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Western University
Mathematics
Mathematics 1229A/B
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Fall
Description
Math 1229A/B
Unit 2:
Products of Vectors
(text reference: Section 1.2)
▯V. Olds 2010 Unit 2 15
2 Products of Vectors
In Unit 1 we learnt about a variety of arithmetic operations we can do with vectors. But the
only kind of multiplication we learnt about is scalar multiplication of a vector, i.e. when we take
a vector, and multipliply it be a scalar (i.e. a number) to obtain 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 diﬀerent kinds of vector products.
The ﬁrst of these is the dot product. This is a product in which two vectors are combined to
produce something new ... but what they produce isn’t a new vector, instead it’s a scalar. So the
inputs to the calculation are 2 vectors, i.e. two elements of ℜ , or of ℜ , but the output of the
calculation is a number, i.e. some element of ℜ. Because this product produces a scalar as its result,
the term scalar product is often used instead of dot product. In Physics, when a force is applied to
a mass (i.e. an object), to displace the object, both the force which is applied and the displacement
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. Again we have two vectors being combined to
produce something new, but this time what they produce is a new vector. (Just like when you take
a product of 2 numbers, what you get is a new number.) The cross product is therefore also known
as the vector product. The new vector that we get is perpendicular to both of the other vectors,
which means it is perpendicular to the whole plane in which those 2 vectors lie. Because of that,
this product is not deﬁned for “vectors in the plane”, i.e. for vectors in ℜ . We must be dealing with
3dimensional vectors, i.e. vectors in ℜ , in order to be able to express a vector that is perpendicular
3
to both. So the cross product is only deﬁned for 2 vectors which are both in ℜ . Again, there’s a
Physics application of this kind of product. This time, think about applying a force to a wrench
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 eﬀect the force has on the bolt, called the moment, is the vector (i.e.
cross) product of those two vectors. Its direction is perpendicular to both the force and the wrench
– the bolt moves up or down. (But don’t worry about the Physics 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 realworld applications.)
The Dot Product
Deﬁnition: Consider any vectors ▯ u and ▯v, either both from ℜ or both from ℜ . The
dot product of ▯ u and v, writtenu •▯v, is the number obtained when the corresponding
components of ▯u and ▯v are multiplied together and then these products are summed.
That is, we have:
2
Ifu,v ∈ ℜ
then u •v▯ = u1 1+ u 2 2
Ifu,v ∈ ℜ 3
then u •v▯ = u v + u v + u v
1 1 2 2 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 ,u ) = (1,2) and ▯ v = (v ,v ) = (3,−1), we calculate the product of the ﬁrst
1 2 1 2
components, and the product of the second components, and add the two products together. That
is,
(1,2) • (3,−1) = [1 × 3] + [2 × (−1)] = 3 + (−2) = 3 − 2 = 1
Likewise, if we consider the vectors to x = (x 1x 2x 3 = (2,−1,4) and ▯ y = (y 1y 2y 3 = (3,2,−2),
we ﬁnd the sum of x ×1y , 1 ×2y an2 x × 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, we just use the formula (recognizing that the order of the vectors 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, we ﬁrst ﬁnd −▯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) •▯, we ﬁrst need to ﬁnd ▯u −▯v, and then dot that vector with w▯. We get
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 product. That is, the order of the components is
of course important, but the order of the vectors isn’t. Whether we calculateu•▯v or calculate v •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 commutative. 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 product
value by that scalar, it all comes out the same. The same with addition. 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, so the dot product of any vector with
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 vector. The only thing missing is the square root sign.
√
So the magnitude of ▯u can be thought of as u •u. (Remember, magnitude must be positive, so we
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.
2 3
Theorem 2.1. Let ▯ u, v and w▯ be any 3 vectors from ℜ , or any 3 vectors from ℜ , and let c be
any scalar. Then the following properties hold:
1. (commutative property) ▯u • v = v •u▯
2. c(u •v) = (cu) • v = u • (v)
3. (distributive property)u • v + ▯) = (u •▯v) + (u •▯)
4. u • 0 = 0
2
5. u •u = u Unit 2 17
For instance, in Example 2.1 (b), we found thatv •u = −10. By property 1. above, we could have
calculatedu • 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 calculated u •u, and then taken the negative (i.e. applied a −1 multiplier)
afterwa√ds. Also, having found that −(u • u) = −13, so that u •u = 13, Property 5. tells us that
u = 13. And in Example 2.1 (c), instead of ﬁnding u −▯v) • ▯, we could have used Property 3.
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 (Property 1.), we don’t need to worry about
whether we have the form ▯x • y + z) (as shown in Property 3.), or the formy + z) •x (as we had
above) when we distribute the dot product over vector addition. 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) •▯. And then when
we have distributed the dot product, we can also, by Property 2., pull the −1 multiplier outside, so
that again we’re adding the negative, ov • ▯ this time, and thus just subtracting.
Also Notice: We learnt in Unit 1 that for any vectou, 0u = 0. That is, if any vector is multiplied
by the scalar value 0, the result is the zero vector. And now, in the above theorem, we are told that
u • 0 = 0. That is, when any vector is dotted with the zero vector, the result of that dot product,
which as always is a scalar, is the number 0. Make sure you understand the use of, and diﬀerence
between, the scalar value 0 and the vector 0, here and elsewhere.
2
Example 2.2. If v = − 3u, and u •v = −6, ﬁnd v.
Solution:
Approach 1: Since v = − 3u, then we have
2 2 2
u •v = u • − ▯u = − u • u = − (u)2
3 3 3
so knowing that u •v = −6 also tells us that −2u = −6, so we get
3
2 2 3
u = −6 ÷ − 3 = −6 × − 2 = 3 × 3 = 9
√
Therefore u = 9 = 3 and so
2 2 2
v =− u= − u = × 3 = 2
3 3 3
Approach 2: (quicker – don’t ﬁnd ▯u ﬁrst) Knowing that ▯v = − 3u also means that we have
2 3 3
u = v ÷ − 3 = v × − 2 = − 2v. And then we have
3 3 3 2
u •v = − 2v •v = − 2▯v • v) = − 2v
and so using the fact thatu •v = −6 we get
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 ℜ or both in ℜ . Thu and v are both directed line
segments which each go from the origin to some endpoint. Their tails, which come together at the
origin, form an angle. Actually, they form 2 angles – the angle formed by “going the short way”, and
the angle formed by “going the long way”, where the 2 angles together form a full ciu and vIf ▯
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, either both in ℜ2 or both in ℜ , the angle
◦
between ▯ u and ▯v means the angle no more than 180 (or π radians) formed by the
directed line segment representations of these vectors. We often use θ to represent this
angle.
For instance, for the vectu and v depicted here, the angle betweu and v is the angle θ shown:
y
6
4 u
3
2
*
1 θ v

0
1 2 3 x
Geometrically, the scalar value that is the dot product of two vectors gives the value of a speciﬁc
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 angle between the vectors.
You have probably taken some trigonometry before, which deals with angles (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 course, so you don’t need to remember any
thing you learned previously about this. We don’t even need to think about what the cosine value
of an angle means or represents. And we certainly don’t want to learn/review enough trigonometry
to understand where the following theorem comes from. However, you should realize that any 2
vectors do form an angle. And the formula in the theorem below expresses how the dot product and
magnitudes of the vectors tell us about this angle.
Theorem 2.2. Consider any vectors ▯u and v, either both in ℜ or both in ℜ . Let θ be the angle
between u and v. Then
u •v = u vcosθ
and provided thatu ▯= 0 andv ▯= 0, this relationship can be rearranged to ﬁnd the value of cosθ as
u •v▯
cosθ =
u v
That is, for any nonzero vectou and ▯v, the angle betweenu and v is the angle θ whose cosine
value is the value given by this formula. (There’s only one angle no bigger than 180 which has that
cosine value.)
Notice: Ifu = 0 orv = 0, then one of the terms in the denominator is 0, so we are trying to divide
by 0, which isn’t possible. It doesn’t make any sense to talk about the angle between the zero vector
and another vector, so in that case θ is undeﬁned and therefore so is cosθ. Unit 2 19
Also Notice: This formula for calculating cosθ, and another formula involving sinθ later in this unit,
are the only times we’ll be using any of the ideas of trigonometry in this course.
Example 2.3. Find the value of cosθ where θ is the angleu andv 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) (cu = (2,3) andv = (4,6)
Solution:
For each of these, we use the formula cosθ =▯, so we need to ﬁnu, v anu •v.
u v
(a) Fou = (5,4) andv = (2,−3) we get:
p 2 2 √ √
u = 5 + 4 = 25 + 16 = 41
p 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 cosθ = √ −2√ = − √ √
( 41)( 13) 41 13
(b) We haveu = (1,2,3) anv = (3,−2,1), so we get:
p √ √
u = 12+ 2 + 3 = 1 + 4 + 9 = 14
p √ √
v = 32+ (−2) + 1 = 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 cosθ = √ √ = =
( 14)( 14) 14 7
(c) When u = (1,2) anv = (2,−1) we get:
p √ √
u = 1 + 2 = 1 + 4 = 5
p √ √
v = 22+ (−1) = 4 + 1 = 5
u •v▯ = (1,2) • (2,−1) = (1)(2) + (2)(−1) = 2 + (−2) = 2 − 2 = 0
0 0
so cosθ = √ √ = = 0
( 5)( 5) 5
(d) Foru = (2,3) anv = (4,6) we have:
p 2 2 √ √
u = 2 + 3 = 4 + 9 = 13
p 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 cosθ = ( 13)( 52) = √ 13 × 52= √ 13 × 13 × 4 √ 13 × 22= 13 × 2 26= 1
Notice: Since (4,6) = 2(2,3) we v = u and sou and v have the same direction. That means
that the angle between them is 0 (i.e. 0 or 0 radians), and cos0 = 1. S▯ = (−4,−6)f w
we have ▯ = −2u and so▯ has the opposite directiou. In this case, the calculations are the
same as those shown above except that the numerator is negative so the ﬁnal answer is cosθ = −1.
Whenever u and v have opposite directions, the angle between them is θ = 180 (or π 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 nonzero vectors
whose dot product was 0, which meant that also the cosine of the angle between them was 0. What
angle has cosine value 0? Do you remember?
You probably don’t (and we said earlier that you wouldn’t need to remember such things), so let’s
look at a picture of these vectors. Thu = (1,2) anv = (2,−1). We have:
y
2 6
u
1

−2 −1 0H H 1 2x
v H
HH
−1 j
−2
Let’s see that again, without the axes:
u
HH
H
v H
Hj
It’s a right angu andv are perpendicular! Notice: In ℜ , the vector (b,−a) is always perpen
◦
dicular to the vector (a,b). And it’s true ... the angle (smaller than 180 ) whose cosine value is
zero is θ = 90 (orradians). But when we’re talking about the relationship between two vectors,
2
perpendicular isn’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.e2 radians).
Notice: We would also use the word orthogonal if we were talking about two directed 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 usage is a bit diﬀerent.
Another word that also means the same thing is normal. We use perpendicular when 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 that tu 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 andv be two vectors, either both in ℜ or both in ℜ . Then:
u andv are orthogonal if and onu •v = 0
Notice that fu = (a,b) anv = (b,−a) we get
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, the 2dimensional vectors (a,b) and (b,−a) are
orthogonal to one another. And any scalar multiple of (b,−a) is also orthogonal to (a,b). However,
in ℜ there’s no easy way to recognize that 2 vectors are orthogonal, other than by calculating their
2
dot product. (For ℜ , you’ll want to remember that you can get a vector orthogonal to (a,b) simply
by switching the components and changing the sign of one of them.)
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. But we can calculate ▯u •v:
u •v = (2,0,4) • (1,1,k) = (2)(1) + (0)(1) + 4(k) = 2 + 0 + 4k = 2 + 4k
−2 1
So it must be true that 2 + 4k = 0, which gives 4k = −2 and so k = 4 = − .2
The Cross Product
As previously stated, the Cross Product of two vectors ▯ u and ▯v in ℜ is a new vector in ℜ 3
3
which is orthogonal to both u and ▯v. This vector operation is only deﬁned for vectors in ℜ .
3
Deﬁnition: Let ▯ u = (u 1u 2u 3 and ▯ v = (v 1v 2v 3 be any two vectors in ℜ . The
cross product of ▯ u and v, denoted u × v, is the vector
u × v = (u2 3− u v3 2 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 have u 1 1, u =22 and u = 33 with v = 4,1v = 5 a2d v = 6, a3d so we get:
u × v = (u 2 3 u v3 2 v3 1u v ,1 3 −1 2v ) 2 1
= ((2)(6) − (3)(5),(3)(4) − (1)(6),(1)(5) − (2)(4))
= (12 − 15,12 − 6,5 − 8)
= (−3,6,−3)
You will have noticed that the formula is kind of nasty ... easy to get confused on. Fortunately,
there are some easier ways to remember how to ﬁnd the cross product. That is, procedures that are
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