### Vector Review Part III

```Vector Refresher
Part 3
• Vector Dot Product
Definitions
• Some Properties
• The Angle Between 2 Vectors
• Scalar Projections
• Vector Projections
Dot Product
• One form of vector multiplication
• Yields a SCALAR quantity
• Can be used to find the angle between 2 vectors
• Can also be used to find the projection of a vector in a
given direction
Symbolism
• The dot product is symbolized with a dot between 2
vectors
Symbolism
• The dot product is symbolized with a dot between 2
vectors
• The following means “Vector A dotted with vector B”
A· B
One Definition
The dot product is defined as the sum of the product of
similar components of a vector
One Definition
The dot product is defined as the sum of the product of
similar components of a vector
If we have the following 2 vectors:
U = aiˆ + bjˆ + ckˆ
V = diˆ + ejˆ + fkˆ
One Definition
The dot product is defined as the sum of the product of
similar components of a vector
If we have the following 2 vectors:
U = aiˆ + bjˆ + ckˆ
V = diˆ + ejˆ + fkˆ
U ·V = (ad)+ (be)+ (cf )
One Definition
The dot product is defined as the sum of the product of
similar components of a vector
If we have the following 2 vectors:
U = aiˆ + bjˆ + ckˆ
V = diˆ + ejˆ + fkˆ
U ·V = (ad)+ (be)+ (cf )
NOTE: This is a SCALAR term whose units are the
product of the units of the 2 vectors
Another Definition
The dot product is also related to the angle produced by
arranging 2 vectors tail to tail.
Another Definition
The dot product is also related to the angle produced by
arranging 2 vectors tail to tail.
If we have the following 2 vectors:
U = aiˆ + bjˆ + ckˆ
V = diˆ + ejˆ + fkˆ
U ·V = U V cos(q )
U
θ
V
Properties of the
Dot Product
Commutative: U ·V = V ·U
Properties of the
Dot Product
Commutative: U ·V = V ·U
Associative:
k(U ·V ) = (kU)·V = (kV)·U
Properties of the
Dot Product
Commutative: U ·V = V ·U
Associative:
k(U ·V ) = (kU)·V = (kV)·U
Distributive:
(U +V)·W = (U ·W )+ (V ·W )
The Angle Between 2
Vectors
The dot product is a useful tool in determining the angle
between 2 vectors
U
θ
V
The Angle Between 2
Vectors
The dot product is a useful tool in determining the angle
between 2 vectors
U ·V = U V cos(q )
U
θ
V
The Angle Between 2
Vectors
The dot product is a useful tool in determining the angle
between 2 vectors
U ·V = U V cos(q )
U ·V
= cos(q )
U V
U
θ
V
The Angle Between 2
Vectors
The dot product is a useful tool in determining the angle
between 2 vectors
U ·V = U V cos(q )
U ·V
= cos(q )
U V
æ U ·V ö
÷ =q
cos-1 çç
÷
U
V
è
ø
U
θ
V
The Angle Between 2
Vectors
The dot product is a useful tool in determining the angle
between 2 vectors
U ·V = U V cos(q )
U ·V
= cos(q )
U V
æ U ·V ö
÷ =q
cos-1 çç
÷
U
V
è
ø
U
θ
V
If 2 vectors are orthogonal, their dot product is 0
Scalar Projection
The dot product is also used to determine how much of a
vector is acting in a particular direction.
Scalar Projection
The dot product is also used to determine how much of a
vector is acting in a particular direction.
U
θ
V
Scalar Projection
The dot product is also used to determine how much of a
vector is acting in a particular direction.
If we want to find how much of U
acts in the direction of V , (length of
the green line) we can use the dot
product
U
θ
V
Scalar Projection
The dot product is also used to determine how much of a
vector is acting in a particular direction.
If we want to find how much of U
acts in the direction of V , (length of
the green line) we can use the dot
product
U
θ
V
V
U · uˆV = U ·
V
Scalar Projection
The dot product is also used to determine how much of a
vector is acting in a particular direction.
If we want to find how much of U
acts in the direction of V , (length of
the green line) we can use the dot
product
U
V
θ
U //
V
U // = U · uˆV = U ·
V
Note that this result is a SCALAR
quantity, meaning that it has no
direction associated.
Scalar Projection
The dot product is also used to determine how much of a
vector is acting in a particular direction.
If we want to find how much of U
acts in the direction of V , (length of
the green line) we can use the dot
product
U
V
θ
U //
V
U // = U · uˆV = U ·
V
Note that this result is a SCALAR
quantity, meaning that it has no
direction associated. Thus, this
calculation is the scalar projection
Vector Projection
The scalar projection can be used to determine a vector
projection
We can transform the scalar
projection, in this case U · uˆ , into a
V
vector by multiplying the scalar
projection and the unit vector that
described the direction of interest, in
this case uˆV
U
V
θ
U //
æ
V
ç
U // = (U · uˆV )uˆV = çU ·
V
è
öV
÷
÷V
ø
This is a VECTOR quantity that
describes the vector shown by the green
arrow
Applications of the
Vector Projection
We can use the vector projection to determine the vector
parallel and perpendicular to a given direction
U
V
θ
U //
Applications of the
Vector Projection
We can use the vector projection to determine the vector
parallel and perpendicular to a given direction
A vector can be described as its vector
component parallel to a direction plus its
component perpendicular to a direction
U =U// +U^
U
V
θ
U //
Applications of the
Vector Projection
We can use the vector projection to determine the vector
parallel and perpendicular to a given direction
A vector can be described as its vector
component parallel to a direction plus its
component perpendicular to a direction
U^ =U -U//
U
V
θ
U //
U =U// +U^
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Looking at this formula, we need to
U ·V
= cos(q )
U V
determine the magnitude of each vector
and evaluate the dot product
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can start by finding the magnitude of
U ·V
= cos(q )
U V
vector U
U = (2)2 + (2)2 + (-1)2
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can start by finding the magnitude of
U ·V
= cos(q )
U V
vector U
U = (2)2 + (2)2 + (-1)2
U = 4 + 4 +1 = 9
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can start by finding the magnitude of
U ·V
= cos(q )
U V
vector U
U = (2)2 + (2)2 + (-1)2
U = 4 + 4 +1 = 9
U =3
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Now, we can do the same for vector V
U ·V
= cos(q )
(3) V
V = (-2)2 + (3)2 + (6)2
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Now, we can do the same for vector V
U ·V
= cos(q )
(3) V
V = (-2)2 + (3)2 + (6)2
V = 4 + 9 + 36 = 49
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Now, we can do the same for vector V
U ·V
= cos(q )
(3)(7)
V = (-2)2 + (3)2 + (6)2
V = 4 + 9 + 36 = 49
V =7
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Next, we’ll take the dot product to
-4
= cos(q )
(3)(7)
complete the formula.
U ·V = (2)(-2)+ (2)(3)+ (-1)(6)
U ·V = -4 + 6 - 6 = -4
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Now, we can use the inverse cosine
function to find the angle
-4
= cos(q )
(3)(7)
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Now, we can use the inverse cosine
function to find the angle
-4
= cos(q )
(3)(7)
-1 æ -4 ö
cos ç ÷ = q
è 21 ø
101° = q
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
To find the projection of U onto V, we
101° = q
U// = U · uˆV
need to use the formula to the left, which
means we need the unit vector that
describes the direction of V
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We already calculated the magnitude of
101° = q
U// = U · uˆV
V. We’ll use that to find the unit vector
V
uˆV =
V
V =7
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We already calculated the magnitude of
101° = q
U// = U · uˆV
V. We’ll use that to find the unit vector
V
uˆV =
V
V =7
2ˆ 3 ˆ 6 ˆ
uˆV = - i + j + k
7 7
7
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Now, we can take the dot product to find
101° = q
U// = U · uˆV
the scalar projection.
V
uˆV =
V
V =7
2ˆ 3 ˆ 6 ˆ
uˆV = - i + j + k
7 7
7
æ -2 ö
æ 3ö
æ6ö
ˆ
U · uV = (2)ç ÷ + (2)ç ÷ + (-1)ç ÷
è 7ø
è 7ø
è 7ø
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Now, we can take the dot product to find
101° = q
U// = U · uˆV
the scalar projection.
V
uˆV =
V
V =7
2ˆ 3 ˆ 6 ˆ
uˆV = - i + j + k
7 7
7
æ -2 ö
æ 3ö
æ6ö
ˆ
U · uV = (2)ç ÷ + (2)ç ÷ + (-1)ç ÷
è 7ø
è 7ø
è 7ø
U · uˆV =
-4
7
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
To find the vector projection, we’ll apply
101° = q
-4
U // =
7
the scalar projection to the unit vector
that describes the direction of V.
U// = U// uˆV
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
To find the vector projection, we’ll apply
101° = q
-4
U // =
7
the scalar projection to the unit vector
that describes the direction of V.
U// = U// uˆV
æ -4 öæ 2 ˆ 3 ˆ 6 ˆ ö
U// = ç ÷ç - i + j + k ÷
è 7 øè 7 7
7 ø
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
To find the vector projection, we’ll apply
101° = q
-4
U // =
7
the scalar projection to the unit vector
that describes the direction of V.
U// = U// uˆV
æ -4 öæ 2 ˆ 3 ˆ 6 ˆ ö
U// = ç ÷ç - i + j + k ÷
è 7 øè 7 7
7 ø
8 ˆ 12 ˆ 24 ˆ
U// = i j- k
49 49
49
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Finally, we can subtract the component
101° = q
-4
U // =
7
8 ˆ 12 ˆ 24 ˆ
U// = i j- k
49 49
49
of U parallel to V from U to get the part
of U that is perpendicular to V.
U^ =U -U//
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
Finally, we can subtract the component
101° = q
-4
U // =
7
of U parallel to V from U to get the part
of U that is perpendicular to V.
U^ =U -U//
æ
æ -24 öö ˆ
8 ö ˆ æ æ -12 öö ˆ æ
U^ = ç 2 - ÷ i + ç 2 - ç
÷÷ j + ç -1- ç
÷÷ k
è 49 ø è è 49 øø è
è 49 øø
90 ˆ 110 ˆ 25 ˆ
U^ = i +
j- k
49
49
49
8 ˆ 12 ˆ 24 ˆ
U// = i j- k
49 49
49
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can check our work with the
101° = q
U // =
-4
7
90 110 ˆ 25 ˆ
U^ = iˆ +
j- k
49
49
49
U// =
8 ˆ 12 ˆ 24 ˆ
ij- k
49 49
49
following formula
2
U = U// + U^
2
because the parallel and perpendicular
components of U form a right triangle,
with U as the hypotenuse.
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can check our work with the
101° = q
U // =
-4
7
U// =
following formula
8 ˆ 12 ˆ 24 ˆ
ij- k
49 49
49
2
U = U// + U^
90 110 ˆ 25 ˆ
U^ = iˆ +
j- k
49
49
49
æ 90 ö æ 110 ö æ -25 ö
U^ = ç ÷ + ç
÷ +ç
÷
è 49 ø è 49 ø è 49 ø
2
2
2
2
because the parallel and perpendicular
components of U form a right triangle,
with U as the hypotenuse.
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can check our work with the
101° = q
U // =
-4
7
U// =
following formula
8 ˆ 12 ˆ 24 ˆ
ij- k
49 49
49
2
U = U// + U^
90 110 ˆ 25 ˆ
U^ = iˆ +
j- k
49
49
49
æ 90 ö æ 110 ö æ -25 ö
U^ = ç ÷ + ç
÷ +ç
÷
è 49 ø è 49 ø è 49 ø
2
2
2
2
because the parallel and perpendicular
components of U form a right triangle,
with U as the hypotenuse.
æ 8100 ö æ 12100 ö æ 625 ö
U^ = ç
÷+ç
÷+ç
÷
è 2401 ø è 2401 ø è 2401 ø
æ 20825 ö
U^ = ç
÷
è 2401 ø
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can check our work with the
101° = q
U // =
-4
7
U// =
8 ˆ 12 ˆ 24 ˆ
ij- k
49 49
49
90 110 ˆ 25 ˆ
U^ = iˆ +
j- k
49
49
49
æ 20825 ö
U^ = ç
÷
è 2401 ø
-4
U // =
7
following formula
2
U = U// + U^
2
because the parallel and perpendicular
components of U form a right triangle,
with U as the hypotenuse.
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can check our work with the
101° = q
U // =
-4
7
U// =
8 ˆ 12 ˆ 24 ˆ
ij- k
49 49
49
90 110 ˆ 25 ˆ
U^ = iˆ +
j- k
49
49
49
-4
U // =
7
æ 20825 ö
U^ = ç
÷
è 2401 ø
2
U =
æ 20825 ö æ -4 ö
ç
÷ +ç ÷
è 2401 ø è 7 ø
2
following formula
2
U = U// + U^
2
because the parallel and perpendicular
components of U form a right triangle,
with U as the hypotenuse.
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can check our work with the
101° = q
U // =
-4
7
U// =
8 ˆ 12 ˆ 24 ˆ
ij- k
49 49
49
90 110 ˆ 25 ˆ
U^ = iˆ +
j- k
49
49
49
-4
U // =
7
æ 20825 ö
U^ = ç
÷
è 2401 ø
2
U =
æ 20825 ö æ -4 ö
ç
÷ +ç ÷
è 2401 ø è 7 ø
2
following formula
2
U = U// + U^
2
because the parallel and perpendicular
components of U form a right triangle,
with U as the hypotenuse.
20825 16
U =
+
2401 49
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can check our work with the
101° = q
U // =
-4
7
U// =
8 ˆ 12 ˆ 24 ˆ
ij- k
49 49
49
90 110 ˆ 25 ˆ
U^ = iˆ +
j- k
49
49
49
-4
U // =
7
æ 20825 ö
U^ = ç
÷
è 2401 ø
2
U =
æ 20825 ö æ -4 ö
ç
÷ +ç ÷
è 2401 ø è 7 ø
2
following formula
2
U = U// + U^
2
because the parallel and perpendicular
components of U form a right triangle,
with U as the hypotenuse.
20825 16
U =
+
2401 49
U= 9
Example Problem
If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2
vectors, the projection of U onto V , and the component of U that is
perpendicular to V .
We can check our work with the
101° = q
U // =
-4
7
U// =
8 ˆ 12 ˆ 24 ˆ
ij- k
49 49
49
90 110 ˆ 25 ˆ
U^ = iˆ +
j- k
49
49
49
-4
U // =
7
æ 20825 ö
U^ = ç
÷
è 2401 ø
2
U =
æ 20825 ö æ -4 ö
ç
÷ +ç ÷
è 2401 ø è 7 ø
2
following formula
2
U = U// + U^
2
because the parallel and perpendicular
components of U form a right triangle,
with U as the hypotenuse.
20825 16
U =
+
2401 49
U= 9
U =3
```