The Law of Sines

c

C

b

B

a

A sinsinsin

Law of SinesIf A, B, and C are the measures of the angles of a

triangle, and a, b, and c are the lengths of the sides opposite these angles, then

The ratio of the length of the side of any triangle to

the sine of the angle opposite that side is the same for all three sides of the triangle.

Solution We begin by drawing a picture of triangle ABC and labeling it with the given information. The figure shows the triangle that we must solve. We begin by finding B.

A B C 180º The sum of the measurements of a triangle’s interior angles is 180º.

Add.83.5º B 180º

50º B 33.5º 180º

A = 50º and C = 33.5º.

B 96.5º Subtract 83.5º from both sides.

a

cA B

C

b 33.5º

50º

Text Example

• Solve triangle ABC if A 50º, C 33.5º, and b 76.

Solve triangle ABC if A 50º, C 33.5º, and b 76.

Solution Keep in mind that we must be given one of the three ratios to apply the Law of Sines. In this example, we are given that b 76 and we found that B 96.5º. Thus, we use the ratio b/sin B, or 76/sin96.5º, to find the other two sides. Use the Law of Sines to find a and c.

The solution is B 96.5º, a 59, and c 42.

Find a: Find c: This is the known ratio.

a

cA B

C

b 33.5º

50º

Text Example cont.

asin A

bsin B

a

sin50 76

sin96.5

a 76sin 50

sin 96.5 59

csinC

bsin B

c

sin33.5 76

sin 96.5

c 76sin 33.5

sin 96.5 42

Consider a triangle in which a, b, and A are given. This information may result in:Consider a triangle in which a, b, and A are given. This information may result in:

Two Triangles One Triangle

No Triangle One Right Triangle

b a

A

h = b sin A

a is less than h and not long

enough to form a triangle.

b

A

a h = b sin A

a = h and is just the right

length to form a right triangle.

a ab

A

h = b sin A

a is greater than h and a is less than b. Two distinct triangles are formed.

b a

A

a is greater than h and a is greater

than b. One triangle is formed.

The Ambiguous Case (SSA)

Example

ab

c

• Solve the triangle shown with A=36º, B=88º and c=29 feet.

Solution:

A=36º, B=88º so 180-88-36=56º C=56º

61.2011.1783.29

83.59.

94.342983.29

83.129

83.159.29

56sin88sin36sin

aasoa

bbsob

ba

ba

Area of An Oblique Triangle

BacCabAbcArea sin2

1sin

2

1sin

2

1

C

A Bc

b ah

D

• The area of a triangle equals one-half the product of the lengths of two sides times the sine of their included angle. In the following figure, this wording can be expressed by the formulas

Area 1/2 (24)(10)(sin 62º) 106

the area of the triangle is approximately 106 square meters.

Solution The triangle is shown in the following figure. Its area is half the product of the lengths of the two sides times the sine of the included angle.

b = 10 meters

c = 24 metersA B

C

62º

Text Example

• Find the area of a triangle having two sides of lengths 24 meters and 10 meters and an included angle of 62º.

Example

• Find the area of a triangle having two sides of lengths 12 ft. and 20 ft. and an included angle of 57º.

Solution:

..8.10084.*120

57sin)20)(12(2

1sin

2

1

ftsq

AbcArea

The Law of Cosines

The Law of CosinesIf A, B, mid C are the measures of the angles of a

triangle, and a, b, and c are the lengths of the sides opposite these angles, then

a2 b2 c2 2bc cos A

b2 a2 c2 2ac cos B

c2 a2 b2 2ab cos C.

The square of a side of a triangle equals the sum of the squares of the other two sides minus twice their product times the cosine of their included angle.

Solving an SAS Triangle

1. Use the Law of Cosines to find the side opposite the given angle.

2. Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute.

3. Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180º.

Step l Use the Law of Cosines to find the side opposite the given angle. Thus, we will find a.

Solution We are given two sides and an included angle. Therefore, we apply the threestep procedure for solving a SAS triangle.

a2 202 302 2(20)(30) cos 60º b = 20, c = 30, and A = 60°.

400 900 1200(0.5) 700 Perform the indicated operations.

a2 b2 c2 2bc cos A Apply the Law of Cosines to find a.

Take the square root of both sides and solve for a.

a 700 26

ab = 20

c =30

A B

C

60º

Text Example

• Solve the triangle shown with A = 60º, b = 20, and c = 30.

ab = 20

c = 30

A B

C

60º

Find sin-10.6547 using a calculator.

41B

Text Example cont.

Solve the triangle shown with A = 60º, b = 20, and c = 30.

Step 2 Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute. The shorter of the two given sides is b 20. Thus, we will find acute angle B.

Solution

Apply the Law of Sines.

We are given b = 20 and A = 60°. Use

the exact value of a, 700, from step 1.

Cross multiply.

Divide by square root of 700 and solve for sin B.

b

sin B

a

sin A

20sin B

700sin60

700 sin B 20sin 60

sin B 20sin60

7000.6547

B 41

Solution We are given two sides and an included angle. Therefore, we apply the threestep procedure for solving a SAS triangle.

Step 3 Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180º.

C 180º A B 180º 60º 41º 79º

The solution is a 26, B 41º, and C 79º.

ab = 20

c = 30

A B

C

60º

Text Example cont.Solve the triangle shown with A = 60º, b = 20, and c = 30.

Solving an SSS Triangle

1. Use the Law of Cosines to find the angle opposite the longest side.

2. Use the Law of Sines to find either of the two remaining acute angles.

3. Find the third angle. Subtract the measures of the angles found in steps 1 and 2 from 180º.

Solution After two hours. the plane flying at 325 miles per hour travels 325 · 2 miles, or 650 miles. Similarly, the plane flying at 300 miles per hour travels 600 miles. The situation is illustrated in the figure.

Let b the distance between the planes after two hours. We can use a northsouth line to find angle B in triangle ABC. Thus,

B 180º 66º 26º 88º.

We now have a 650, c 600, and B 88º.

Text Example• Two airplanes leave an airport at the same time on

different runways. One flies at a bearing of N66ºW at 325 miles per hour. The other airplane flies at a bearing of S26ºW at 300 miles per hour. How far apart will the airplanes be after two hours?

b2 a2 c2 2ac cos B Apply the Law of Cosines.

Solution We use the Law of Cosines to find b in this SAS situation.

b2 6502 6002 2(650)(600) cos 88º Substitute: a= 650, c =600, and B= 88°. 755,278 Use a calculator.

After two hours, the planes are approximately 869 miles apart.

b 869 Take the square root and solve for b.

Text Example cont.Two airplanes leave an airport at the same time on different runways. One flies at a bearing of N66ºW at 325 miles per hour. The other airplane flies at a bearing of S26ºW at 300 miles per hour. How far apart will the airplanes be after two hours?

Heron’s Formula

)(2

1

))()((

cbas

csbsassArea

The area of a triangle with sides a, b, and c is

Example• Use Heron’s formula to find the area of the

given triangle:a=10m, b=8m, c=4m

Solution:

11)22(2

1

)4810(2

1

)(2

1

s

s

cbas

..231)7)(3)(1(11

)411)(811)(1011(11

))()((

msq

csbsassArea

Polar Coordinates

The Sign of r and a Point’s Location in Polar Coordinates.

• The point P (r, ) is located |r| units from the pole. If r > 0, the point lies on the terminal side of . If r < 0 the point lies along the ray opposite the terminal side of . If r 0. the point lies at the pole, regardless of the value of .

Solution

a. To plot the point (r, ) (2, 135°), begin with the 135° angle. Because 135° is a positive angle, draw 135° counterclockwise from the polar axis. Now consider r 2. Because r > 0, plot the point by going out two units on the terminal side of . Figure (a) shows the point.

(2, 135°)

(2, 135°)

135º

Text Example

• Plot the points with the following polar coordinates: a. (2, 135°)

Multiple Representation of Points

If n is any integer, the point (r, ) can be represented as

(r, ) (r, 2n) or (r, ) (r, r 2n )

Relations between Polar and Rectangular Coordinates

P = (r, ) = (x, y)

x r cos y r sin x2 y2 r2

tan y/x

Converting a Point from Rectangular to Polar Coordinates

(r > 0 and 0 < < 2)1. Plot the point (x, y).

2. Find r by computing the distance from the origin to (x, y).

3. Find using tan = y/x with lying in the same quadrant as (x, y).

Example

)1.1,52(

1.1

22

4tan

522042 2222

x

y

yxr

• Find the polar coordinates of a point whose rectangular coordinates are (2, 4)

Solution:

Example

sincos2

1

1)sincos2(

1sincos2

12

r

r

rr

yxSolution:

• Convert 2x-y=1 to a polar equation.

Graphs of Polar Equations

Recall that a polar equation is an equation whose variables are r and è. The graph of a polar equation is the set of all points whose polar coordinates satisfy the equation. We use polar grids like the one shown to graph polar equations. The grid consists of circles with centers at the pole. This polar grid shows five such circles. A polar grid also shows lines passing through the pole, In this grid, each fine represents an angle for which we know the exact values of the trigonometric functions.

0

6˝4

32

32

43

65

67

45

34

35 4

7 6

11

23

42

Using Polar Grids to Graph Polar Equations

˝ 0

64

˝32

32

43

65

67

45

34

35 4

7 6

11

23

42

Solution We construct a partial table of coordinates using multiples of 6. Then we plot the points and join them with a smooth curve, as shown.

(-4, )4 cos ˝ = 4(-1) = -4˝

(-3.5,5 /6)4 cos5 /6 = 4(- 3/2)=-2 3=-3.55/6

(-2, 2 /3)4 cos2 /3 = 4(- 1/2) = -22 /3

(0, /2)4 cos /2 = 4 • 0 = 0 /2

(2, /3)4 cos /3 = 4 • 1/2 = 4 /3

(3.5, /6)4 cos /6 = 4 •3/2=2 3=3.5 /6

(4, 0)4 cos 0 = 4 • 1 = 40

(r, )r 4 cos

(0, /2)

(2, /3)

(3.5, /6)

(4, 0) or (-4, )

(-3.5, 5 /6)

(-2, 2 /3)

Text Example

• Graph the polar equation r 4 cos with in radians.

The graphs ofr = a cos and r = a sin

Are circles.r = a cos r = a sin

The graphs ofr = a cos and r = a sin

Are circles.r = a cos r = a sin

˝ 0

/2

3 /2

˝ 0

/2

3 /2

a

a

Circles in Polar Coordinates

Solution We apply each of the tests for symmetry.

Polar Axis: Replace by in r 1 cos :r 1 cos () Replace by in r 1 cos .

r 1 cos The cosine function is even: cos ( ) cos .

Because the polar equation does not change when is replaced by , the graph is symmetric with respect to the polar axis.

Text Example

• Check for symmetry and then graph the polar equation: r 1 cos .

The Line 2: Replace (r, ) by (r, ) in r 1 cos :r 1 cos() Replace r by r and by – in r 1 cos( ).

r 1 – cos cos( ) cos .

r cos 1 Multiply both sides by 1.

Because the polar equation r 1 cos changes to r cos 1 when (r, ) is replaced by (r, ), the equation fails this symmetry test. The graph may

of may not be symmetric with respect to the line 2.

Solution

The Pole: Replace r by r in r 1 cos :r 1 – cos Replace r by –r.

r cos 1 Multiply both sides by 1.

Because the polar equation r 1 cos changes to r cos 1 when r is replaced by r, the equation fails this symmetry test. The graph may or may not be symmetric with respect to the pole.

Text Example cont.

Now we are ready to graph r 1 cos . Because the period of the cosine function is 2r, we need not consider values of beyond 2. Recall that we discovered the graph of the equation r 1 cos has symmetry with respect to the polar axis. Because the graph has symmetry, we may be able to obtain a complete graph without plotting points generated by values of from 0 to 2. Let's start by finding the values of r for values of from 0 to .

Solution

21.871.501.000.500.130r

5/62/32/3/60

The values for r and are in the table. Examine the graph. Keep in mind that the graph must be symmetric with respect to the polar axis.

0

64

3˝2

32

43

65

67

45

34

35 4

7 6

11

23

21

Text Example cont.

Thus, if we reflect the graph from the last slide about the polar axis, we will obtain a complete graph of r 1 cos , shown below.

Solution

0

64

32

32

43

65

67

45

34

35 4

7 6

11

23

1 2

Text Example cont.

The graphs ofr a b sin , r a b sin , r a b cos , r a b cos , a > 0, b > 0

are called limacons. The ratio ab determines a limacon's shape.

Inner loop if ab < 1 Heart shaped if ab 1 Dimpled with no inner No dimple and no inner

and called cardiods loop if 1< ab < 2 loop if ab 2.

The graphs ofr a b sin , r a b sin , r a b cos , r a b cos , a > 0, b > 0

are called limacons. The ratio ab determines a limacon's shape.

Inner loop if ab < 1 Heart shaped if ab 1 Dimpled with no inner No dimple and no inner

and called cardiods loop if 1< ab < 2 loop if ab 2.

0

2

23

0

2

23

0

2

23

0

2

23

Limacons

Example• Graph the polar equation

y= 2+3cos

Example• Graph the polar equation

y= 2+3cosSolution:

The graphs ofr a sin n and r a cos n, a does not equal 0,

are called rose curves. If n is even, the rose has 2n petals. If n is odd, the rose has n petals.

r a sin 2 r a cos 3 r a cos 4 r a sin 5 Rose curve Rose curve Rose curve Rose curve

with 4 petals with 3 petals with 8 petalswith 5 petals

The graphs ofr a sin n and r a cos n, a does not equal 0,

are called rose curves. If n is even, the rose has 2n petals. If n is odd, the rose has n petals.

r a sin 2 r a cos 3 r a cos 4 r a sin 5 Rose curve Rose curve Rose curve Rose curve

with 4 petals with 3 petals with 8 petalswith 5 petals

n = 2

a

0

2

23

0

2

23

n = 5

a

0

2

23

n = 4

a

0

2

23

n = 3

a a

Rose Curves

Example

• Graph the polar equation y=3sin2

Example

• Graph the polar equation y=3sin2

Solution:

Lemniscate:

r2 = a2 cos 2

Lemniscates

• The graphs of r2 = a2 sin 2 and r2 = a2 cos 2 are called lemniscates

Complex Numbers in Polar Form;

DeMoivre’s Theorem

We know that a real number can be represented as a point on a number line. By contrast, a complex number z = a + bi is represented as a point (a, b) in a coordinate plane, shown below. The horizontal axis of the coordinate plane is called the real axis. The vertical axis is called the imaginary axis. The coordinate system is called the complex plane. Every complex number corresponds to a point in the complex plane and every point in the complex plane corresponds to a complex number.

Imaginary axis

Real axisa

b

z = a + bi

The Complex Plane

Plot in the complex plane: a. z = 3 + 4i b. z = -1 – 2i c. z = -3 d. z = -4i

-5 -4 -3 -2 1 2 3 4 5

54321

-3-4-5

• The complex number z = -1 – 2i corresponds to the point (-1, -2) in the rectangular coordinate system. Plot the complex number by moving one unit to the left on the real axis and two units down parallel to the imaginary axis.

Solution

• We plot the complex number z = 3 + 4i the same way we plot (3, 4) in the rectangular coordinate system. We move three units to the right on the real axis and four units up parallel to the imaginary axis.

z = 3 + 4i

z = -1 – 2i

Text Example

Plot in the complex plane: a. z = 3 + 4i b. z = -1 – 2i c. z = -3 d. z = -4i

Solution

• Because z = -3 = -3 + 0i, this complex number corresponds to the point (-3, 0). We plot –3 by moving three units to the left on the real axis.

-5 -4 -3 -2 1 2 3 4 5

54321

-3-4-5

z = 3 + 4i

z = -1 – 2i

z = -3

• Because z = -4i = 0 – 4i, this complex number corresponds to the point (0, -4). We plot the complex number by moving three units down on the imaginary axis.

z = -4i

Text Example cont.

The Absolute Value of a Complex Number

22 babiaz

• The absolute value of the complex number a + bi is

Example

5220

164)4(2 22

22

babiaz

• Determine the absolute value of z=2-4i

Solution:

Polar Form of a Complex Number

The complex number a + bi is written in polar form as

z = r (cos + i sin )

where a = r cos , b = r sin , and

tan =b/a The value of r is called the modulus

(plural: moduli) of the complex number z, and the

angle is called the argument of the complex

number z, with 0 < < 2

r a2 b2

Plot z = -2 – 2i in the complex plane. Then write z in polar form.

Solution The complex number z = -2 – 2i, graphed below, is in rectangular form a + bi, with a = -2 and b = -2. By definition, the polar form of z is r(cos + i sin ). We need to determine the value for r and the value for , included in the figure below.

2

-2

-2

è

r

Imaginaryaxis

Realaxis2

z = -2 – 2i

Text Example

Since tan 4 = 1, we know that lies in quadrant III. Thus,

SolutionText Example cont.

r a2 b2 ( 2)2 ( 2)2 4 4 8 2 2

tan b

a

2

21

4

44

4

54

z r(cos i sin ) 2 2(cos54

i sin54

)

Product of Two Complex Numbers in Polar Form

Let z1 = r1 (cos 1+ i sin 1) and z2 = r2 (cos 2 + i sin 2) be two complex numbers in polar form. Their product, z1z2, is

z1z2 = r1 r2 (cos ( 1 + 2) + i sin ( 1 + 2))

To multiply two complex numbers, multiply moduli and add arguments.

Find the product of the complex numbers. Leave the answer in polar form.z1 = 4(cos 50º + i sin 50º) z2 = 7(cos 100º + i sin 100º)

Solutionz1z2

= [4(cos 50º + i sin 50º)][7(cos 100º + i sin 100º)]Form the product of the given numbers.

= (4 · 7)[cos (50º + 100º) + i sin (50º + 100º)]

= 28(cos 150º + i sin 150º)

Multiply moduli and add arguments.

Simplify.

Text Example

Quotient of Two Complex Numbers in Polar Form

Let z1 = r1 (cos 1 + i sin 1) and z2 = r2 (cos 2

+ i sin 2) be two complex

numbers in polar form. Their quotient, z1/z2, is

To divide two complex numbers, divide moduli and subtract arguments.

)]sin()[cos( 21212

1

2

1 ir

r

z

z

DeMoivre’s Theorem

• Let z = r (cos + i sin ) be a complex numbers in polar form. If n is a positive integer, z to the nth power, zn, is

)sin(cos

)]sin(cos[

ninr

irzn

nn

Find [2 (cos 10º + i sin 10º)]6. Write the answer in rectangular form a + bi.

Solution By DeMoivre’s Theorem,[2 (cos 10º + i sin 10º)]6

= 26[cos (6 · 10º) + i sin (6 · 10º)]Raise the modulus to the 6th power and multiply the argument by 6.

= 64(cos 60º + i sin 60º) Simplify.

Text Example

Write the answer in rectangular form.

Multiply and express the answer in a + bi form.

641

2 i

3

2

32 32 3i

1,...,3,2,1,0

360sin

360cos

nkwhere

n

ki

n

krz n

k

DeMoivre’s Theorem for Finding Complex Roots

• Let =r(cos+isin) be a complex number in polar form. If 0, has n distinct complex nth roots given by the formula

Example

• Find all the complex fourth roots of 81(cos60º+isin60º)

Solution:

)15sin15(cos3

4

0*36060sin

4

0*36060cos81

360sin

360cos

4

i

i

n

ki

n

krz n

k

Example cont.

• Find all the complex fourth roots of 81(cos60º+isin60º)

Solution:

)285sin285(cos3

)195sin195(cos3

)105sin105(cos3

4

1*36060sin

4

1*36060cos814

i

i

i

i

Vectors

A line segment to which a direction has been assigned is called a directed line segment. The figure below shows a directed line segment form P to Q. We call P the initial point and Q the terminal point. We denote this directed line segment by PQ.

The magnitude of the directed line segment PQ is its length. We denote this by || PQ ||. Thus, || PQ || is the distance from point P to point Q. Because distance is nonnegative, vectors do not have negative magnitudes.

Geometrically, a vector is a directed line segment. Vectors are often denoted by a boldface letter, such as v. If a vector v has the same magnitude and the same direction as the directed line segment PQ, we write

v = PQ.

P

Q

Initial pointTerminal point

Directed Line Segments and Geometric Vectors

Vector Multiplication

If k is a real number and v a vector, the vector kv is called a scalar multiple of the vector v. The magnitude and direction of kv are given as follows:

The vector kv has a magnitude of |k| ||v||. We describe this as the absolute value of k times the magnitude of vector v.

The vector kv has a direction that is: the same as the direction of v if k > 0, and opposite the direction of v if k < 0

A geometric method for adding two vectors is shown below. The sum of u + v is called the resultant vector. Here is how we find this vector.

1. Position u and v so the terminal point of u extends from the initial point of v.

2. The resultant vector, u + v, extends from the initial point of u to the terminal point of v.

Initial point of u

u + vv

u

Resultant vector

Terminal point of v

The Geometric Method for Adding Two Vectors

The difference of two vectors, v – u, is defined as v – u = v + (-u), where –u is the scalar multiplication of u and –1: -1u. The difference v – u is shown below geometrically.

v

u-u

-u

v – u

The Geometric Method for the Difference of Two Vectors

1

1

i

j

Ox

y

The i and j Unit Vectors

• Vector i is the unit vector whose direction is along the positive x-axis. Vector j is the unit vector whose direction is along the positive y-axis.

Representing Vectors in Rectangular Coordinates

Vector v, from (0, 0) to (a, b), is represented asv = ai + bj.The real numbers a and b are called the scalar

components of v. Note that a is the horizontal component of v, and b is the vertical component of v.The vector sum ai + bj is called a linear combination

of the vectors i and j. The magnitude of v = ai + bj is given by

v a2 b2

Sketch the vector v = -3i + 4j and find its magnitude.Solution For the given vector v = -3i + 4j, a = -3 and b = 4. The vector, shown below, has the origin, (0, 0), for its initial point and (a, b) = (-3, 4) for its terminal point. We sketch the vector by drawing an arrow from (0, 0) to (-3, 4). We determine the magnitude of the vector by using the distance formula. Thus, the magnitude is

-5 -4 -3 -2 -1 1 2 3 4 5

5

4

3

2

1

-1-2

-3

-4-5

Initial point

Terminal point

v = -3i + 4j

v a2 b2

( 3)2 42

9 16

25 5

Text Example

Representing Vectors in Rectangular Coordinates

• Vector v with initial point P1 = (x1, y1) and terminal point P2 = (x2, y2) is equal to the position vector

• v = (x2 – x1)i + (y2 – y1)j.

Adding and Subtracting Vectors in Terms of i and j

• If v = a1i + b1j and w = a2i + b2j, then • v + w = (a1 + a2)i + (b1 + b2)j• v – w = (a1 – a2)i + (b1 – b2)j

If v = 5i + 4j and w = 6i – 9j, find: a. v + w b. v – w.

Solution

• v + w = (5i + 4j) + (6i – 9j) These are the given vectors.= (5 + 6)i + [4 + (-9)]j Add the horizontal components. Add

the vertical components.= 11i – 5j Simplify.

• v + w = (5i + 4j) – (6i – 9j) These are the given vectors.= (5 – 6)i + [4 – (-9)]j Subtract the horizontal components.

Subtract the vertical components.= -i + 13j Simplify.

Text Example

Scalar Multiplication with a Vector in Terms of i and j

• If v = ai + bj and k is a real number, then the scalar multiplication of the vector v and the scalar k is

• kv = (ka)i + (kb)j.

Example

ji

jiv

ji

jiv

96

)3*3()2*3(3

1510

)3*5()2*5(5

• If v=2i-3j, find 5v and -3v

Solution:

The Zero Vector

• The vector whose magnitude is 0 is called the zero vector, 0. The zero vector is assigned no direction. It can be expressed in terms of i and j using

• 0 = 0i + 0j.

Properties of Vector Addition and Scalar Multiplication

If u, v, and w are vectors, then the following properties are true.

Vector Addition Properties 1. u + v = v + u Commutative Property 2. (u + v) + w = v + (u + w) Associative Property 3. u + 0 = 0 + u = u Additive Identity 4. u + (-u) = (-u) + u = 0 Additive Inverse

Properties of Vector Addition and Scalar Multiplication

If u, v, and w are vectors, and c and d are scalars, then the following properties are true.

Scalar Multiplication Properties 1. (cd)u = c(du) Associative Property 2. c(u + v) = cv + cu Distributive Property 3. (c + d)u = cu + du Distributive Property 4. 1u = u Multiplicative Identity 5. 0u = 0 Multiplication Property 6. ||cv|| = |c| ||v||

Finding the Unit Vector that Has the Same Direction as a Given Nonzero Vector v

• For any nonzero vector v, the vector

• is a unit vector that has the same direction as v. To find this vector, divide v by its magnitude.

v

v

Example

jiv

v

v

65

7

65

4

654916

)7(4 22

• Find a unit vector in the same direction as v=4i-7j

Solution:

The Dot Product

Definition of a Dot ProductIf v=a1i+b1j and w = a2i+b2j are vectors, the dot product is defined as

2121 bbaawv The dot product of two vectors is the sum of the products of their horizontal and vertical components.

If v = 5i – 2j and w = -3i + 4j, find: a. v · w b. w · v c. v · v.

Solution To find each dot product, multiply the two horizontal components, and then multiply the two vertical components. Finally, add the two products. a. v · w = 5(-3) + (-2)(4) = -15 – 8 = -23

b. w · v = (-3)(5) + (4)(-2) = -15 – 8 = -23

c. v · v = (5)(5) + (-2)(-2) = 25 + 4 = 29

Text Example

Properties of the Dot Product

If u, v, and w, are vectors, and c is a scalar, then

1. u · v = v · u 2. u · (v + w) = u · v + u · w 3. 0 · v = 0 4. v · v = || v ||2

5. (cu) · v = c(u · v) = u · (cv)

Alternative Formula for the Dot Product

• If v and w are two nonzero vectors and is the smallest nonnegative angle between them, then

• v · w = ||v|| ||w|| cos.

Alternative Formula for the Dot Product

• If v and w are two nonzero vectors and is the smallest nonnegative angle between them, then

coswvwv

Formula for the Angle between Two Vectors

wv

wvand

wv

wv

1coscos

• If v and w are two nonzero vectors and is the smallest nonnegative angle between v and w, then

Example

• Find the angle between v=2i-4j and w=3i+2j.

Solution:

1.9765

1cos

652

2cos

1320

86cos

23*)4(2

2*43*2cos

cos

1

11

2222

1

1

wv

wv

The Dot Product and Orthogonal Vectors

• Two nonzero vectors v and w are orthogonal if and only if v•w=o. Because 0•v=0, the zero vector is orthogonal to every vector v.

Example

0549

2*23*3

wv

• Are the vectors v=3i-2j and w=3i+2j orthogonal?

Solution:

The vectors are not orthogonal.

The Vector Projection of v Onto w

• If v and w are two nonzero vectors, the vector projection of v onto w is

ww

wvvprojw 2

Example

jiji

ji

ww

wvvprojw

29

70

29

28)52(

29

14

)52(])5(2[

5*42*322

2

• If v=3i+4j and w=2i-5j, find the projection of v onto w

Solution:

The Vector Components of v• Let v and w be two nonzero vectors. Vector v can

be expressed as the sum of two orthogonal vectors v1 and v2, where v1 is parallel to w and v2 is orthogonal to w.

1221 , vvvww

wvvprojv w

Thus, v = v1 + v2. The vectors v1 and v2 are called the vector components of v. The process

of expressing v as v1 and v2 is called the

decomposition of v into v1 and v2.

Example• Let v=3i+j and w=2i-3j. Decompose v into

two vectors, where one is parallel to w and the other is orthogonal to w.

Solution:

ji

jiv

jiji

jiv

vvvww

wvv

13

30

13

33

)13

93()

13

63(

13

9

13

6)32(

13

3

)32()3(2

3*12*3

,

2

221

1221

Definition of Work

• The work W done by a force F in moving an object from A to B is

• W = F · AB.