we have 0 = (0,0,0,0,0,0,0,0,0,0,0,0).
If we haveu = (1,2,c,3,−1) and ▯v = (a,2,8,d,−1) and we know that ▯u = v, then we must have
a = 1, c = 8 and d = 3.
Deﬁnition: The distance between two m-vectors ▯ u and v is given by the distance
formula:
p
d(u,v) = (v1− u1) + (v2− u 2 + ... + (m − um)2
Also, the magnitude or length of an m-vector, sometimes referred to as the norm of
the vector, is
p
||u|| = (u1) + (u2) + ...(m )2
An m-vector whose magnitude is 1 is called a unit vector.
Note: These are precisely analogous to the notation, terminology and formulas we had for the dis-
2 3
tance between 2 vectors, and the magnitude of a vector, in ℜ and in ℜ .
Example 4.1.
(a) Show that (0,0,0,0,−1,0,0,0) is a unit vector.
(b) Show that there are no real numbers a and b for whiu = (1,1,a,b) is a unit vector.
Solution:
(a) We have
p
||(0,0,0,0,−1,0,0,0)||= 02 + 0 + 0 + 0 + (−1) + 0 + 0 + 0 2
√
= 0 + 0 + 0 + 0 + 1 + 0 + 0 + 0
√
= 1 = 1
so this vector is a unit vector.
(b) We ﬁnd the magnitude of u:
p p p
|u|| = ||(1,1,a,b)|| =1 + 1 + a + b =2 1 + 1 + a + b = 2 + a + b2
2 2 2 2 √ 2 2 √
Since a ≥ 0 and b ≥ 0 for √ll real numbers a and b, then 2+a +b ≥ 2 and so 2 + a + b ≥ 2.
But then we see that u|| ≥ 2 > 1, so u||▯=1 and u cannot be a unit vector. Unit 4 53
Example 4.2. Ifu = (1,2,3,4) andv = (0,−1,12,7), ﬁnd the distance betweeu and v and also
ﬁnd the magnitude of each.
Solution:
The distance betweenu andv is
p p
du,v) = (0 − 1)+ (−1 − 2) + (12 − 3) + (7 − 4) = (−1)2+ (−3) + (9) + (3)2
√ √
= 1 + 9 + 81 + 9 = 100 = 10
For the magnitudes ou and v we get
p √ √
|u|| = ||(1,2,3,4)|| =1 + 2 + 3 + 4 = 1 + 4 + 9 + 16 = 30
p √ √
and |v|| = ||(0,−1,12,7)|| =02 + (−1) + (12) + 7 = 0 + 1 + 144 + 49 = 194
Example 4.3. How long is the 5-vector (1,1,−1,0,−1)?
Solution:
We are being asked to ﬁnd the length, i.e. the magnitude, of this vector. We get:
p √ √
||(1,1,−1,0,−1)|| = 12+ 1 + (−1) + 0 + (−1) = 1 + 1 + 1 + 0 + 1 =4 = 2
We see that the vector (1,1,−1,0,−1) is 2 units long.
Deﬁnition: For any scalar c and any m-vectu, the scalar multiple ou by c is
cu = (cu1,cu2,...,mu )
For the scalar −1, the scalar multiple u ∈ ℜy ▯by −1 is called the negativeu,f ▯
denoted −u, so that
−▯u = (−u1,−u2,...,−m )
If two m-vectors are scalar multiples of one another, we say that they are parallel.
Note: Again, this is the same terminology and notation, and similar calculations, as we had in ℜ
3
and ℜ . To ﬁnd a scalar multiple of an m-vector, we multiply each component of that vector by
the scalar. And to ﬁnd the negative of an m-vector, we multiply each component by −1, i.e. switch
the sign of each component. Also, we use the term parallel to describe two m-vectors which have
the same property that would lead us to conclude that they were parallel if they were vectors with
fewer components.
Example 4.4. Foru = (1,2,−1,2),v = (2,0,3,0,1) an▯ = (1,0,1,0,0,3,2), ﬁndu, 5v and −2▯.
Solution:
We simply need to multiply the components of each vector by the corresponding scalar:
−▯u = −(1,2,−1,2) = (−1,−2,−(−1),−2) = (−1,−2,1,−2)
5v = 5(2,0,3,0,1) = (5 × 2,5 × 0,5 × 3,5 × 0,5 × 1) = (10,0,15,0,5)
−2▯ = −2(1,0,1,0,0,3,2)
= (−2 × 1,−2 × 0,−2 × 1,−2 × 0,−2 × 0,−2 × 3,−2 × 2)
= (−2,0,−2,0,0,−6,−4)
Of course, we could have founu more quickly by realizing that we just needed to switch the sign
of each component:
−▯u = −(1,2,−1,−2) = (−1,−2,1,2) 54 Unit 4
Deﬁnition: Two m-vectors are said to be collinear if and only if each is a scalar
multiple of the other. That iu and v ∈ ℜm are collinear if and only if there is some
scalar c such thu = cv. If two non-zero m-vectou and v are collinear, so u = cv,
then they are said to have the same direction if c > 0 and are said to have opposite
directions if c < 0.
Examples: The vectors (1,2,3,4) and (2,4,6,8) are collinear because (2,4,6,8) = 2(1,2,3,4). These
vectors have the same direction. Also,u = (1,0,−1,0,1) and v = (−10,0,10,0,−10), v = −10u
and so u andv are collinear, with directions opposite to one another.
Theorem 4.1. For any m-vector ▯u and any scalar c,
||cu|| = |cu||▯
That is, the magnitude of the scalar multipu is simply the magnitude ofu times the absolute
value of the scalar multiplier.
Example 4.5. Find the magnitude of (4,8,−20,12).
Solution:
The arithmetic is easier if we realize that (4,8,−20,12) = 4(1,2,−5,3). We get
p 2 2 2 2 √ √
||(4,8,−20,12)|| = ||4(1,2,−5,3)|| = |4|||(1,2,−5,3)||1 + 2 + (−5) + 3 = 4 1 + 4 + 25 + 9 = 439
Example 4.6. Ifu = (1,−1,2,−2,3,−3) and ▯v = −5u, ﬁnd |v||.
Solution:
It will be easier to ﬁnd the magnitudeu and use that to ﬁnd the magnitude v, rather than to
ﬁnd the magnitude ofv directly. We get:
|v|| = |p −u|| = | − 5u|| = 5||(1,−1,2,−2,3,−3)||
= 5 1 + (−1) + 2 + (−2) + 3 + (−3) 2
√
= 5 1 + 1 + 4 + 4 + 9 + 9
√ √ √ √ √ √
= 5 28 = 5 4 × 7 = 5 4 7 = 5(2) 7 = 10 7
Example 4.7. Ifu = (1,−1,1,0), ﬁnd a unit vector in the opposite directiu. to ▯
Solution:
Let v be a unit vector in the opposite directiu. Then ▯v is a scalar multipleu, i.ev = cu,
for some scalar value c < 0, such thv|| = 1. Sincv = u, then we get
||v|| = |u|| = |cu||▯
and so |v|| = 1 gives |u|| = 1, which means that we must have |c| |▯|| (Notice that we have
2 3
previously observed this in ℜ and ℜ .) Fu = (1,−1,1,0), we have
p √ √
||u|| = 12+ (−1) + 1 + 0 =2 1 + 1 + 1 + 0 = 3
so we see that we need |c| = in order fov = cu to be a unit vector, and we need c < 0 in order
3 1
forv = cu to have the opposite directionu. Therefore we need c = −√ 3. So we see that a unit
vector in the opposite directionu is▯
1 1 1 1
v = cu = − √ (1,−1,1,0) = − √ ,√ ,− √ ,0
3 3 3 3 Unit 4 55
m
Deﬁnition: Consider any ▯u andv in ℜ . The vector sum of the vectors is
u + v = (u1,u2,...mu ) + 1v 2v ,.m.,v ) 1 (u 1 v2,u 2 v ,.m.+ vm)
and the vector diﬀerence of u and v is
u−▯ v =u+(−▯ v) = (u1,u2,...,m )+(−v 1−v 2...,−vm) = (u1−v 1u 2v ,.2.,u m −v m
Note: This says that, as we do in ℜ and ℜ , we ﬁnd the sum of 2 vectors as the vector whose com-
ponents are the sums of the corresponding components of the 2 vectors, and we ﬁnd the diﬀerence
of 2 vectors, which can be considered to be the sum of the ﬁrst vector and the negative of the second
vector, as the vector whose components are the diﬀerences of the corresponding components.
Example 4.8. Ifu = (1,2,3,4) andv = (0,−1,3,−2), ﬁnd u + v and u −v.
Solution:
We get u + v = (1,2,3,4) + (0,−1,3,−2) = (1 + 0,2 + (−1),3 + 3,4 + (−2)) = (1,1,6,2)
and u − v = (1,2,3,4) − (0,−1,3,−2) = (1 − 0,2 − (−1),3 − 3,4 − (−2)) = (1,3,0,6)
Example 4.9. If u = (1,0,1,0,−1,0) and ▯v = (0,1,1,0,−1,−1), ﬁnd w▯ where it is known that
▯ = 3u − 2v.
Solution:
We simply carry out the speciﬁed arithmetic operations. We get:
w▯ = 3u − v▯
= 3(1,0,1,0,−1,0)− 2(0,1,1,0,−1,−1)
= (3,0,3,0,−3,0)− (0,2,2,0,−2,−2)
= (3 − 0,0 − 2,3 − 2,0 − 0,−3 − (−2),0 − (−2))

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