dot product angles-projection

1. Dot product and Angles
* Vector Dot Product and Angels
* The Unitized–Vector (Unit Directional–Vector)
* Orthogonal Projections

2. Dot product and Angles
The Dot Product
Let u = <a, b> and v = <c, d> be two unit vectors, i.e.
vectors of length 1, and  be the angle between them,
then cos() = ac + bd.

3. Dot product and Angles
The Dot Product
Let u = <a, b> and v = <c, d> be two unit vectors, i.e.
vectors of length 1, and  be the angle between them,
then cos() = ac + bd.

u = <a, b>
v = <c, d>
1

4. Dot product and Angles
The Dot Product
Let u = <a, b> and v = <c, d> be two unit vectors, i.e.
vectors of length 1, and  be the angle between them,
then cos() = ac + bd.

u = <a, b>
v = <c, d>
1
cos( ) = ac + bd

5. Dot product and Angles
The Dot Product
Let u = <a, b> and v = <c, d> be two unit vectors, i.e.
vectors of length 1, and  be the angle between them,
then cos() = ac + bd. This sum is denoted as u • v
and it’s the dot product of u and v.

u = <a, b>
v = <c, d>
1
cos( ) = ac + bd

6. Dot product and Angles
This actually is the “cosine of
difference–of–angles” formula.

u = <a, b>
v = <c, d>
The Dot Product
Let u = <a, b> and v = <c, d> be two unit vectors, i.e.
vectors of length 1, and  be the angle between them,
then cos() = ac + bd. This sum is denoted as u • v
and it’s the dot product of u and v.
1
cos( ) = ac + bd

7. Dot product and Angles

u = <a, b>
v = <c, d>
AThis actually is the “cosine of
difference–of–angles” formula.
Let u = <a, b> and v = <c, d> with
angles A and B be as shown.
The Dot Product
Let u = <a, b> and v = <c, d> be two unit vectors, i.e.
vectors of length 1, and  be the angle between them,
then cos() = ac + bd. This sum is denoted as u • v
and it’s the dot product of u and v.
1
cos( ) = ac + bd

8. Dot product and Angles

u = <a, b>
v = <c, d>
A
B
This actually is the “cosine of
difference–of–angles” formula.
Let u = <a, b> and v = <c, d> with
angles A and B be as shown.
The Dot Product
Let u = <a, b> and v = <c, d> be two unit vectors, i.e.
vectors of length 1, and  be the angle between them,
then cos() = ac + bd. This sum is denoted as u • v
and it’s the dot product of u and v.
1
cos( ) = ac + bd

9. Dot product and Angles
The Dot Product
Let u = <a, b> and v = <c, d> be two unit vectors, i.e.
vectors of length 1, and  be the angle between them,
then cos() = ac + bd. This sum is denoted as u • v
and it’s the dot product of u and v.

u = <a, b>
v = <c, d>
A
B
This actually is the “cosine of
difference–of–angles” formula.
Let u = <a, b> and v = <c, d> with
angles A and B be as shown.
We have  = A – B, so
cos() = cos(A – B)
= cos(A)cos(B) + sin(A)sin(B)
= ac + bd = u • v.
1
cos( ) = ac + bd

10. Dot product and Angles
The Dot Product
Let u = <a, b> and v = <c, d> be two unit vectors, i.e.
vectors of length 1, and  be the angle between them,
then cos() = ac + bd. This sum is denoted as u • v
and it’s the dot product of u and v.

u = <a, b>
v = <c, d>
A
B
This actually is the “cosine of
difference–of–angles” formula.
Let u = <a, b> and v = <c, d> with
angles A and B be as shown.
We have  = A – B, so
cos() = cos(A – B)
= cos(A)cos(B) + sin(A)sin(B)
= ac + bd = u • v.
Let’s generalize the dot product to all vectors.
1
cos( ) = ac + bd

11. Dot Product of Vectors

12. Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd

13. Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors

14. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors

15. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors

16. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
= 5*4 + (-6)*(-5)
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors

17. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors

18. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
b. 3u  v = 3<5, -6>  <4, -5>
Dot Product of Vectors

19. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
b. 3u  v = 3<5, -6>  <4, -5>
= <15, -18>  <4, -5>
Dot Product of Vectors

20. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
b. 3u  v = 3<5, -6>  <4, -5>
= <15, -18>  <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150
Dot Product of Vectors

21. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
b. 3u  v = 3<5, -6>  <4, -5>
= <15, -18>  <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150 = 3(u  v)
Dot Product of Vectors

22. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
b. 3u  v = 3<5, -6>  <4, -5>
= <15, -18>  <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150 = 3(u  v)
Facts: 1. u•v = v•u
Dot Product of Vectors

23. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
b. 3u  v = 3<5, -6>  <4, -5>
= <15, -18>  <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150 = 3(u  v)
Facts: 1. u•v = v•u
2. For any scalar λ, λu • v = λ(u • v)
Dot Product of Vectors

24. Example A: Let u = <5, -6>, v = <4, -5>
a. u  v = <5, -6>  <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u  v = <a, b>  <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
b. 3u  v = 3<5, -6>  <4, -5>
= <15, -18>  <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150 = 3(u  v)
Facts: 1. u•v = v•u
2. For any scalar λ, λu • v = λ(u • v)
3. u•(v + w) = u•v + u•w

25. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Unitized Vectors

26. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
Unitized Vectors

27. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
Unitized Vectors

28. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
Unitized Vectors

29. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is1
5
1
5
Unitized Vectors

30. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>|
1
5
1
5
3
5
4
5
1
5
Unitized Vectors

31. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>| = * 5 = 1
1
5
1
5
3
5
4
5
1
5
1
5
Unitized Vectors

32. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>| = * 5 = 1
1
5
1
5
3
5
4
5
1
5
1
5
Vectors of length 1
are called unit vectors.
Unitized Vectors

33. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>| = * 5 = 1
1
5
1
5
3
5
4
5
1
5
1
5
Vectors of length 1
are called unit vectors.
<3/5, 4/5>
r = 1
Unit vectors
Unitized Vectors

34. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>| = * 5 = 1
1
5
1
5
3
5
4
5
1
5
1
5
Vectors of length 1
are called unit vectors.
In the above example,
3
5
4
5
< , > is a unit vector.
<3/5, 4/5>
r = 1
Unit vectors
Unitized Vectors

35. Unitized Vectors

36. Unitized Vectors
In the above example b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.3
5
4
5,

37. Unitized Vectors
In the above example b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that the “unitized v” is < >.
3
5
4
5,
3
5
4
5,

38. Unitized Vectors
From this example, we see the formula for
the unitized v is since | v | = 1.1
|v|
1
|v| |v| =
1
|v|
v
In the above example b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that the “unitized v” is < >.
3
5
4
5,
3
5
4
5,

39. Unitized Vectors
Geometrically, unitizing v
corresponds to shrinking
or stretching v to be a
unit vector.
From this example, we see the formula for
the unitized v is since | v | = 1.1
|v|
1
|v| |v| =
1
|v|
v
In the above example b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that the “unitized v” is < >.
3
5
4
5,
3
5
4
5,

40. Unitized Vectors
From this example, we see the formula for
the unitized v is since | v | = 1.1
|v|
1
|v| |v| =
1
|v|
v
Geometrically, unitizing v
corresponds to shrinking
or stretching v to be a
unit vector.
v
r=1
unitized v
In the above example b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that the “unitized v” is < >.
3
5
4
5,
3
5
4
5,
=<3, 4>
1
|v|
v = <3/5,4/5>
unitized v
=
Unitized v is the only unit vector
that has the same direction so it’s
the unit directional vector of v.

41. Unitized Vectors
From this example, we see the formula for
the unitized v is since | v | = 1.1
|v|
1
|v| |v| =
1
|v|
v
Geometrically, unitizing v
corresponds to shrinking
or stretching v to be a
unit vector.
v
r=1
unitized v
In the above example b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that the “unitized v” is < >.
3
5
4
5,
3
5
4
5,
=<3, 4>
1
|v|
v = <3/5,4/5>
unitized v
=
Unitized v is the only unit vector
that has the same direction so it’s
the unit directional vector of v.

42. Dot product and Angles

43. Dot product and Angles
The following theorem put all the above concepts
together.

44. Dot product and Angles
Dot Product Theorem:
The following theorem put all the above concepts
together.

45. Dot product and Angles
Dot Product Theorem:
a. Let u and v be two unit vectors and  be
the angle between them, then cos() = u•v.
The following theorem put all the above concepts
together.

46. Dot product and Angles
Dot Product Theorem:
a. Let u and v be two unit vectors and  be
the angle between them, then cos() = u•v.
b. Let u and v be any two vectors and  be
the angle between them
cos() = (unitized u)  (unitized v)
The following theorem put all the above concepts
together.

47. Dot product and Angles
Dot Product Theorem:
a. Let u and v be two unit vectors and  be
the angle between them, then cos() = u•v.
b. Let u and v be any two vectors and  be
the angle between them
cos() = (unitized u)  (unitized v)
=  v
|v|
u
|u|
Easy to memorize
The following theorem put all the above concepts
together.

48. Dot product and Angles
Dot Product Theorem:
a. Let u and v be two unit vectors and  be
the angle between them, then cos() = u•v.
b. Let u and v be any two vectors and  be
the angle between them
cos() = (unitized u)  (unitized v)
=  =v
|v|
u
|u|
u•v
|u|*|v|
Easy to memorize Easier way to compute
The following theorem put all the above concepts
together.

49. Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.

50. uv=10,
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.

51. uv=10, |u|=10,
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.

52. uv=10, |u|=10, |v|=20,
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.

53. uv=10, |u|=10, |v|=20, hence,
cos() = =
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20

54. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20

55. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2)
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20

56. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20

57. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.

58. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>,

59. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>

60. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10

61. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50,

62. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50, |u – v| = 10

63. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50, |u – v| = 10
So cos(A) = -10/500

64. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50, |u – v| = 10
So cos(A) = -10/500  -0.447

65. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So  = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle 
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50, |u – v| = 10
So cos(A) = -10/500  -0.447
A = cos-1(-0.447)  117o.

66. Dot product and Angles
Theorem: Given u, v two nonzero vectors such that
u•v=0, then u and v are perpendicular (orthogonal).

67. Dot product and Angles
Theorem: Given u, v two nonzero vectors such that
u•v=0, then u and v are perpendicular (orthogonal).
If  is the angle between u and v and u•v = 0,
then cos() = = 0.
Hence  = cos-1(0) = 90o
u•v
|u|*|v|

68. Example D: <5, -3>•<3, 5> = 15 – 15 = 0,
hence they are orthogonal.
Dot product and Angles
Theorem: Given u, v two nonzero vectors such that
u•v=0, then u and v are perpendicular (orthogonal).
If  is the angle between u and v and u•v = 0,
then cos() = = 0.
Hence  = cos-1(0) = 90o
u•v
|u|*|v|

69. Example D: <5, -3>•<3, 5> = 15 – 15 = 0,
hence they are orthogonal.
Dot product and Angles
Theorem: Given u, v two nonzero vectors such that
u•v=0, then u and v are perpendicular (orthogonal).
If  is the angle between u and v and u•v = 0,
then cos() = = 0.
Hence  = cos-1(0) = 90o
u•v
|u|*|v|
From this calculation, we see that if u = <a, b>, then
the vectors <–b, a> and <b, –a> are orthogonal to u.

70. Example D: <5, -3>•<3, 5> = 15 – 15 = 0,
hence they are orthogonal.
Dot product and Angles
Theorem: Given u, v two nonzero vectors such that
u•v=0, then u and v are perpendicular (orthogonal).
If  is the angle between u and v and u•v = 0,
then cos() = = 0.
Hence  = cos-1(0) = 90o
u•v
|u|*|v|
From this calculation, we see that if u = <a, b>, then
the vectors <–b, a> and <b, –a> are orthogonal to u.
The vector <–b, a> and <b, –a> are u rotated 90o
counter–clockwise, and 90o clockwise respectively.
Exercise: Draw them to convince yourself that it is so.

71. Projections

72. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).

73. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).

74. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)

75. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)
Let  < 1800 be the angle between u and v,
the signed length of the projv(u) is |u |*cos().

76. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)
If  is less than 90o then
cos() > 0, so |u|cos() is
also positive.
Let  < 1800 be the angle between u and v,
the signed length of the projv(u) is |u |*cos().

77. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)
If  is less than 90o then
cos() > 0, so |u|cos() is
also positive. This means
the shadow is cast in the
direction of v.
Let  < 1800 be the angle between u and v,
the signed length of the projv(u) is |u |*cos().

78. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)
If  is less than 90o then
cos() > 0, so |u|cos() is
also positive. This means
the shadow is cast in the
direction of v.
u
v
|u|cos() is positive means the
projv(u) is in the direction of v

 < 90o
Let  < 1800 be the angle between u and v,
the signed length of the projv(u) is |u |*cos().

79. Projections
If  is more than 90o then
cos() < 0, so |u|cos() is
negative.

80. Projections
If  is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.

81. Projections
If  is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.

 >90o

82. Projections
If  is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.

 >90o

83. Projections
If  is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.

Example E: Given that the
angle between u and v is
67o and |u|=23, |v|=28.
Find the length of the
shadow projv(u). Draw.
 >90o

84. Projections
If  is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.

Example E: Given that the
angle between u and v is
67o and |u|=23, |v|=28.
Find the length of the
shadow projv(u). Draw.
|u|=23
67o
|v|=28
 >90o

85. Projections
If  is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.

Example E: Given that the
angle between u and v is
67o and |u|=23, |v|=28.
Find the length of the
shadow projv(u). Draw.
|u|=23
67o
|v|=28
The length of projv(u) = |u|cos() = 23cos(67)  8.99
 >90o

86. Projections
Example F: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.

87. Projections
Example F: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28

88. Projections
Example F: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28

89. Projections
Example F: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
The length of proju(v) = |v|cos()

90. Projections
Example F: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
The length of proju(v) = |v|cos()
= 28cos(117)  -12.7

91. Projections
Example F: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
The length of proju(v) = |v|cos()
= 28cos(117)  -12.7
Note the negative answer indicates the shadow is
cast in the opposite direction of u.

92. Projections
Example F: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
The length of proju(v) = |v|cos()
= 28cos(117)  -12.7
Note the negative answer indicates the shadow is
cast in the opposite direction of u.
Your Turn: Given the angle
between u and v is 135o and
|u|=36, |v|=18. Draw and find
the length of the projv(u).
135o
|u|=36
v
Ans: Signed length of projv(u) = –36/2  –25.4
projv(u)

93. Signed Length of Standard Projection

94. Theorem: Given two vectors u and v in the standard
position and  be the angle between them,
the signed length of projv(u) = |u|cos() = u•
or it’s u dotted with unitized v.
Signed Length of Standard Projection
v
|v|

95. Theorem: Given two vectors u and v in the standard
position and  be the angle between them,
the signed length of projv(u) = |u|cos() = u•
or it’s u dotted with unitized v.
Signed Length of Standard Projection
v
|v|
x
y
u
v


96. Example G:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
Signed Length of Standard Projection
Theorem: Given two vectors u and v in the standard
position and  be the angle between them,
the signed length of projv(u) = |u|cos() = u•
or it’s u dotted with unitized v.
v
|v|
x
y
u
v


97. x
y Example G:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
Projv(u)
u=<3, -3>
v=<-1, -4>
Signed Length of Standard Projection
Theorem: Given two vectors u and v in the standard
position and  be the angle between them,
the signed length of projv(u) = |u|cos() = u•
or it’s u dotted with unitized v.
v
|v|


98. x
y Example G:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
uv = <3,-3><-1,-4> = 9
|v|= 17,Projv(u)
u=<3, -3>
v=<-1, -4>
Signed Length of Standard Projection
Theorem: Given two vectors u and v in the standard
position and  be the angle between them,
the signed length of projv(u) = |u|cos() = u•
or it’s u dotted with unitized v.
v
|v|


99. x
y Example G:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
uv = <3,-3><-1,-4> = 9
|v|= 17, hence the length of
projv(u) is =
Projv(u)
u=<3, -3>
v=<-1, -4>
u•v
|v|
9
17
 2.18.
Signed Length of Standard Projection
Theorem: Given two vectors u and v in the standard
position and  be the angle between them,
the signed length of projv(u) = |u|cos() = u•
or it’s u dotted with unitized v.
v
|v|


100. Projections of Standard Vectors

101. Projections of Standard Vectors
For any vector v, v = (its's length)*(unitized v)

102. Projections of Standard Vectors
For any vector v, v = (its's length)*(unitized v)
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)

103. Projections of Standard Vectors
For any vector v, v = (its's length)*(unitized v)
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)= vu•v
|v|2

104. Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)=
Projections of Standard Vectors
For any vector v, v = (its's length)*(unitized v)
Example H:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
vu•v
|v|2

105. Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)=
Projections of Standard Vectors
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example H:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
vu•v
|v|2

106. Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)=
Projv(u)
Projections of Standard Vectors
Projv(u)
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example H:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
vu•v
|v|2

107. Projv(u)
Projections of Standard Vectors
Projv(u)
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example H:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
We've u•v = 9 and |v|2 =17.
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)= vu•v
|v|2

108. Projv(u)
Projections of Standard Vectors
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example H:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
We've u•v = 9 and |v|2 =17.
So proju(v) = vu•v
|v|2 *
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)= vu•v
|v|2

109. Projv(u)
Projections of Standard Vectors
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example H:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
We've u•v = 9 and |v|2 =17.
So proju(v) = vu•v
|v|2 *
9
17
* <-1, -4>=
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)= vu•v
|v|2
9
17
<-1, -4>=

110. HW
Given u=<-1,2>, v=<3,-2>, and w=<4,2>
find the signed length and the vectors for
each of the following:
1. proju(v) 2. projv(u) 3. projv(w)
4. proju(u+v) 5. projw(v – u)
6. proju+v (u – w) 7. projv+w (v + u)