1.3. DOT PRODUCT
1.3
19
Dot Product
1.3.1
De…nitions and Properties
The dot product is the …rst way to multiply two vectors. The de…nition we will
give below may appear arbitrary. But it is not. It is motivated by applications,
in particular projections.
De…nition 34 (Dot Product) The dot product, also called inner product,
is denoted with the symbol : We can only take the dot product of two vectors
having the same dimension. The result is a number, obtained by multiplying the
corresponding coordinates of the two vectors, then adding all the products. If
~u = ha; bi and ~v = hc; di are two vectors, then ~u ~v = ac + bd. You will note that
the answer is a number, not a vector. We have a similar de…nition for vectors in
higher dimensions. In general, if ~u = hx1 ; x2 ;
; xn i and ~v = hy1 ; y2 ;
; yn i,
n
P
then ~u ~v =
xi yi .
i=1
Example 35 If ~u = h1; 1; 1i and ~v = h1; 2; 3i then ~u ~v = (1) (1) + (1) (2) +
(1) (3) = 6.
Proposition 36 The dot product satis…es the following properties:
1. ~u ~u = k~uk
2
2. ~u ~0 = ~0 ~u = 0
3. ~u ~v = ~v ~u
4.
~u
~v =
(~u ~v ) for any scalars
and .
5. ~u (~v + w)
~ = ~u ~v + ~u w
~
6. If is the angle (between 0 and ) between two non-zero vectors ~u and ~v ,
then ~u ~v = k~uk k~v k cos .
7. ~u ? ~v if and only if ~u ~v = 0. Again, we assume that ~u and ~v are two
non-zero vectors.
8. j~u ~v j
k~uk k~v k
Proof. We give an idea of a proof for some of these properties.
1. We see why this is true for a 3-D vector. A general proof is similar.
Suppose that ~u = ha; b; ci. Then, by de…nition
~u ~u = a2 + b2 + c2
Also by de…nition,
k~uk =
p
a2 + b2 + c2
20
CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE
Thus
2
k~uk
=
a2 + b2 + c2
= ~u ~u
2. Left as an exercise.
3. Left as an exercise.
4. Left as an exercise.
5. Left as an exercise.
6. Consider …gure 6. Using the law of cosines, we see that
k~u
2
2
2
~v k = k~uk + k~v k
2 k~uk k~v k cos
Thus
2 k~uk k~v k cos
=
=
=
=
2 k~uk k~v k cos
=
k~u
2
2
2
~v k + k~uk + k~v k
~v ) + ~u ~u + !
v !
v
!
!
(~u ~u 2~u ~v + v v ) + ~u ~u + !
v !
v
~u ~u + 2~u ~v !
v !
v + ~u ~u + !
v !
v
(~u
~v ) (~u
2~u ~v
Hence
~u ~v = k~uk k~v k cos
7. If the vectors are perpendicular, then the angle between them is 90 .
Thus, cos = 0. The result follows from property 6. Conversely, if ~u and
~v are two non-zero vectors such that ~u ~v = 0 then k~uk k~v k cos = 0. Since
k~uk 6= 0 and k~v k 6= 0 (~u and ~v are two non-zero vectors), it must be that
cos = 0 thus = 90 hence ~u ? ~v .
1.3. DOT PRODUCT
21
8. From 6, we see that
j~u ~v j
= jk~uk k~v k cos j
=
k~uk k~v k jcos j
k~uk k~v k
since jcos j
1:
Remark 37 We can also use the dot product to prove the triangle inequality
2
k~u + ~v k k~uk + k~v k. Using the fact that ~u ~u = k~uk , we have
2
k~u + ~v k
=
(~u + ~v ) (~u + ~v )
= ~u ~u + 2~u ~v + ~v ~v using properties of the dot product
2
2
= k~uk + 2~u ~v + k~v k
2
2
k~uk + 2 j~u ~v j + k~v k property of absolute value
2
2
k~u + ~v k
2
k~uk + 2 k~uk k~v k + k~v k by property 8 above
(k~uk + k~v k)
2
Since both k~u + ~v k and k~uk + k~v k are positive, we can take the square root on
both side of the last equality to obtain the result.
Remark 38 Property 6 in proposition 36 can also be used to determine how
close two non-zero vectors are to pointing in the same direction. Given two
non-zero vectors ~u and ~v , consider P, the plane passing through the origin and
perpendicular to ~u. ~v will be on the same side of P as ~u if and only if ~u ~v > 0
because in that case, the angle between the two vectors will be less than 90 and
its cosine will be positive.
Remark 39 Property 7 in proposition 36 is also very important, it gives us a
quick way to check if two vectors are perpendicular. Another term for perpendicular is orthogonal. We de…ne the zero vector ~0 = (0; 0; :::) to be orthogonal
to every vector.
Remark 40 It is important to understand that when we talk about the angle
between two vectors, we are talking about the smallest positive angle between
them in other words an angle such that 0
.
We illustrate some of these properties with examples.
Example 41 Suppose that the length of a vector !
u is 2, the length of a vector
!
v.
v is 3 and they make an angle of 135 as shown in …gure 1.11. Find ~u !
~u !
v = k~uk k~v k cos 135
p !
2
= (2) (3)
2
p
=
3 2
22
CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE
Figure 1.11: Compute !
u !
v if k!
u k = 2 and k!
vk=3
Example 42 What is the angle in degrees between !
u = h1; 1; 1i and !
v =
h2; 1; 0i?
From the formula ~u ~v = k~uk k~v k cos , we can compute cos and therefore .
First, let us compute the various quantities needed.
~u ~v
=
(2) (1) + (1) (1) + (1) (0)
=
3
k~uk
=
=
k!
vk
p
p
12 + 12 + 12
3
p
22 + 1 2
p
5
=
=
It follows that
=
cos
1
=
cos
1
=
39:23
~u ~v
k~uk k~v k
3
p
15
1.3. DOT PRODUCT
23
Figure 1.12: proj ~a~b and orth ~a~b
1.3.2
Projections
It often happens that we need to decompose a vector into components that
are parallel and perpendicular to a given vector. Or, as shown in …gure 1.12,
given two non-zero vectors ~a and ~b, one needs to …nd the projection of ~b onto ~a
and the projection of ~b onto a vector perpendicular to ~a. Let us denote proj ~a~b
the projection of ~b onto ~a, comp~a~b the component of ~b along ~a (also known as
the scalar projection of ~b onto ~a) and orth ~a~b the projection of ~b onto a vector
perpendicular to ~a.Let us call the angle between ~a and ~b. Then, we have:
cos
=
comp~a~b
~b
Therefore
comp~a~b = ~b cos
Using Property 6 in proposition 36, we can write
comp~a~b =
It follows that
~b ~a ~b
k~ak ~b
~a ~b
comp~a~b =
k~ak
(1.4)
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CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE
To …nd proj ~a~b, we simply multiply comp~a~b by a unit vector in the direction of
~a
~a. Such a vector is
and therefore
k~ak
~a ~b
a
proj ~a~b =
2~
k~ak
Finally, orth ~a~b = ~b
(1.5)
proj ~a~b and therefore
~a ~b
orth ~a~b = ~b
a
2~
k~ak
(1.6)
!
Example 43 Find the scalar projection and vector projection of b = h1; 1; 2i
onto !
a = h 2; 3; 1i.
Let us …rst compute the various quantities involved.
!
!
a b
=
=
j!
aj =
=
q
p
2+3+2
3
2
( 2) + 32 + 12
14
Therefore
3
comp~a~b = p
14
and
1.3.3
3
h 2; 3; 1i
proj ~a~b =
14
Applications
The dot product has many applications including:
1. Use it to see if two non-zero vectors are perpendicular. Recall that two
non-zero vectors are perpendicular if and only if their dot product is 0.
2. Use it to compute the angle between two vectors. From the formula ~u ~v =
k~uk k~v k cos , if we are given ~u and ~v , we have
= cos
1
~u ~v
k~uk k~v k
3. Use it to quickly determine if the angle between two vectors is acute (between 0 and 90 degrees) or obtuse (between 90 and 180 degrees). Given ~u
and ~v , if the angle between them is acute, then ~u ~v > 0 and if it is obtuse,
~u ~v < 0.
1.3. DOT PRODUCT
25
4. Given a vector and a plane perpendicular to it, use it to determine if a
second vector is on the same side of the plane or the opposite side. If the
two vectors are on the same side of the plane, the angle between them will
be acute. If they are on opposite sides, it will be obtuse. So, we can use
the technique describe above.
5. Use it to …nd the projection of a vector onto another one.
6. Use it to compute the work done by a force acting on an object and
!
displacing it. In physics, the work W done by a force F moving an
!
object along the displacement vector r is de…ned to be the product of
!
the component of F along !
r by the magnitude of the displacement (see
…gure 6). This gives us:
W
! !
comp !
r F jr j
! !
F r !
=
jrj
j!
rj
!
= F !
r
=
We illustrate that last application with an example.
!
Example 44 A crate is displaced by 8m up a ramp by a constant force F of
200N applied at an angle of 25 degrees to the ramp. Find W , the work done.
26
CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE
!
!
Let D be the displacement vector. Then D = 8 and F = 200:
W
! !
F D
!
= D F cos 25
=
=
(8) (200) cos 25
1450N:m
1450J
Note that in the metric system, distances are in meters (m), forces are in Newton
(N), work is expressed in Joules (J).
1.3.4
Problems
1. Prove that the normal to the line ax + by + c = 0 (in 2D) is ha; bi (hint:
Find 2 points on the line, deduce what a vector parallel to the line is.
From there, …nd a vector perpendicular to it).
2. For each problem below, …nd the following:
!
v !
u , k!
v k, k!
uk
The cosine of the angle between !
u and !
v.
!
comp! u
v
!
proj!
v u
!
!
!
! p !
u = 2 i +4j
(a) !
v =2i
4 j + 5 k and !
!
!
!
!
!
(b) !
v = 10 i + 11 j
2 k and !
u =3j +4k
!
!
! ! !
(c) !
v =5j
3 k and !
u = i + j + k
! !
! p !
(d) !
v = 5 i + j and !
u = 2 i + 17 j
p !
5k
! !
!
! !
3. Find the angle between !
u = 2 i + j and !
v = i +2j
k.
p
p
!
!
!
! !
4. Find the angle between !
u = 3i
7 j and !
v = 3i + j
2k.
5. Find the measures of the angles of the triangle whose vertices are A =
( 1; 0), B = (2; 1) and C = (1; 2).
6. Let , , and
!
!
!
be the direction angles of !
v =a i +bj +ck
a
b
c
, cos = !
, and cos = !
and cos2 +
vk
k!
kvk
kvk
cos2 + cos2 = 1. These are called the direction cosines.
!
!
!
(b) Show if !
v = a i + b j + c k is a unit vector then a, b, c are the
direction cosines.
(a) Show that cos
=
1.3. DOT PRODUCT
27
7. Show that the diagonals of a rhombus (parallelogram with sides of equal
length) are perpendicular.
8. Prove that a parallelogram is a rectangle if and only if its diagonals are
equal in length.
9. A gun with muzzle velocity of 1200f t=s is …red at an angle of 8 above the
horizontal. Find the horizontal and vertical components of the velocity.
10. Find an equation of the line through P (2; 1) perpendicular to !
v = h1; 2i.
11. Find an equation of the line through P ( 2; 7) perpendicular to !
v =
h 2; 1i.
12. Find an equation of the line through P ( 2; 1) parallel to !
v = h1; 1i.
13. Find an equation of the line through P (1; 2) parallel to !
v = h 1; 2i.
!
!
14. Find the work done by the force F = 5 i when it moves an object from
the origin to the point (1; 1) where distances are in meters.
15. Using the fact that the acute angle between intersecting lines which
are not perpendicular is the same as the angle between their normals, …nd
the acute angle between the lines 3x + y = 5 and 2x y = 4.
16. Using the fact that the acute angle between intersecting lines which
are not perpendicular is the same as
p normals, …nd
p the angle between their
3y = 1.
the acute angle between the lines 3x y = 2 and x