1. The Law of Sines

2. Law of Sines
If 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.

3. Text Example Solve triangle ABC if A = 50º, C = 33.5º, and b = 76.
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 c A B C
b =76
33.5º
50º
A + B + C = 180º
The sum of the measurements of a triangle’s interior angles is 180º.
50º + B º = 180º
A = 50º and C = 33.5º.
Add.
83.5º + B = 180º
B = 96.5º
Subtract 83.5º from both sides.

4. Solve triangle ABC if A = 50º, C = 33.5º, and b = 76.
Text Example cont.
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. a c A B C
b =76
33.5º
50º
Find a: Find c:
This is the known ratio.
The solution is B = 96.5º, a  59, and c  42.

5. The Ambiguous Case (SSA)
Consider a triangle in which a, b, and A are given. This information may result in:
No Triangle One Right Triangle b A a
h = b sin A
a = h and is just the right length to form a right triangle. b a A
h = b sin A
a is less than h and not long enough to form a triangle.
Two Triangles One Triangle a b 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.

6. Example Solve the triangle shown with A=36º, B=88º and c=29 feet.
Solution:
A=36º, B=88º so =56º C=56º

7. Area of An Oblique Triangle
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 C A B c b a h D

8. Text Example
Find the area of a triangle having two sides of lengths 24 meters and 10 meters and an included angle of 62º.
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 meters A B C
62º

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

10. The Law of Cosines

11. The Law of Cosines
If 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.

12. Solving an SAS Triangle
Use the Law of Cosines to find the side opposite the given angle.
Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute.
Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180º.

13. Text Example a
b = 20
c =30 A B C
60º
Solve the triangle shown with A = 60º, b = 20, and c = 30.
Solution We are given two sides and an included angle. Therefore, we apply the three-step procedure for solving a SAS triangle.
Step l Use the Law of Cosines to find the side opposite the given angle. Thus, we will find a.
a2 = b2 + c2 - 2bc cos A Apply the Law of Cosines to find a.
a2 = (20)(30) cos 60º b = 20, c = 30, and A = 60°.
= (0.5) = 700 Perform the indicated operations.
Take the square root of both sides and solve for a.
a = ÷  26

14. Solve the triangle shown with A = 60º, b = 20, and c = 30.
Text Example cont. a
b = 20
c = 30 A B C
60º
Solve the triangle shown with A = 60º, b = 20, and c = 30.
Solution
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.
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.
Find sin using a calculator.

15. Text Example cont. a
b = 20
c = 30 A B C
60º
Solve the triangle shown with A = 60º, b = 20, and c = 30.
Solution We are given two sides and an included angle. Therefore, we apply the three-step 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.

16. Solving an SSS Triangle
Use the Law of Cosines to find the angle opposite the longest side.
Use the Law of Sines to find either of the two remaining acute angles.
Find the third angle. Subtract the measures of the angles found in steps 1 and 2 from 180º.

17. 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?
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 north-south line to find angle B
in triangle ABC. Thus,
B = 180º - 66º - 26º = 88º.
We now have a = 650, c = 600, and B = 88º.

18. 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?
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 = (650)(600) cos 88º Substitute: a= 650, c =600, and B= 88°.
 755, Use a calculator.
b  Take the square root and solve for b.
After two hours, the planes are approximately 869 miles apart.

19. Heron’s Formula
The area of a triangle with sides a, b, and c is

20. Example
Use Heron’s formula to find the area of the given triangle: a=10m, b=8m, c=4m
Solution:

21. Polar Coordinates

22. 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 .

23. Text Example
Plot the points with the following polar coordinates: a. (2, 135°)
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°)
135º

24. Multiple Representation of Points
If n is any integer, the point (r, ) can be represented as
(r, ) = (r,  + 2n) or (r, ) = (-r,  + r + 2n )

25. Relations between Polar and Rectangular Coordinates
x = r cos 
y = r sin 
x2 + y2 = r2
tan  = y/x
P = (r, ) = (x, y)

26. Converting a Point from Rectangular to Polar Coordinates (r > 0 and 0 <  < 2)
Plot the point (x, y).
Find r by computing the distance from the origin to (x, y).
Find  using tan = y/x with  lying in the same quadrant as (x, y).

27. Example
Find the polar coordinates of a point whose rectangular coordinates are (2, 4)
Solution:

28. Example
Convert 2x-y=1 to a polar equation.
Solution:

29. Graphs of Polar Equations

30. Using Polar Grids to Graph 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.  6 ˝ 4 3 2 2
3 
5 
7 
4 
11   

31. Text Example Graph the polar equation r = 4 cos  with  in radians.
Solution We construct a partial table of coordinates using multiples of p6. Then we plot the points and join them with a smooth curve, as shown.
(2,  /3) ˝ 6  4 3 2
2 
3 
5 
7 
4 
11 
(-4, )
4 cos ˝  = 4(-1) = -4 ˝
(-3.5,5 /6)
4 cos5  /6 = 4(- 3/2)=-2 3=-3.5
5/6
(-2, 2  /3)
4 cos2  /3 = 4(- 1/2) = -2
2 /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 = 4
(r, )
r = 4 cos   
(0,  /2)
(3.5,  /6)
(4, 0) or (-4, )
(-2, 2  /3)
(-3.5, 5 /6)

32. Circles in Polar Coordinates
The graphs of
r = a cos  and r = a sin 
Are circles.
r = a cos  r = a sin  ˝
 /2
3  /2
/2 a

33. Text Example
Check for symmetry and then graph the polar equation: r = 1 - cos .
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.

34. Text Example cont. Solution
The Line  = p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  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  = p2.
The Pole: Replace r by -r in r = 1 - cos  :
-r = 1 – cos  Replace r by –r.
r = cos  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.

35. Text Example cont. Solution
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 2p. 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 2p. Let's start by finding the values of r for values of  from 0 to p. p 6 4 3 ˝ 2
2 p
3 p
5 p
7 p
4 p
11 p 1 2
1.87
1.50
1.00
0.50
0.13 r p
5p/6
2p/3
p2
p/3
p/6  p
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.

36. Text Example cont. Solution
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. p 6 4 3 2
2 p
3 p
5 p
7 p
4 p
11 p 1

37. Limacons The graphs of r = a + b sin , r = a - b sin ,
r = a + b cos , r = a - b cos , a > 0, b > 0
are called limacons. The ratio ab determines a limacon's shape.
Inner loop if ab < 1 Heart shaped if ab = 1 Dimpled with no inner No dimple and no inner
and called cardiods loop if 1< ab < 2 loop if ab  2. p 2 3p p 2 3p p 2 3p p 2 3p

38. Example
Graph the polar equation y= 2+3cos

39. Example
Graph the polar equation y= 2+3cos
Solution:

40. Rose Curves The graphs of
r = 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 petals with 5 petals
n = 2 a p 2 3p p 2 3p
n = 3 a p 2 3p
n = 4 a p 2 3p
n = 5 a

41. Example
Graph the polar equation y=3sin2

42. Example
Graph the polar equation y=3sin2
Solution:

43. Lemniscates
The graphs of r2 = a2 sin 2 and r2 = a2 cos 2 are called lemniscates
Lemniscate:
r2 = a2 cos 2

44. Complex Numbers in Polar Form; DeMoivre’s Theorem

45. The Complex Plane
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 axis a b
z = a + bi

46. Text Example Plot in the complex plane:
a. z = 3 + 4i b. z = -1 – 2i c. z = -3 d. z = -4i
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. -5 -4 -3 -2 1 2 3 4 5
z = 3 + 4i
z = -1 – 2i
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.

47. Text Example cont. Plot in the complex plane:
a. z = 3 + 4i b. z = -1 – 2i c. z = -3 d. z = -4i
Solution
Because z = -3 = i, 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
z = 3 + 4i
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
z = -1 – 2i

48. The Absolute Value of a Complex Number
The absolute value of the complex number a + bi is

49. Example
Determine the absolute value of z=2-4i
Solution:

50. 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

51. Text Example
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.
Imaginary
axis 2 è
Real
axis -2 2 r -2
z = -2 – 2i

52. Text Example cont. Solution
Since tan p4 = 1, we know that  lies in quadrant III. Thus,

53. 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.

54. z1 = 4(cos 50º + i sin 50º) z2 = 7(cos 100º + i sin 100º)
Text Example
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)
Solution
z1z2
= [4(cos 50º + i sin 50º)][7(cos 100º + i sin 100º)]
Form the product of the given numbers.
Multiply moduli and add
arguments.
= (4 · 7)[cos (50º + 100º) + i sin (50º + 100º)]
= 28(cos 150º + i sin 150º)
Simplify.

55. 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.

56. 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

57. Text Example
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.
Write the answer in rectangular form.
Multiply and express the answer in a + bi form.

58. 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

59. Example Find all the complex fourth roots of 81(cos60º+isin60º)
Solution:

60. Example cont. Find all the complex fourth roots of 81(cos60º+isin60º)
Solution:

61. Vectors

62. Directed Line Segments and Geometric 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. P Q
Terminal point
Initial point
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.

63. 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

64. The Geometric Method for Adding Two Vectors
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.
Resultant vector
u + v v
Terminal point of v u
Initial point of u

65. The Geometric Method for the Difference of 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
v – u

66. 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. 1 i j O x y

67. Representing Vectors in Rectangular Coordinates
Vector v, from (0, 0) to (a, b), is represented as
v = 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

68. Text Example 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
Initial point
Terminal point
v = -3i + 4j

69. 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.

70. 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

71. Text Example 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.

72. 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.

73. Example
If v=2i-3j, find 5v and -3v
Solution:

74. 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.

75. 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

76. 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||

77. 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.

78. Example
Find a unit vector in the same direction as v=4i-7j
Solution:

79. The Dot Product

80. Definition of a Dot Product
If v=a1i+b1j and w = a2i+b2j are vectors, the dot product is defined as
The dot product of two vectors is the sum of the products of their horizontal and vertical components.

81. Text Example 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) = = 29

82. 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)

83. 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.

84. Alternative Formula for the Dot Product
If v and w are two nonzero vectors and  is the smallest nonnegative angle between them, then

85. Formula for the Angle between Two Vectors
If v and w are two nonzero vectors and  is the smallest nonnegative angle between v and w, then

86. Example
Find the angle  between v=2i-4j and w=3i+2j.
Solution:

87. 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.

88. Example Are the vectors v=3i-2j and w=3i+2j orthogonal?
Solution:
The vectors are not orthogonal.

89. The Vector Projection of v Onto w
If v and w are two nonzero vectors, the vector projection of v onto w is

90. Example If v=3i+4j and w=2i-5j, find the projection of v onto w
Solution:

91. 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.
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.

92. 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:

93. Definition of Work
The work W done by a force F in moving an object from A to B is
W = F · AB.