# Vectors in 1D - Kwantlen Polytechnic mikec/P1170_Notes/Chapter02/4 - The dot product

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### Transcript of Vectors in 1D - Kwantlen Polytechnic mikec/P1170_Notes/Chapter02/4 - The dot product

DOT PRODUCT

Objective:

Students will be able to use

the dot product to:

a) determine an angle

between two vectors, and,

b) determine the projection

of a vector along a

specified line.

APPLICATIONS

For this geometry, can you

determine angles between the

pole and the cables?

For force F at Point A, what

component of it (F1) acts along

the pipe OA? What component

(F2) acts perpendicular to the

pipe?

The dot product of vectors π¨ and π© is defined as π¨β’π© = A B cos ο±.

Angle ο± is the smallest angle between the two vectors and is

always in a range of 0ΒΊ to 180ΒΊ.

Dot Product Characteristics:

1. The result of the dot product is a scalar (a positive or

negative number).

2. The units of the dot product will be the product of the units

of the A and B vectors.

DEFINITION

DOT PRODUCT DEFINITON

(continued)

Examples: π β’ π = 0, π β’ π = 1, and so on.

As a result, the dot product is easy to evaluate if you

have vectors in Cartesian form.

π¨ β’ π© = (Ax π + Ay π + Az π ) β’ (Bx π + By π + Bz π )

= Ax Bx + AyBy + AzBz

The dot product, π¨β’π© = A B cosο±, is easy to evaluate

for π , π , and π .

USING THE DOT PRODUCT TO DETERMINE THE

ANGLE BETWEEN TWO VECTORS

If we know two vectors in Cartesian form, finding ο± is easy since we have two

methods of doing the dot product.

ABcosο± = π΄π₯π΅π₯ + π΄π¦π΅π¦+ π΄π§π΅π§

cosο± = π΄π₯π΅π₯ + π΄π¦π΅π¦+ π΄π§π΅π§

π΄ π΅

cosο± = π΄π₯π΅π₯ + π΄π¦π΅π¦+ π΄π§π΅π§

π΄π₯ 2 + π΄π¦

2 + π΄π§ 2 π΅π₯

2 + π΅π¦ 2 + π΅π§

2

More usually just written or cosο± = π΄ β π΅

π΄ π΅ cosο± = π’ π΄ β π’ π΅

2D Example β Find ο±

π΄ = 4π + 7π

π΅

10 2 ο±

First π’ π΅ = 10

104 π +

2

104 π and π’ π΄ =

4

65 π +

7

65 π

cosο± = π’ π΄ β π’ π΅

= 4

65

10

104 +

7

65

2

104

= 54

65 104

So ο± = 48.95Β°

Projection

π¨ Β· π© = π¨π©π©

If we divide both sides in the of the definition above by the

magnitude B, we can get the magnitude of the projection of

π¨ in that direction.

π¨π© = π¨ Β· π©

π© = π¨ Β· π π©

A dot product finds how much of π¨ is in the same direction

as π© and then multiplies it by the magnitude of B

It is easy to find Aο as well, once we have A and AB;

π΄ο = π΄ 2 β π΄π΅

2

It is also easy to find π΄ B. We already found the magnitude of the vector π¨π© = π¨ Β· π π© And we know which way it points, π π©. So

π¨π© = π¨π©π π©

π¨π© = π¨οπ π© π π©

This looks a little odd because of the two unit vectors.

Once you have π¨ and π΄ B, it is also easy to find π΄ ο.

π΄ ο = π΄ β π΄ π΅

Projection (cont)

Projection (cont)

A dot product is commutative π¨ Β· π© = π© Β· π¨

π©π¨ = π© Β· π¨

π¨ = π© Β· π π¨

Note!

Vectors are not fixed to a point is space. Can do dot

product on two vectors that are not touching and can find

the angle between them and any projection of one vector

along the other.

ο±

What if B is in the other direction?

Vectors are not fixed to a point is space. Can do dot

product on two vectors that are not touching and can find

the angle between them and any projection of one vector

along the other.

ο±

2D Example β Find π΄ B and π΄ ο π΄ = 4π + 7π

π΅

10 2 ο±

First π’ π΅ = 10

104 π +

2

104 π .

π¨π© = π¨οπ π© π π©

= 4π + 7π β 10

104 π +

2

104 π

10

104 π +

2

104 π

= 40

104 +

14

104

10

104 π +

2

104 π

= 540

104 π +

108

104 π

π΄ ο = π΄ β π΄ π΅ = 4π + 7π β 540

104 π +

108

104 π = β

124

104 π +

624

104 π

EXAMPLE

Given: The force acting on

the pole

Find: The angle between

the force vector and

the pole, and the

magnitude of the

projection of the

force along the pole

OA.

A

Plan:

1. Get rOA

2. ο± = cos-1{(F β’ rOA)/(F rOA)}

3. FOA = F β’ uOA or F cos ο±

EXAMPLE

(continued)

A

rOA = {2 i + 2 j β 1 k} m

rOA = (2 2 + 22 + 12)1/2 = 3 m

F = {2 i + 4 j + 10 k}kN

F = (22 + 42 + 102)1/2 = 10.95 kN

ο± = cos-1{(F β’ rOA)/(F rOA)}

ο± = cos-1 {2/(10.95 * 3)} = 86.5Β°

uOA = rOA/rOA = {(2/3) i + (2/3) j β (1/3) k}

FOA = F β’ uOA = (2)(2/3) + (4)(2/3) + (10)(-1/3) = 0.667 kN

Or FOA = F cos ο± = 10.95 cos(86.51Β°) = 0.667 kN

F β’ rOA = (2)(2) + (4)(2) + (10)(-1) = 2 kNΒ·m

CONCEPT QUIZ

1. If a dot product of two non-zero vectors is 0, then the

two vectors must be _____________ to each other.

A) parallel (pointing in the same direction)

B) parallel (pointing in the opposite direction)

C) perpendicular

D) cannot be determined.

2. If a dot product of two non-zero vectors equals -1, then the

vectors must be ________ to each other.

A) parallel (pointing in the same direction)

B) parallel (pointing in the opposite direction)

C) perpendicular

D) cannot be determined.

Example

Given: The force acting on

the pole.

Find: The angle between the

force vector and the

pole, the magnitude of

the projection of the

force along the pole

AO, as well as FAO (F||)

and Fο.

Plan:

1. Get rAO

2. ο± = cos-1{(F β’ rAO)/(F rAO)}

3. FAO = F β’ uAO or F cos ο±

4. FAO = F|| = FOA uAO and Fο = F ο F||

rAO = {-3 i + 2 j β 6 k} ft.

rAO = (3 2 + 22 + 62)1/2 = 7 ft.

F = {-20 i + 50 j β 10 k}lb

F = (202 + 502 + 102)1/2 = 54.77 lb

ο± = cos-1{(F β’ rAO)/(F rAO)}

ο± = cos-1 {220/(54.77 Γ 7)} = 55.0Β°

F β’ rAO = (-20)(-3) + (50)(2) + (-10)(-6) = 220 lbΒ·ft

uAO = rAO/rAO = {(-3/7) i + (2/7) j β (6/7) k}

FAO = F β’ uAO = (-20)(-3/7) + (50)(2/7) + (-10)(-6/7) = 31.4 lb

Or FAO = F cos ο± = 54.77 cos(55.0Β°) = 31.4 lb

uAO = {-3 i + 2 j β 6 k}/7.

F|| = FAO uAO

= (31.4 lb) uAO

= {-13.46 i + 8.97 j β 44.86k}lb

F ο = F β F||

= {-20 i + 50 j β 10 k}lb

β {-13.46 i + 8.97 j β 44.86k}lb

= {-6.54 i + 41.03 j + 34.86 k}lb

QUIZ

1. The Dot product can be used to find all of the following

except ____ .

A) sum of two vectors

B) angle between two vectors

C) component of a vector parallel to another line

D) component of a vector perpendicular to another line

2. Find the dot product of the two vectors P and Q.

P = {5 i + 2 j + 3 k} m

Q = {-2 i + 5 j + 4 k} m

A) -12 m B) 12 m C) 12 m2

D) -12 m2 E) 10 m2