6.4Vectors and Dot Products

Day 1 2015

6.4

Vectors and Dot Products

The Definition of the Dot Product of Two Vectors:

The dot product of u = and v = is

u1,u2

v1,v2

1 1 2 2u v u v u v

The dot product of two vectors is a scalar.

6.4

Vectors and Dot Products

1 1 2 2u v u v u v

Examples: Find each dot product.

4 2 5 3 23

2 1 1 2 0

0 4 3 2 6

) 4,5 2,3a

) 2, 1 1,2b

) 0,3 4, 2c

Properties of the Dot Product

Let u, v, and w be vectors in the plane or in space and let c be a scalar.

2

0 0u v v u

vu v w u v u w

v v v

c u v cu v u cv

Let

u 1,3 , v 2, 4 , and w 1, 2

Find

uv w First, find u . v

uv 1 2 3 4 14

uv w 14 1, 2 14,28

Find u . 2v= = -281,3 4, 8

Examples:

The Angle Between Two Vectors

The Angle Between Two Vectors

• The angle between two nonzerovectors is the angle , 0 ,between their respective standard position vectors, as shown in

• Figure 6.33. This angle can be foundusing the dot product.

Figure 6.33

Example – Finding the Angle Between Two Vectors

Find the angle between u = 4, 3 and v = 3, 5.

Solution: The two vectors and are shown in Figure 6.34.

Figure 6.34

Solution

This implies that the angle between the two vectors is

The Angle Between Two Vectors

cos uvu v

7.13

You try:

Find the angle between 2 32

u i jv i j

The Angle Between Two Vectors

Find if 100, 250, and6

u v

cos uvu v

u v

12,500 3

The Angle Between Two Vectors

Figure 6.35 shows the five possible orientations of two vectors.

Figure 6.35

The terms orthogonal and perpendicular mean essentially the same thing—meeting at right angles.

Note that the zero vector is orthogonal to every vector u, because 0 u = 0.

Vectors in the same or oppositedirection will be multiples of each other.

Orthogonal Vectors (90 degree angles)

The vectors u and v are orthogonal if u . v = 0

Are the vectors orthogonal?

u 2, 3 & v 6,4

Find the dot product of the two vectors.

uv 2, 3 6,4 2 6 3 4 0

Because the dot product is 0, the two vectors are orthogonal.

Determine if the vectors u and v are orthogonal, parallel, or neither

) 1,3 , 2, 2a u v Neither, because the dot product is not = 0, and one vector is not a multiple of the other.

) 8 4 , 2b u i j v i j Parallel, because u = -4v.

Ex: Use vectors to find the interior angles of the triangle with the given vertices: P (-3,0), Q (2,2), & R (0,6)

Hint: Draw and label the triangle. Find vectors , , and the angle between them,Find vectors , , and the angle between them.

PQ

PRQR

QP

between PR and PQ is 41.6 between QR and QP is 85.2 between RP and RQ is 53.2

6.4 Day 1 Homework:

pg. 429 1-37 odd

6.4Vectors and Dot Products

Day 2Vector components and Work

We have seen applications in which two vectors are added to produce a resultant vector. Many applications in physics and engineering pose the reverse problem – decomposing a given vector into the sum of two vector components.

Consider a boat on an inclined ramp

1 2F w w

The force due to gravity F pulls the boat down the ramp and against the ramp. The 2 orthogonal forces and are the vector components of F. The negative of represents the force needed to keep the boat from rolling down the ramp and represents the force that the tires must withstand against the ramp.

1w 2w

1w

2w

Projection of u onto v

Let u and v be nonzero vectors. The projection of u onto v is

projvu uvv 2

v

Ex: Find the projection of onto Then write u as the sum of two orthogonal vectors, one which is projvu. Graph this!

u 3, 5

v 6,2 .

w1 = projvu =

uvv 2

v

840

6,2

65

, 25

w2 = u - w1 =6 2 9 273, 5 , ,5 5 5 5

Check and see:

u = w1 + w2 =

65

, 25

95

, 275

3, 5

Since u = w1 + w2, then:

projvu uvv 2

v

You try:

Find the projection of u onto v. Then write u as the the sum of two orthogonal vectors, one which is projvu.

5,2 , 3, 1u v 1

2 1

39 13,10 10

11 33,10 10

39 13 11 33, , 5,210 10 10 10

w

w u w

u

projvu uvv 2

v

You try:

Find the projection of u onto v. Then write u as the the sum of two orthogonal vectors, one which is projvu.

5, 1 , 1,1u v

1

2 1

2,2

5, 1 2,2 3, 3

3, 3 2,2 5, 1

w

w u w

u

projvu uvv 2

v

1w

2w G

Finding ForceA 200 lb. cart sits on a ramp inclined at 30 degrees. What force is required to keep the cart from rolling down the ramp?

200 0, 200G j

cos30 ,sin 30v

30

To find the force required to keep the cart from rolling down the ramp, project G onto a unit vector v in the direction of the ramp. The projection is .

v

1w

Finding Force with a vector projection

A 200 lb. cart sits on a ramp inclined at 30 degrees. What force is required to keep the cart from rolling down the ramp?

To find the force required to keep the cart from rolling down the ramp, project G onto a unit vector v in the direction of the ramp.

0, 200G 3 1cos30 ,sin 30 ,

2 2v

1 2vG vw proj G vv

3 10, 200 ,2 2

1v

3 11002 2

i j

The magnitude of this force is 100. 100 lbs of force is required to keep the cart from rolling down the ramp.

1w

2wG

Finding Force with Right Triangle Trig.A 200 lb. cart sits on a ramp inclined at 30 degrees. What force is required to keep the cart from rolling down the ramp?

200G j

1sin 30200w

30

Notice similar triangles (same angles so sides are in proportion) .

v

1w

301 200sin 30 100w

The magnitude of this force is 100. 100 lbs of force is required to keep the cart from rolling down the ramp.

We need which is the opposite side from the 30 degree angle in the larger triangle.

Finding Force with Right Triangle Trig.A force of 576 lbs is required to pull a boat and trailer up a ramp inclined at 12 degrees. Find the combined weight of the boat and trailer.

1 576w

1sin12w

weight

576sin12

weight

2770weight lbs

1 576w

2w Weight12

12

v

Work

The work W done by a constant force F acting along the line of motion of an object is given by:

cosW F PQ (magnitude of force)(distance)WF PQ

The work W done by a constant force F not directed along the line of motion of an object is given by:

Ex: A man pushes a broom with a constant force of 40 lbs.

and holds the handle of the broom at an angle of How much work is done pushing the broom 50 ft?

30 .

cosW F PQ

cos30 40 50 1732 ft.lbs.W

You try: A person pulls a wagon with a constant force of

15 lbs. at an angle of for 500 ft. Find the work done by the person.

40

cosW F PQ

cos 40 15 500 5745 ft.lbs.W

You try: To close a barn door, a person pulls a rope with

a constant force of 50 lbs. at an angle Find the work done to move the door 12 ft.

60 .

cosW F PQ

cos60 50 12 300 ft.lbs.W

Example: Find the work done in moving a particle from P

to Q if the magnitude and direction of the force are given by v. 0,0 , 4,7 , 1,4P Q v

32 . .W ft lbs

Find cos , where is the angle between and v.PQ

cosW F PQ

Types of Energy and the units used to measure it

   Many branches of science and technology are involved with energy, and each group originally defined energy using units that they considered useful. It was only much later that these different forms of energy were recognized as different manifestations of the same fundamental quantity. However, the multiplicity of units remains and can be a source of confusion.

One common source of confusion is the difference between mass and weight. Although the two are often used interchangeably, they are fundamentally different quantities. Mass, usually denoted by the letter m, refers to a quantity of matter and is measured in kilograms in the metric system. Weight is the pull of gravity on a mass, and is usually denoted by the symbol w. It is measured in pounds in the British system and Newtons in the Metric system. The two are simply related near the surface of the Earth: to convert mass to weight, you must multiply by the acceleration of gravity, which equals 9.8 m/s2 in the metric system. Thus a mass of m kilograms has a weight of 9.8m Newtons. Note that a body far from the earth may be “weightless,” since the pull of the earth is negligibly small, but its mass has not changed at all from its value at the surface. 

Mechanical energy is defined as the work done by a force acting through a distance, and is measured in foot-pounds or Newton-meters, which are called joules. A foot-pound is the work done when a force of 1 pound acts through a distance of 1 foot, and the Newton-meter (or joule) was defined in a similar way. No mechanical work is done if the force does not move the body. Standing still may be tiring, but no physical work is done.

Homework:

pg. 430: 39, 41, 47, 49, 51-63 all do only on #55 do only 800 meters on #58

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