The Inner Product
Part 1: Elementary Properties
In elementary physics and mathematics courses, vectors are often defined to be
quantities that have magnitude and direction. In reality, however, the magnitude and direction of a vector depend on the choice of an inner product. In this
section, we introduce the standard inner product and how it is used to define
the usual notions of magnitude and angle.
To begin with, if u = u1 , u2 , u3 and v = v1 , v2 , v3 , then their inner
product is defined
u · v = u1 v1 + u2 v2 + u3 v3
For example, if u = 1, 2, 3 and v = 5, 6, 7 , then
u · v = 1 · 5 + 2 · 6 + 3 · 7 = 38
Likewise, if u = −2, 3 and v = 9, 6 , then
u · v = −2 · 9 + 3 · 6 = 0
An inner product is also known as a dot product, and the inner product of
2 vectors is a scalar–i.e., the result is a number. Other properties of the inner
product which follow immediately from the definition include the following:
Theorem 1.2: If u, v and w are vectors and k is a scalar, then
i)
ii)
iii)
iv)
u·v =v·u
(ku) · v = k (u · v) = u· (kv)
u· (v + w) = u · v + u · w
v · v ≥ 0 and v · v = 0 only if v = 0
If v = a, b , then v · v = a2 + b2 , so that in accordance with the Pythagorean
theorem, the length v of the vector v is defined to be
√
v = a2 + b2 = v · v
That is, the length v of a 2 dimensional vector satisfies the identity v2 =
v · v.
1
EXAMPLE 1
Find the length of v = 3, 4
Solution: Our definition implies that the length of v satisfies
2
v = v · v = 32 + 42 = 25
Thus, the length of vector v = 3, 4 is v = 5.
Likewise, we define the length of a 3 dimensional vector v = a, b, c to be
√
v = a2 + b2 + c2 = v · v
It follows that if u, v are vectors and k is a scalar, then the following properties
hold:
i.
u + v ≤ u + v
(triangle inequality)
ii. kv = |k| v
iii. v = 0 if and only if v = 0.
We will prove these in the exercises.
EXAMPLE 2 The distance between two points P1 (x1 , y1 , z1 ) and
P2 (x2 , y2 , z2 ) is defined to be the length of the vector between them.
What is the distance from P1 to P2 ?
Solution: The vector between P1 and P2 is
−−−→
v = P1 P2 = x2 − x1 , y2 − y1 , z2 − z1 Consequently, the distance between P1 and P2 is
√
Dist (P1 , P2 ) = v · v = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2
2
Since a sphere of radius R centered at (a, b, c) is the set of all point (x, y, z) that
are a distance R from (a, b, c) , the distance formula in example 2 implies that
the equation of the sphere of radius R centered at (a, b, c) is
2
2
2
(x − a) + (y − b) + (z − c) = R2
The unit sphere is the sphere with equation x2 + y2 + z 2 = 1.
Check your Reading: How long is the vector v = 2, 2, 1?
Part 2: The Angle Between 2 Vectors
Next, let’s use the standard inner product to define an angle formed by two
vectors. We observed in the last section that if u and v are two dimensional
vectors, then there are angles α and β such that
u = u cos (α) , u sin (α)
and
v = v cos (β) , v sin (β)
As a result, the inner product of u and v is
u·v
= u cos (α) · v cos (β) + u sin (α) · v sin (β)
= u v cos (α) · cos (β) + u v sin (α) sin (β)
= u v (cos (α) cos (β) + sin (α) sin (β))
The difference-of-the-angles identity for the cosine thus leads us to
u · v = u v cos (β − α)
However, θ = β − α is the angle formed when u and v share the same initial
point:
This leads us to the following definition:
3
Definition 1.3: Any two non-zero vectors u and v determine an
angle θ which satisfies
u · v = u v cos (θ)
Moreover, we also use definition 1.3 to define the angle between two 3 dimensional vectors.
EXAMPLE 3 Find the angle in degrees between the vectors u =
−2, 2, 1 and v = 2, 3, 6 .
Solution: To begin with,
√
u = 4 + 4 + 1 = 3,
v =
√
4 + 9 + 36 = 7
Definition 2.3 implies that θ satisfies the equation
u·v
u v
cos (θ) =
(1)
so that u · v = −2 · 2 + 2 · 3 + 1 · 6 = 8 implies that
cos (θ) =
8
8
=
3·7
21
A scientific calculator then reveals that
8
−1
θ = cos
= 1.17997 radians
21
Multiplication of θ = 1. 17997 by 180◦ /π then yields θ = 67. 61◦ .ss
4
Definition 1.3 also implies that if two vectors u and v form a right angle, then
their inner product is 0. In particular, we say that u and v are orthogonal if
they form a right angle, so that definition 1.3 leads to the following:
Theorem 1.4: Vectors u and v are orthogonal only if u · v = 0.
Right angles are essential to the study of vectors, which means that theorem
1.4 is an important result.
EXAMPLE 4 Find a number k such that u = 2, 3, 4 is orthogonal
to v = k, 3, −7 .
Solution: To do so, we compute u · v and set it equal to 0:
u·v = 0
2·k+3·3−4·7 = 0
2k − 19 = 0
19
k =
= 9.5
2
5
Thus, v = 9.5, 3, −7 is orthogonal to u = 2, 3, 4 .
EXAMPLE 5
Show that i, j, and k are mutually orthogonal.
Solution: Since i = 1, 0, 0 and j = 0, 1, 0 , their inner product
satisfies
i · j = 1, 0, 0 · 0, 1, 0 = 1 · 0 + 0 · 1 + 0 · 0 = 0
Since i · j = 0, the vectors i and j are orthogonal. Likewise, j · k = 0
and k · i = 0, showing that j ⊥ k and k ⊥ i.
In fact, the vectors i, j, and k satisfy
i·i = j·j=k·k =1
i·i = j·j=k·k =0
(2)
(3)
Any set of 3 vectors that satisfies (2) and (3) is known as an orthonormal
basis for the vector space of 3-dimensional vectors. Thus, i, j, and k form an
orthonormal basis.
6
In chapter 3, we will encounter different orthonormal bases for the vector space
of 3-dimensional vectors.
Check your Reading: For what value of k is v = −7, k orthogonal to
u = 3, 7 .
Part 3: Unit Vectors and Projections
A vector with a length of 1 is known as a unit vector.
nonzero vector v can be written uniquely in the form
It follows that any
v = v u
(4)
where u is a unit vector. Solving for u in (4) reveals that if v = 0, then
u=
v
v
(5)
Since v is the magnitude of v, the unit vector u is understood to represent
the direction of v.
6
√ EXAMPLE
1, 3 .
Find the direction and decomposition (4) of v =
√
Solution: The length of v is v = 1 + 3 = 2. Thus, the direction
of v is
√ √ 1, 3
v
1
3
u=
=
=
,
v
2
2 2
7
It follows that v can be written in terms of its magnitude and direction as
√ 1
3
v = v u = 2
,
2 2
The projection of a vector v onto a nonzero vector p is defined
v·p
projp (v) =
p
p·p
(6)
Geometrically, the projection of v onto p is parallel to p. Moreover, notice that
v·p
v−
p ·p
(v − projp (v)) · p =
p·p
v·p
= v·p−
p·p
p·p
= v·p−v·p
= 0
That is, v − projp (v) is orthogonal to p. Geometrically, this means that
projp (v) and v − projp (v) form the sides of a right triangle.
EXAMPLE 7 Find the projection of v = 0, 4, 1 onto p = 2, 2, 1 .
Also, find ||projp (v)|| and show that
w = v − projp (v)
is orthogonal to p.
Solution: To do so, we use the formula (6) to obtain
projp (v) =
v·p
0·4+4·4+1·2
18
p=
4, 4, 2 =
4, 4, 2
p·p
4·4+4·4+2·2
36
Thus, projp (v) = 2, 2, 1 and w = 0, 4, 1 − 2, 2, 1 = 2, −2, 0 .
Also,
|v · p|
0·4+4·4+1·2
||projp (v)|| =
=√
=3
||p||
4·4+4·4+2·2
8
and projp (v) · w = 2 (2) + 2 (−2) = 0.
Projections are important in many different applications, as we will see in the
exercises, in the worksheet, and in several sections of the text.
Check your reading: What is ||P|| if P = 2, 2, 1? How is this related to
example 7?
Part 4: Direction Angles and Direction Cosines
Given a vector v, let us let α denote the angle between v and the x-axis, let
us let β denote the angle between v and the y-axis, and let us let γ denote the
angle between v and the z-axis.
9
The angles α, β, and γ are called the direction angles for v.
If v = a, b, c = ai + bj + ck, then
v·i =a i·i+b i·j+c i·k =a
Similarly, b = v · j and c = v · k, so that
a = ||v|| cos (α) , b = ||v|| cos (β) , c = ||v|| cos (γ)
When combined with v = a, b, c , this leads to
v = ||v|| cos (α) , cos (β) , cos (γ)
That is, the direction vector for v is the unit vector given by
u = cos (α) , cos (β) , cos (γ)
The components of u are the direction cosines for v.
EXAMPLE 8
for
Determine the direction cosines and direction angles
√ v = 1, 1, 2
Solution: Since ||v||2 = 1 + 1 + 2 = 4, the magnitude is ||v|| = 2.
Equation (7) implies that
√
1 = 2 cos (α)
1 = 2 cos (β)
2 = 2 cos (γ)
cos (α) = 12
cos (β) = 12
cos (γ) = √22
γ = π3
γ = π3
γ = π4
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(7)
as is shown below:
Exercises
Sketch the two vectors. Then find the length of each of the two vectors and their
direction. Finally, use (1) to find the angle between them.
1.
3.
5.
7.
9.
u = 3, 4 , v = 0, 2
u = −1, 5 , v = 2, 9
u = 6, 8 , v = 8, −6
u = 2, 2, 1 , v = 6, 0, 2
u = 0, 0, 1 , v = 1, 0, 0
2.
4.
6.
8.
10.
u = 5, 12 , v = 5, 0
u = 1, 11 , v = 3, 8
u = 3, 4 , v = 6, 8
u = 3, 1, 4 , v = 2, 0, 2
u = 0, 0, 1 , v = 2, 0, 2
Find the value(s) of the parameter k for which the two vectors are orthogonal.
11.
13.
15.
17.
u = 2, k , v = 6, 1
u = 3, 1, 2 , v = −2, k, 2
u = k, 2, k , v = k, 3, 1
u = −0.1, −2.3, 7.5 , v = 2.3, 3.5, k
11
12.
14.
16.
18.
u = k, 7 , v = −1, 5
u = 0, 2, 1 , v = 6, k, 2
u = 0.2, 3k, 1 , v = 6, 9.1, 2
u = 0.2, 3k, 1 , v = 6, 9.1, 2
Find the direction cosines and direction angles for the following vectors.
√ √ √ 19. v = √
1, √
3
20. v = 3,
√ 3
21. v =
2, 2, 2
22. v = 1, √2, 1
23. v = 1, 0, 1
24. v = 1, 3, 0
Find the projection of the vector v onto the vector p. Then show that
w = v − projp (v)
is perpendicular to p.
25.
27.
29.
p = 1, 7 , v = 2, 11
p = 1, 2, 1 , v = 2, 2, 7
p = j, v = 2i − 3j + 4k
26.
28.
30.
p = 3, 4 , v = 7, 4
p = 3, 5, 9 , v = 2, 11, 6
p = k, v = 2i − 3j + 4k
31. Find a 2-dimensional unit vector u which forms an angle of 30◦ with the
vector v = 3, 4 .
32. Find a 2-dimensional vector u with length 4 which forms an angle of 135◦
with the vector v = 4, 1 .
33. Is there a value of k such that u = 3, 0, 2 is perpendicular to v =
−2, k, 2? Can you explain geometrically why v cannot be perpendicular to u
for any value of k?
34.
Find the projection of the vector v = 3, 1, 4 onto the vector p =
2, −10, 1
35. A frictionless bead of mass 1 kg slides down a wire which is parallel to the
vector v = 1, 2 .ssxx
Find projv (Fg ) , which is the part of the gravitational force in the direction of
motion of the bead. (Hint: use −9.8 m/ sec2 for the acceleration due to gravity
near the earth’s surface).
36. Repeat exercise 35 if v forms an angle of 30◦ with the horizontal.
37. A 10 kg block is initially at rest on a board with a coefficient of static
friction of µs = 0.2 between them. The board runs 7 feet horizontally and rises
3 feet vertically. The force of gravity acting on the block is Fg = 0, −98 (i.e.,
the product of the mass and an acceleration of −9.8 m/ sec2 ).
12
The force parallel to the incline is Finc , which is the projection of Fg onto the
vector 7, 3 parallel to the board. The force normal Fn is perpendicular to
Finc . Specifically, Fn is the negative of
w = Fg − Finc
Finally, the force of static friction is the force in the opposite direction of Finc
with a magnitude of µs ||Fn || (that is,
Finc
Ff ric = −µs ||Fn ||
||Finc ||
What are Finc , Fn , and Ff ric ?
38. The block in exercise 37 will begin to slide if ||Finc || ≥ µs ||Fn || . Does this
block begin to slide?
39. Repeat exercises 37 and 38 for the block shown below left.
Exercise 39
Exercise 40
40. Suppose a board with a coefficient of static friction of µs = 0.2 is lifted
from horizontal as shown above right. Use exercise 38 to determine the angle θ
at which the block begins to slide down the board. (Note: this is also a method
for measuring µs ).
41. Use the properties of the inner product to show that
u + v2 = u2 + 2u · v + v2
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42. Use the properties of the inner product to show that
2
2
2
u − v = u − 2u · v + v
and then use the result to prove the law of cosines,
c2 = a2 + b2 − 2ab cos (θ)
43. Substitute the vectors u = u1 , u2 , u3 and v = v1 , v2 , v3 into the following
identities to prove theorem 2.1.
i)
ii)
iii)
iv)
u·v =v·u
(ku) · v = k (u · v) = u· (kv)
u· (v + w) = u · v + u · w
v · v ≥ 0 and v · v = 0 only if v = 0
44. We call u · v = u1 v1 + u2 v2 + u3 v3 the standard inner product because it
is possible to define other inner products. Indeed, show that if u = u1 , u2 , u3 and v = v1 , v2 , v3 , then
u ◦ v = 3u1 v1 + 4u2 v2 + 5u3 v3
also satisfies the properties in theorem 2.1.
45. Let v = a, b, c and let k be a scalar.
1. (a) Compute kv .
(b) Show that the result in (a) reduces to |k| v .
46. Let u and v be vectors.
1. (a) Explain why 2u · v ≤ 2 u v.
(b) Use (a) and the result in problem 45 to show that
u + v2 ≤ u2 + 2 u v + v2
and then explain why this implies the triangle inequality u + v ≤
u + v.
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47. What is the angle between a diagonal of a cube and the sides of the cube
it is adjacent to?
48. Suppose we have a pyramid whose base is a square and whose sides are
equilateral triangles.
What is the angle between an edge of the triangle and an adjacent diagonal of
the base?
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