2
+ j + 12
(g) Find the unit vector v .
v
Solution: v
(2, 1,
, 1 , −1
v
=
1
−1) = 2
22+12+(−1)2
6
6
6
14
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
We can now easily prove Theorem 1.1 from the previous section. The distance d between
two points P = ( x , y , z ) and Q
, y , z ) in
1
1
1
= ( x 2 2 2
R3 is the same as the length of the vector w−v,
where the vectors v and w are defined as v = ( x , y , z ) and w
, y , z ) (see Figure 1.2.8).
1
1
1
= ( x 2 2 2
So since w −v = ( x
, y
, z
), then d
)2
)2
)2 by
2 − x 1
2 − y 1
2 − z 1
= w−v =
( x 2 − x 1 + ( y 2 − y 1 + ( z 2 − z 1
Theorem 1.2.
z
P( x 1, y 1, z 1)
w − v
Q( x 2, y 2, z 2)
v
w
y
0
x
Figure 1.2.8
Proof of Theorem 1.2: d = w − v
Exercises
A
1. Let v = (−1,5,−2) and w = (3,1,1).
(a) Find v − w.
(b) Find v + w.
(c) Find v .
(d) Find
1 (v
v
2
− w) .
(e) Find
1 (v
2
+ w) .
(f) Find −2v + 4w.
(g) Find v − 2w.
(h) Find the vector u such that u + v + w = i.
(i) Find the vector u such that u + v + w = 2j + k.
(j) Is there a scalar m such that m(v + 2w) = k? If so, find it.
2. For the vectors v and w from Exercise 1, is v − w = v − w ? If not, which quantity
is larger?
3. For the vectors v and w from Exercise 1, is v + w = v + w ? If not, which quantity
is larger?
B
4. Prove Theorem 1.5(f) for R3.
5. Prove Theorem 1.5(g) for R3.
C
6. We know that every vector in R3 can be written as a scalar combination of the vectors i,
j, and k. Can every vector in R3 be written as a scalar combination of just i and j, i.e. for
any vector v in R3, are there scalars m, n such that v = m i + n j? Justify your answer.
1.3 Dot Product
15
1.3 Dot Product
You may have noticed that while we did define multiplication of a vector by a scalar in the
previous section on vector algebra, we did not define multiplication of a vector by a vector.
We will now see one type of multiplication of vectors, called the dot product.
Definition 1.6. Let v = ( v , v , v ) and w
, w , w ) be vectors in
1
2
3
= ( w 1
2
3
R3.
The dot product of v and w, denoted by v · w, is given by:
v · w = v w
w
w
(1.6)
1
1 + v 2
2 + v 3
3
Similarly, for vectors v = ( v , v ) and w
, w ) in
1
2
= ( w 1
2
R2, the dot product is:
v · w = v w
w
(1.7)
1
1 + v 2
2
Notice that the dot product of two vectors is a scalar, not a vector. So the associative law
that holds for multiplication of numbers and for addition of vectors (see Theorem 1.5(b),(e)),
does not hold for the dot product of vectors. Why? Because for vectors u, v, w, the dot
product u · v is a scalar, and so (u · v) · w is not defined since the left side of that dot product
(the part in parentheses) is a scalar and not a vector.
For vectors v = v i
j
k and w
i
j
k in component form, the dot product
1
+ v 2 + v 3
= w 1 + w 2 + w 3
is still v · w = v w
w
w .
1
1 + v 2
2 + v 3
3
Also notice that we defined the dot product in an analytic way, i.e. by referencing vector
coordinates. There is a geometric way of defining the dot product, which we will now develop
as a consequence of the analytic definition.
Definition 1.7. The angle between two nonzero vectors with the same initial point is the
smallest angle between them.
We do not define the angle between the zero vector and any other vector. Any two nonzero
vectors with the same initial point have two angles between them: θ and 360◦ − θ. We will
always choose the smallest nonnegative angle θ between them, so that 0◦ ≤ θ ≤ 180◦. See
Figure 1.3.1.
θ
360◦ − θ
θ
θ
360◦ − θ
360◦ − θ
(a) 0◦ < θ < 180◦
(b) θ = 180◦
(c) θ = 0◦
Figure 1.3.1
Angle between vectors
We can now take a more geometric view of the dot product by establishing a relationship
between the dot product of two vectors and the angle between them.
16
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
Theorem 1.6. Let v, w be nonzero vectors, and let θ be the angle between them. Then
v · w
cos θ =
(1.8)
v
w
Proof: We will prove the theorem for vectors in R3 (the proof for R2 is similar). Let v =
( v , v , v ) and w
, w , w ). By the Law of Cosines (see Figure 1.3.2), we have
1
2
3
= ( w 1
2
3
v − w 2 = v 2 + w 2 − 2 v
w cos θ
(1.9)
(note that equation (1.9) holds even for the “degenerate” cases θ = 0◦ and 180◦).
z
v − w
v
θ
w
y
0
x
Figure 1.3.2
Since v − w = ( v
, v
, v
), expanding v
1 − w 1
2 − w 2
3 − w 3
− w 2 in equation (1.9) gives
v 2 + w 2 − 2 v
w cos θ = ( v
)2
)2
)2
1 − w 1
+ ( v 2 − w 2 + ( v 3 − w 3
= ( v 2 − 2 v w
) + ( v 2 − 2 v w
) + ( v 2 − 2 v w
)
1
1
1 + w 2
1
2
2
2 + w 2
2
3
3
3 + w 2
3
= ( v 2 + v 2 + v 2) + ( w 2 + w 2 + w 2) − 2( v w
w
w )
1
2
3
1
2
3
1
1 + v 2
2 + v 3
3
= v 2 + w 2 − 2(v · w) , so
−2 v
w cos θ = −2(v · w) , so since v = 0 and w = 0 then
v · w
cos θ =
, since v > 0 and w > 0.
QED
v
w
Example 1.5. Find the angle θ between the vectors v = (2,1,−1) and w = (3,−4,1).
Solution: Since v · w = (2)(3) + (1)(−4) + (−1)(1) = 1, v = 6, and w = 26, then
v · w
1
1
cos θ =
=
=
≈ 0.08 =⇒ θ = 85.41◦
v
w
6 26
2 39
Two nonzero vectors are perpendicular if the angle between them is 90◦. Since cos 90◦ =
0, we have the following important corollary to Theorem 1.6:
Corollary 1.7. Two nonzero vectors v and w are perpendicular if and only if v · w = 0.
We will write v ⊥ w to indicate that v and w are perpendicular.
1.3 Dot Product
17
Since cos θ > 0 for 0◦ ≤ θ < 90◦ and cos θ < 0 for 90◦ < θ ≤ 180◦, we also have:
Corollary 1.8. If θ is the angle between nonzero vectors v and w, then
> 0
for 0◦ ≤ θ < 90◦
v · w is
0
for θ = 90◦
< 0 for 90◦ < θ ≤ 180◦
By Corollary 1.8, the dot product can be thought of as a way of telling if the angle be-
tween two vectors is acute, obtuse, or a right angle, depending on whether the dot product
is positive, negative, or zero, respectively. See Figure 1.3.3.
w
w
w
90◦ < θ ≤ 180◦
θ = 90◦
0◦ ≤ θ < 90◦
v
v
v
(a) v · w > 0
(b) v · w < 0
(c) v · w = 0
Figure 1.3.3
Sign of the dot product & angle between vectors
Example 1.6. Are the vectors v = (−1,5,−2) and w = (3,1,1) perpendicular?
Solution: Yes, v ⊥ w since v · w = (−1)(3) + (5)(1) + (−2)(1) = 0.
The following theorem summarizes the basic properties of the dot product.
Theorem 1.9. For any vectors u, v, w, and scalar k, we have
(a) v · w = w · v
Commutative Law
(b) ( kv) · w = v · ( kw) = k(v · w)
Associative Law
(c) v · 0 = 0 = 0 · v
(d) u · (v + w) = u · v + u · w
Distributive Law
(e) (u + v) · w = u · w + v · w
Distributive Law
(f) |v · w| ≤ v
w
Cauchy-Schwarz Inequality5
Proof: The proofs of parts (a)-(e) are straightforward applications of the definition of the
dot product, and are left to the reader as exercises. We will prove part (f).
(f) If either v = 0 or w = 0, then v· w = 0 by part (c), and so the inequality holds trivially. So
assume that v and w are nonzero vectors. Then by Theorem 1.6,
v · w = cos θ v
w , so
|v · w| = |cos θ| v
w , so
|v · w| ≤ v
w
since |cos θ| ≤ 1.
QED
5Also known as the Cauchy-Schwarz-Buniakovski Inequality.
18
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
Using Theorem 1.9, we see that if u·v = 0 and u·w = 0, then u·( kv+ lw) = k(u·v)+ l(u·w) =
k(0) + l(0) = 0 for all scalars k, l. Thus, we have the following fact:
If u ⊥ v and u ⊥ w, then u ⊥ ( kv + lw) for all scalars k, l.
For vectors v and w, the collection of all scalar combinations kv + lw is called the span
of v and w. If nonzero vectors v and w are parallel, then their span is a line; if they are
not parallel, then their span is a plane. So what we showed above is that a vector which is
perpendicular to two other vectors is also perpendicular to their span.
The dot product can be used to derive properties of the magnitudes of vectors, the most
important of which is the Triangle Inequality, as given in the following theorem:
Theorem 1.10. For any vectors v, w, we have
(a) v 2 = v · v
(b) v + w ≤ v + w
Triangle Inequality
(c) v − w ≥ v − w
Proof: (a) Left as an exercise for the reader.
(b) By part (a) and Theorem 1.9, we have
v + w 2 = (v + w) · (v + w) = v · v + v · w + w · v + w · w
= v 2 + 2(v · w) + w 2 , so since a ≤ | a| for any real number a, we have
≤ v 2 + 2|v · w| + w 2 , so by Theorem 1.9(f) we have
≤ v 2 + 2 v
w + w 2 = ( v + w )2 and so
v + w ≤ v + w after taking square roots of both sides, which proves (b).
(c) Since v = w+(v−w), then v = w+(v−w) ≤ w + v−w by the Triangle Inequality, so subtracting w from both sides gives v − w ≤ v − w .
QED
v
The Triangle Inequality gets its name from the fact that in any triangle,
+ w
no one side is longer than the sum of the lengths of the other two sides (see
w
Figure 1.3.4). Another way of saying this is with the familiar statement “the
v
shortest distance between two points is a straight line.”
Figure 1.3.4
Exercises
A
1. Let v = (5,1,−2) and w = (4,−4,3). Calculate v · w.
2. Let v = −3i − 2j − k and w = 6i + 4j + 2k. Calculate v · w.
For Exercises 3-8, find the angle θ between the vectors