Class Notes
(806,624)

Canada
(492,358)

Western University
(47,599)

Mathematics
(261)

Mathematics 1229A/B
(198)

Vicki Olds
(116)

Lecture

# Math 1229A Ch1.2 - Product of Vectors .pdf

Unlock Document

Western University

Mathematics

Mathematics 1229A/B

Vicki Olds

Fall

Description

Math 1229A/B
Unit 2:
Products of Vectors
(text reference: Section 1.2)
▯V. Olds 2010 Unit 2 15
srotce Vfostcu d 2rP
In Unit 1 we learnt about a variety of arithmetic operations wecandowihcr. tte
only kind of multiplication we learnt about is scalar multip lication of a vector, i.e. when we take
avo,dm ilibeasl.i. mb)oobt ain a new vector. We didn’t have
anything that was like multiplying two vectors together, i. e. forming a product of two vectors. In
this Unit, we learn about two di▯erent kinds of vector products.
The ﬁrst of these is the dot product.iietv
produce something new ... but what they produce isn’t a new vector, instead it’s a scalar.oe
2 3
inputs to the calculation are 2 vectors, i.e. two elements of ▯ ,rf ▯ ,btteoutfhe
calculation is a number, i.e. some element of ▯cshioutousaalsit,l
the term scalar product is often used instead of dot product. In Physics, when a force is applied to
amass(i.e. anobject),todisplacetheobject,boththeforc ewhichisappliedandthedisplacement
of the object (mass) are vectors. The amount of work done is a scalar quantity, calculated as the
dot product of the two vectors.
The second product is called the cross product.Aieetemit
produce something new, but this time what they produce is a newvector. (Justlikewhenyoutake
a product of 2 numbers, what you get is a new number.) The cross product is therefore also known
as the vector productarti both of the other vectors,
which means it is perpendicular to the whole plane in which those 2 vectors lie. Because of that,
2
this product is not deﬁned for “vectors in the plane”, i.e. forvector▯ .W emtediliw
3-dimensional vectors, i.e. vectors in ▯ ,inordertobeabletoexpressavectorthatisperpendicular
to both. So the cross product is only deﬁned for 2 vectors which are both in ▯ .A,ra
Physics application of this kind of product. This time, thinkabtapigafeoaw nh
in order to turn a bolt. The force applied is a vector, and the wrench represents another vector
(actually, the vector runs from the centre of the bolt to the point on the wrench at which the force
is applied). And the turning e▯ect the force has on the bolt, called the moment,i tect.i.
cross) product of those two vectors. Its direction is perpendicular to both the force and the wrench
–theboltmovesupordown.(Butdon’tworryaboutthePhysics of these things. We’re not going
to be talking about Physics at all. They’re mentioned here just to show you that these products do
have real-world applications.)
The Dot Product
Deﬁnition: Consider any vectors ▯ u and ▯v,ihehfrom ▯ or both from ▯ .T3 he
dot product of ▯u and ▯v,written u •v,isthenumberobtainedwhenthecorresponding
components of ▯u and ▯v are multiplied together and then these products are summed.
That is, we have:
2
Ifu,▯v ▯▯
then u •v▯ = u 1 1 u v2 2
3
Ifu,▯v ▯▯
then u •v▯ = u 1 1 u v2 2u v 3 3
Example 2.1.
(a) Calculate (1,2) • (3,▯1) and (2,▯1,4) • (3,2,▯2).
(b) Ifu =(0 ,3,▯2) and ▯v =(1 ,0,5), ﬁnd v • u and (▯▯u) •u.
(c) Ifu =(1 ,2),v =(2 ,▯1) and w ▯ =(1 ,▯2), ﬁnd (u ▯ v) • ▯. 16 Unit 2
Solution:
(a) For u =( u ,1 )21 ( ,2) and ▯v =( v 1v )23 ( ,▯1), we calculate the product of the ﬁrst
components, and the product of the second components, and addthetwoproductstogether. That
is,
(1,2) • (3,▯1) = [1 ▯ 3] + [2 ▯ (▯1)] = 3 + (▯2) = 3 ▯ 2=1
Likewise, if we consider the vectors to x =( x ,x ,x )=(2 ,▯1,4) and ▯ y =( y ,y ,y )=(3 ,2,▯2),
1 2 3 1 2 3
we ﬁnd the sum ofx ▯ 1 , x1▯ 2 and2x ▯ y :3 3
(2,▯1,4) • (3,2,▯2) = (2)(3) + (▯1)(2) + (4)(▯2) = 6 + (▯2) + (▯8) = 6 ▯ 2 ▯ 8= ▯4
(b) For v•▯u,wejustusetheformula(recognizingthattheorderoftheve ctors is reversed this time).
▯v •u =(1 ,0,5) • (0,3,▯2) = (1)(0) + (0)(3) + (5)(▯2) = 0 + 0 + (▯10) = ▯10
For (▯▯u) •u,weﬁrstﬁnd ▯▯u = ▯(0,3,▯2) = (0,▯3,2) and then take the dot product asked for:
(▯u) • u =(0 ,▯3,2) • (0,3,▯2) = (0)(0) + (▯3)(3) + (2)(▯2) = 0 + (▯9) + (▯4) = ▯9 ▯ 4= ▯13
(c) For (u ▯ v) •▯,weﬁrstneedtoﬁnd u ▯▯v,andthendotthatvectorwith ▯.W et
u ▯ v =1( ,2) ▯ (2,▯1) = (1 ▯ 2,2 ▯ (▯1)) = (▯1,3)
so (u ▯▯v) • ▯ =( ▯1,3) • (1,▯2) = (▯1)(1) + (3)(▯2) = ▯1+( ▯6) = ▯7
Because all we’re doing is multiplying and adding, and order isn’t important in those operations,
then order also isn’t important in calculating a dot producththedrfeomonsi
of course important, but the order of the vectors isn’t. Whet her we calculatu•▯v or calculatev •u,
the same numbers are being multiplied together, giving the same products being added up, so the
answer is the same. That is, the dot product operation is comm utative. Likewise, it doesn’t matter
when, or to what, a scalar multiplier is applied. Whether we multiply ▯u by a scalar before forming
the dot product, or multiply ▯v by that scalar instead, or even wait and multiply the dot pro uct
value by that scalar, it all comes out the same. The same wd ithitiaon. We can distribute dot
product over addition of vectors, or we can “factor out” a dot product from a sum of dot products
(as long as the same vector is dotted) and the answer never changes. Of course, if there are 0’s in all
the products that are being added, the ﬁnal result is just 0, sothedotprodyvectihwt
the 0 vector is just 0. And compare the dot product formula, when dotting a vector with itself,to
the formula for ﬁnding the magnitude of a ve▯tor. The only thing missing is the square root sign.
So the magnitude of ▯u can be thought of as u • u.Rmmb,mgiemuteo,owe
take the positive square root.) Or the dot product of u with itself can be thought of as the square
of the magnitude of u ... whichever way you want to look at it. These ideas are summarized in the
following theorem.
Theorem 2.1. Let ▯ u, v and ▯ be any 3 vectors from▯ , or any 3 vectors from ▯ ,a 3 dlt c be
any scalar. Then the following properties hold:
1. (commutative property) ▯u •v = v • u
2. c(u •v)=( c▯u) •v = ▯u • (v)
3. (distributive propertyu • (v + ▯)=( ▯u •▯v)+( u • ▯)
▯
4. u • 0=0
5. u •u = ||u|| Unit 2 17
For instance, in Example 2.1 (b), we found▯ v•aut= ▯10. By property 1. above, we could have
calculated u •v instead, and we would have got the same answer:
u •▯v =(0 ,3,▯2) • (1,0,5) = (0)(1) + (3)(0) + (▯2)(5) = 0 + 0 + (▯10) = ▯10
Also, we found that (▯▯ u) •u = ▯13. Property 2. tells us that we didn’t need to ﬁnd ▯▯ u ﬁrst.
We could instead have calcula▯ utedu,andheaenhenvi.i.ilda ▯1m lilr)
afterwa▯ds. Also, having found that ▯(▯u •u)= ▯13, so that ▯ u •u =13,Property5. tellsusthat
|u|| = 13. And in Example 2.1 (c), instead of ﬁnding u ▯▯v) • ▯,wecouldhaveusedProperty3.
as follows:
(u▯▯v)•w▯ =( u•w▯)▯(▯v•w▯)=[(1 ,2)•(1,▯2)]▯[(2,▯1)•(1,▯2)] = (1+(▯4))▯(2+2) = 1▯4▯4= ▯7
Notice: In that calculation, we actually used other properties, of dot product and of vector arith-
metic, too. Because the dot product is commutative (Property1) .,edo’ needorryabout
whether we have the form ▯x • y + z)(asshowninProperty3.),ortheform( y + z) •x (as we had
above) when we distribute the dot product over vector additi on. Also, because we know that vector
subtraction is really just the same as addition of the negative of the vector, we can also distribute
the dot product over vector subtraction. That is, (u ▯ v) •▯ =( ▯u +( ▯1)▯v) •▯.Adewe
we have distributed the dot product, we can also, by Propert2 y., pull the ▯1multiplieroutside,so
that again we’re adding the negative, ofv •▯ this time, and thus just subtracting.
Also Notice: We learnt in Unit 1 that for any vec ut0ru = 0. That is, if any vector is multiplied
by the scalar value 0, the result itshe zero vectndo,ieaveom,erolat
▯
u • 0=0. Thatis,whenanyvectorisdottedwith the zero vectorherlofhatdotp,oduct
which as always is aalr ,ihenuber0. esurouunandheus,anddi ▯erence
between, the scalar value 0 and the vector0, here and elsewhere.
Example 2.2. If v = ▯ ▯u,and u • v = ▯6, ﬁnd ||v||.
3
Solution:
Approach 1: Since v = ▯ 2u,thenwehave
3
▯ ▯ ▯ ▯
2 2 2 2
u • v = u • ▯ 3u = ▯ 3 u •u = ▯ 3(|u||)
2 2
so knowing that u • v = ▯6alsotellsusthat ▯ 3 |u|| = ▯6, so we get
▯ ▯
2 2 3
||u|| = ▯6 ÷ ▯ = ▯6 ▯▯ =3 ▯ 3=9
3 2
▯
Therefore |u|| = 9=3andso
▯▯ 2 ▯▯ ▯ 2▯ 2
|v|| = ▯▯ u ▯▯= ▯ ▯||u|| = ▯ 3=2
▯▯ 3 ▯▯ ▯ 3▯ 3
Approach 2: (quicker – don’t ﬁnd ||▯u|| ﬁrst) Knowing that ▯ v = ▯ 2u also means that we have
2 3 3 3
u = v ÷ ▯ 3 = v ▯▯ 2 = ▯ 2v.Andenweve
▯ ▯
3 3 3 2
u •▯v = ▯ v •▯v = ▯ (▯v •v)= ▯ ||▯ v||
2 2 2
and so using the fact thatu •v = ▯6weget
3 2 2 2 ▯
▯ 2 |v|| = ▯6 ▯ |v|| = ▯6 ▯▯ 3 =4 ▯ |v|| = 4=2 18 Unit 2
The angle between 2 vectors
Let u and v be any 2 vectors, either both in ▯ or both in ▯ .n ▯u and v are both directed line
segments which each go from the origin to some endpoint. Their tails,wihcometogetheratthe
origin, form an angle. Actually, they form 2 angles – the angl eformedby“goingtheshortway”,and
the angle formed by “going the long way”, where the 2 angles together form a full circleu and v
are opposite in direction, then “the short way” and “the long way” are the same distance, halfway
around a circle.) When we say the angle between the vectors, we mean the angle the 2 vectors form
at the origin, “going the short way”.
Deﬁnition: For any vectors u and ▯v,ioti ▯ 2 or both in ▯ , the angle
between ▯ u and ▯v means the angle no more than 180 (or ▯ radians) formed by the
directed line segment representations of these vectors. We often use ▯ to represent this
angle.
For instance, for the vectourasnd v depicted here, the angle betweeu and ▯v is the angle ▯ shown:
y
▯
4 ▯▯
3 ▯
▯
2 ▯
1 ▯ v
▯▯ ▯
▯ ▯
0 12 3 x
Geometrically, the scalar value that is the dot product of twovectorsgivesthevalueofaspeciﬁc
calculation involving the magnitudes of the vectors and the angle between them. And we can use
this fact to express a formula for the cosine value of the anglebetweenthevectors.
You have probably taken some trigonometry before, whils chwidhaangles (particularly in
triangles). Since our focus in this course is primarily on vectors expressed in component form, we
are not particularly concerned with trigonometry in this co urse, so you don’t need to remember any-
thing you learned previously about this. We don’t even need toikaboutwhattielesn
of an angle means or represents. And we certainly don’t want t olearn/reviewenoughtrigonometry
to understand where the following theorem comes from. However, you should realize that any 2
vectorsdo form an angle. And the formula in the theorem below expresseshow the dot product and
magnitudes of the vectors tell us about this angle.
Theorem 2.2. Consider any vectors ▯ u and v,eerthi ▯ or both in ▯ .Lt ▯ be the angle
between▯u and v.Tn
u •v = |u|| |v||cos▯
and provided thatu ▯= 0 and v ▯= 0,tltioaneraandtoﬁndtleof cos▯ as
u •v▯
cos▯ =
||u|| v||
That is, for any non-zero vectorsu and v,heal wen u and ▯v is the angle ▯ whose cosine
value is the value given by this formula. (There’s only ongleanno bigger than 180 which has that
cosine value.)
Notice: Ifu = 0or ▯v = 0, then one of the terms in the denominator is 0, so we are tryingtodivide
by 0, which isn’t possible. It doesn’t make any sense to talout the angle between the zero vector
and another vector, so in that case ▯ is undeﬁned and therefore so is cos▯. Unit 2 19
Also Notice: This formula for calculating cos▯,andanotherformulainvo▯ later in this unit,
are the only times we’ll be using any of the ideas of trigonome try in this course.
Example 2.3. Find the value of cos▯ where ▯ is the angleu and v in each of the following:
(a)u =(5 ,4) anv =(2 ,▯3) (b)u =(1 ,2,3) anv =(3 ,▯2,1)
(c)u =(1 ,2) anv =(2 ,▯1) (c)u =(2 ,3) anv =(4 ,6)
Solution:
For each of these, we use the formulaos=u •v▯ ,soweneedtoﬁnd |u||,v|| andu •v.
|u||v||▯
(a) Foru =(5 ,4) anv =(2 ,▯3) we get:
▯ 2 2 ▯ ▯
|u|| = 5 +4 = 25 + 16 = 41
▯ 2 2 ▯ ▯
|v|| = 2 +( ▯3) = 4+9= 13
and u •v▯ =5( ,4) • (2,▯3) = (5)(2) + (4)(▯3) = 10 + (▯12) = 10 ▯ 12 = ▯2
so cos▯ = ▯ ▯2▯ = ▯ ▯ 2▯
( 41)( 13) 41 13
(b) We haveu =(1 ,2,3) anv =(3 ,▯2,1), so we get:
▯ ▯ ▯
||u|| = 12+2 +3 2 = 1+4+9= 14
▯ ▯ ▯
|v|| = 32+( ▯2) +1 2 = 9+4+1= 14
u •v▯ =1( ,2,3) • (3,▯2,1) = (1)(3) + (2)(▯2) + (3)(1) = 3 + (▯4) + 3 = 6 ▯ 4=2
2 2 1
and cos ▯ = ▯ ▯ = =
( 14)( 14) 14 7
(c) When u =(1 ,2) anv =(2 ,▯1) we get:
▯ ▯ ▯
|u|| = 1 +2 2= 1+4= 5
▯ ▯ ▯
|v|| = 22 +( ▯1) = 4+1= 5
u •v▯ =1( ,2) • (2,▯1) = (1)(2) + (2)(▯1) = 2 + (▯2) = 2 ▯ 2=0
0 0
so cos▯ = ▯ ▯ = =0
( 5)( 5) 5
(d) Foru =(2 ,3) anv =(4 ,6) we have:
▯ 2 2 ▯ ▯
|u|| = 2 +3 = 4+9= 13
▯ 2 2 ▯ ▯
|v|| = 4 +6 = 16 + 36 = 52
and u •v▯ =2( ,3) • (4,6) = (2)(4) + (3)(6) = 8 + 18 = 26
26 26 26 26 26 26
so cos▯ = ( 13)( 52)= ▯13 ▯ 52= ▯ 13 ▯ 13 ▯ 4 ▯ 13 ▯ 22 = 13 ▯ 2= 26 =1
Notice: Since (4,6) = 2(2,3) we v =2 u and sou andv have the same direction. That means
that the angle between them is 0 (i.e. 0 or 0 radians), and cos0 = 1. Si▯ =( ▯4,▯6) w
we have ▯ = ▯2u and so ▯ has the opposite directiouIhle▯
same as those shown above except that the numerator is negative so the ﬁnal answer is cos▯ = ▯1.
Whenever u and v have opposite directions, the angle between them 0s ▯ (or ▯ radians),
which has cosine value ▯1. 20 Unit 2
Let’s think some more about what we had in part (c) of that example. We had two non-zero vectors
whose dot product was 0, which meant that also the cosine of theanglebetweenthemwas0. What
angle has cosine value 0? Do you remember?
You probably don’t (and we said earlier that you wouldd n’treeerhtigs)’tse
look at a picture of these vectors. Theyu =(1 ,2) andv =(2 ,▯1). We have:
y
2 ▯
▯▯
u ▯
1 ▯
▯
▯
▯ ▯
▯2 ▯10 ▯ ▯ 1 2 x
v ▯ ▯
▯1 ▯▯
▯2
Let’s see that again, without the axes:
▯▯
u ▯
▯
▯
▯ ▯
▯▯▯
v ▯ ▯
▯
It’s a right anglu and v are perpendicular! Notice: In hetr( b,▯a)i lysperpen-
dicular to the vector (a,b). And it’s true ... the angle (smaller than 180 ) whose cosine value is
▯ ▯
zero is ▯ =90 (or2 radians). But when we’re talking about the relationship bet ween two vectors,
perpendicularisn’t the word we usually use. We say that two lines are perpendicular, or that two
planes are perpendicular, but we say that two vectors are orthogonal.
Deﬁnition: Two vectors are said to be orthogonal if the angle between them is 90
(i.e.▯radians).
2
Notice: We would also use the word orthogonal if we were talkingtaoodutected line segments.
Two directed line segments are orthogonal if the vectors obtained by translating them to the origin
are orthogonal.
Also Notice: Orthogonal means the same thing as perpendicular, but the us age is a bit di▯erent.
Another word that also means the same thing is normalues perpendicularwhen comparing
two lines or planes,orthogonal when comparing two vectors (or directed line segments), and normal
when comparing a vector to a line or a plane. That is, we say evort u is normal to a
particular line or plane if the vector, or the corresponding directed line segment when translated to
some point on that line or plane, meets the line or plane at an angle of 90 .
Theorem 2.3. Let ▯u and v be two vectors, either botn ior both in ▯ n:
u and v are orthogonal if and onlu •v =0
Notice that fou =( a,b)and v =( b,▯a)weget
u •v =( a,b) • (b,▯a)=( a)(b)+( b)(▯a)= ab +( ▯ba)= ab ▯ ab =0 Unit 2 21
So as mentioned above, for any real values of a and b,the2-dimensionalvectors( a,b)and( b,▯a)are
orthogonal to one another. And any scalar multiple of ( b,▯a)isalsoorthogonalto( a,b). However,
3
in ▯ there’s no easy way to recognize that 2 vectors are orthogonal, other than by calculating their
dot product. (For ▯ ,you’llwanttorememberthatyoucangetavectororthogonal to (a,b)simply
by switching the components and changing the sign of one of e th .)
Example 2.4. Prove that ▯ u =(1 ,3,▯2) and ▯ v =(5 ,1,4) are orthogonal.
Solution:
u •v =(1 ,3,▯2) • (5,1,4) = (1)(5) + (3)(1) + (▯2)(4) = 5 + 3 + (▯8) = 0
Since u •▯v =0,then u and v must be orthogonal.
Example 2.5. Let ▯ u =(2 ,0,4) and ▯v =(1 ,1,k ). Ifu and v are orthogonal, what is the value of k?
Solution:
Ifu and ▯v are orthogonal, then it must be true that ▯u •v =0. Butwecancalculate u • v:
u •v =(2 ,0,4) • (1,1,k )=(2)(1)+(0)(1)+4( k)=2+0+4 k =2+4 k
▯2 1
So it must be true that 2 + 4k =0,whichgives4 k = ▯2andso k = 4 = ▯ 2
The Cross Product
3 3
As previously stated, the Cross Product of two vectors ▯ u and ▯v in ▯ is a new vector in ▯
which is orthogonal to both ▯u and ▯v.hioreini only deﬁned for vectors in ▯ .
Deﬁnition: Let ▯ u =( u 1u ,2 )3 d v =( v 1v 2v 3b eaytovcsi ▯ .Th
cross product of ▯ u and ▯v,denoted u ▯ v,isthevector
u ▯ ▯v =( u2 3▯ u 3 2u v3 1u v 1 3v ▯1 2v ) 2 1
Example 2.6. Find the cross product of ▯ u =(1 ,2,3) and ▯v =(4 ,5,6).
Solution:
We haveu =11 u =2an2 u 33,with v1=4, v =2and v3=6,andsoweget:
u ▯ v =( u v2 3u v ,3 2 ▯3 1v ,u 1 3 u1 2) 2 1
)6(2(( ▯ (3)(5),(3)(4) ▯ (1)(6),(1)(5) ▯ (2)(4))
21( ▯ 15,12 ▯ 6,5 ▯ 8)
=( ▯3,6,▯3)
You will have noticed that the formula is kind of nastyy.to ..genfssedoortun,tely
there are some easier ways to remember how to ﬁnd the cross product. That is, procedures that are
less easily confused that will get you to the answer. The text shows one such procedure, involving
something called a determinant,o np.7. usol eaokath. r,’kat
another procedure that the text doesn’t show, and which you m ight ﬁnd easier (until later in th

More
Less
Related notes for Mathematics 1229A/B