Transcript Chapter 6 - James Bac Dang

The Law of Sines
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
sin A sin B sin C


a
b
c
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.
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.
C
A + B + C  180º
b 76
a
83.5º + B  180º
B  96.5º
50º
A
50º + B + 33.5º  180º
33.5º
c
B
The sum of the measurements of a
triangle’s interior angles is 180º.
A = 50º and C = 33.5º.
Add.
Subtract 83.5º from both sides.
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.
C
33.5º
b 76
a
50º
A
c
B
Find a:
a
b

sin A sin B
a
76

sin 50
sin 96.5
76sin 50
a
 59
sin 96.5
Find c:
This is the known ratio.
The solution is B  96.5º, a  59, and c  42.
c
b

sin C sin B
c
76

sin 33.5 sin 96.5
76sin 33.5
c
 42
sin 96.5
The Ambiguous Case (SSA)
Consider a triangle in which a, b, and A are given. This information may result in:
No Triangle
a is less than h
and not long
enough to form
a triangle.
One Right Triangle
a
b
h = b sin A
A
b
a
h = b sin A
A
Two Triangles
a is greater than h
and a is less than
b. Two distinct
triangles are
formed.
A
a = h and is just
the right length
to form a right
triangle.
b
a
One Triangle
a h = b sin A
a is greater than h
and a is greater
than b. One
triangle is formed.
b
A
a
Example
• Solve the triangle shown with A=36º, B=88º

and c=29 feet.
b
a


c
Solution:
A=36º, B=88º
so
180-88-36=56º
C=56º
sin 36 sin 88 sin 56


a
b
29
.59 1 .83
 
a b 29
1 .83

so .83b  29
b  34.94
b 29
.59 .83

so .83a  17.11
a  20.61
a
29
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
1
1
1
Area  bc sin A  ab sin C  ac sin B
2
2
2
C
b
A
a
h
D
B
c
Text Example
• Find the area of a triangle having two sides of lengths 24
meters and 10 meters and an included angle of 62º.
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.
Area  1/2 (24)(10)(sin 62º)  106
the area of the triangle is approximately 106 square meters.
C
b = 10 meters
62º
A
c = 24 meters
B
Example
• Find the area of a triangle having two sides
of lengths 12 ft. and 20 ft. and an included
angle of 57º.
Solution:
1
1
Area  bc sin A  (12)( 20) sin 57
2
2
 120 * .84  100.8sq. ft.
The Law of Cosines
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.
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.
Text Example
C
b = 20
• Solve the triangle shown with A = 60º, b = 20, and c = 30. A
a
60º
c =30
B
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
a2  202 + 302 - 2(20)(30) cos 60º
 400 + 900 - 1200(0.5)  700
a   700  26
Apply the Law of
Cosines to find a.
b = 20, c = 30, and A = 60°.
Perform the indicated operations.
Take the square root of both sides and solve for a.
Text Example cont.
C
b = 20
a
Solve the triangle shown with A = 60º, b = 20, and c = 30. A 60º
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.
b
a

sin B sin A
20
700

sin B sin 60
700 sin B  20sin 60
sin B 

B
B  41
41
20sin 60
 0.6547
700
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-10.6547 using a calculator.
B
Text Example cont.
C
Solve the triangle shown with A = 60º, b = 20, and c = 30.
b = 20
A
a
60º
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.
B
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.
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.
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?
Solution
We use the Law of Cosines to find b in this SAS situation.
b2  a2 + c2 - 2ac cos B
Apply the Law of Cosines.
b2  6502 + 6002 - 2(650)(600) cos 88º
Substitute: a= 650, c =600, and B= 88°.
 755,278
Use a calculator.
b  869
Take the square root and solve for b.
After two hours, the planes are approximately 869 miles apart.
Heron’s Formula
The area of a triangle with sides a, b, and c is
Area  s ( s - a )( s - b)( s - c)
1
s  (a + b + c)
2
Example
• Use Heron’s formula to find the area of the
given triangle:
a=10m, b=8m, c=4m
Solution:
1
s  (a + b + c)
2
1
s  (10 + 8 + 4)
2
1
s  (22)  11
2
Area  s( s - a)( s - b)( s - c)
 11(11 - 10)(11 - 8)(11 - 4)
 11(1)(3)(7)  231 sq.m.
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 .
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°)
(2,
135°)
135º
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
x  r cos 
y  r sin 
x 2 + y 2  r2
y
tan   /x
P = (r, ) = (x, y)
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
• Find the polar coordinates of a point whose
rectangular coordinates are (2, 4)
Solution:
r  x 2 + y 2  2 2 + 4 2  20  2 5
tan  
y 4
 2
x 2
  1.1
(2 5 , 1.1)
Example
• Convert 2x-y=1 to a polar equation.
Solution:
2x - y  1
2r cos  - r sin   1
r (2 cos  - sin  )  1
1
r
2 cos  - sin 
Graphs of Polar
Equations
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.
2
3 3
4
5
6

2

4 
˝
6

7
6
5
4 4
3

3
2
3
2
4
0
11 
6
7
5 4
3
Text Example
• Graph the polar equation r  4 cos  with  in radians.
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.

r  4 cos 
(r, )
0
4 cos 0 = 4 • 1 = 4
(4, 0)
 /6
4 cos  /6 = 4 •3/2=2 3=3.5
(3.5,  /6)
 /3
4 cos  /3 = 4 • 1/2 = 4
(2,  /3)
 /2
4 cos  /2 = 4 • 0 = 0
(0,  /2)
2 /3
4 cos2  /3 = 4(- 1/2) = -2
(-2, 2  /3)
5/6
4 cos5  /6 = 4(- 3/2)=-2 3=-3.5
(-3.5,5 /6)
˝
4 cos ˝  = 4(-1) = -4
(-4, )
(2,  /3)
(0,  /2)
2
3 3
4
5
6

2

3
˝
4 
6
(3.5,  /6)
0
˝
2 4
(4, 0) or (-4, )
7
11 
6
6
5
7
(-2, 2  /3)
4 4
5 4
3

3
3
2
(-3.5, 5 /6)
Circles in Polar Coordinates
The graphs of
r = a cos  and
r = a sin 
Are circles.
r = a cos 
r = a sin 
 /2
/2
a
˝
0
˝
0
a
3  /2
3  /2
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.
Text Example cont.
Solution
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.
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.
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 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 .

r
0
0
 /6
 /3
2
2  /3
0.13
0.50
1.00
1.50
5  /6

1.87
2
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.

2
3 3
4

˝
2
3
4
5
6

6

7
6
5
4 4
3

1
0
2
11 
6
3
2
7
5 4
3
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.

2
3 3
4

2
3
4
5
6

6

7
6
5
4 4
3

1
0
2
11 
6
3
2
7
5 4
3
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
loop if 1< ab < 2
and called cardiods


2

3
2


2
0
3
2
loop if ab  2.

2
0
No dimple and no inner

2
0
3
2

0
3
2
Example
• Graph the polar equation
y= 2+3cos
Example
• Graph the polar equation
y= 2+3cos
Solution:
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
with 4 petals
Rose curve
with 3 petals
Rose curve
with 8 petals

Rose curve
with 5 petals



2
2
2
2
n=4
a
a
n=3

0

0
a

0

0
a
n=2
3
2
3
2
3
2
a
3
2
n=5
Example
• Graph the polar equation y=3sin2
Example
• Graph the polar equation y=3sin2
Solution:
Lemniscates
• The graphs of r2 = a2 sin 2 and r2 = a2
cos 2 are called lemniscates
Lemniscate:
r2 = a2 cos 2
Complex Numbers
in Polar Form;
DeMoivre’s Theorem
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
b
z = a + bi
a
Real axis
Text Example
Plot in the complex plane:
a. z = 3 + 4i
b. z = -1 – 2i
Solution
•
•
c. z = -3
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.
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.
d. z = -4i
5
4
3
2
1
-5 -4 -3 -2
z = -1 – 2i
z = 3 + 4i
1 2 3 4 5
-3
-4
-5
Text Example cont.
Plot in the complex plane:
a. z = 3 + 4i
b. z = -1 – 2i
Solution
•
•
c. z = -3
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.
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.
d. z = -4i
z = -3
5
4
3
2
1
-5 -4 -3 -2
z = -1 – 2i
z = 3 + 4i
1 2 3 4 5
-3
-4
-5
z = -4i
The Absolute Value of a Complex
Number
• The absolute value of the complex number
a + bi is
2
2
z  a + bi  a + b
Example
• Determine the absolute value of z=2-4i
Solution:
z  a + bi  a + b
2
2
 2 + (-4)  4 + 16
2
2
 20  2 5
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  , r  a 2 + b 2 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
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
2
-2
r
-2
z = -2 – 2i
Real
axis
Solution
Text Example cont.
r  a 2 + b 2  (-2)2 + (-2)2  4 + 4  8  2 2
tan  
b -2

1
a -2
Since tan 4 = 1, we know that  lies in quadrant III. Thus,

4  5
  + +
+ 
4
4
4
4
5
5
z  r(cos + i sin  )  2 2(cos
+ i sin
)
4
4
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.
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
z1 z 2
Form the product of the given
= [4(cos 50º + i sin 50º)][7(cos 100º + i sin 100º)]
numbers.
= (4 · 7)[cos (50º + 100º) + i sin (50º + 100º)]
Multiply moduli and add
arguments.
= 28(cos 150º + i sin 150º)
Simplify.
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
z1 r1
 [cos(1 -  2 ) + i sin( 1 -  2 )]
z 2 r2
To divide two complex numbers, divide
moduli and subtract arguments.
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
z  [r (cos  + i sin  )]
n
 r n (cos n + i sin n )
n
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)]
= 64(cos 60º + i sin 60º)
1
3 

 64
 + i

2 
2
 32 + 32 3i
Raise the modulus to the 6th power and multiply
the argument by 6.
Simplify.
Write the answer in rectangular form.
Multiply and express the answer in a + bi form.
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
   + 360k 
  + 360k 
zk  r cos
 + i sin 

n
n



 
where k  0,1,2,3,..., n - 1
n
Example
• Find all the complex fourth roots of
81(cos60º+isin60º)
Solution:
   + 360k 
  + 360k 
z k  n r cos
+
i
sin



n
n



 
  60 + 360 * 0 
 60 + 360 * 0 
 4 81 cos
+
i
sin



4
4



 
 3(cos 15 + i sin 15 )
Example cont.
• Find all the complex fourth roots of
81(cos60º+isin60º)
Solution:
  60 + 360 *1 
 60 + 360 *1 
 81 cos
 + i sin 

4
4



 
4
 3(cos 105 + i sin 105 )
 3(cos 195 + i sin 195 )
 3(cos 285 + i sin 285 )
Vectors
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.
Q
Initial point
Terminal point
P
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.
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
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
u
Initial point of u
v
Terminal point of v
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.
-u
v–u
-u
v
u
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.
y
1
j
O
i
1
x
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
2
2
v  a +b
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
v  a +b
2
2
 (-3) + 4
 9 + 16
 25  5
2
Terminal point
5
4
3
2
v = -3i + 4j
2
1
-5 -4 -3 -2
-1
-1
-2
-3
-4
-5
1
2
3 4
5
Initial point
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
Text Example
If v = 5i + 4j and w = 6i – 9j, find:
a. v + w
b. v – w.
Solution
•
v + w = (5i + 4j) + (6i – 9j)
= (5 + 6)i + [4 + (-9)]j
= 11i – 5j
•
v + w = (5i + 4j) – (6i – 9j)
= (5 – 6)i + [4 – (-9)]j
= -i + 13j
These are the given vectors.
Add the horizontal components. Add the
vertical components.
Simplify.
These are the given vectors.
Subtract the horizontal components. Subtract
the vertical components.
Simplify.
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
• If v=2i-3j, find 5v and -3v
Solution:
5v  (5 * 2)i + (5 * -3) j
 10i - 15 j
- 3v  (-3 * 2)i + (-3 * -3) j
 -6i + 9 j
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
v
v
• is a unit vector that has the same direction
as v. To find this vector, divide v by its
magnitude.
Example
• Find a unit vector in the same direction as
v=4i-7j
Solution:
v  4 + (-7)
2
2
 16 + 49  65
v
4
7

ij
v
65
65
The Dot Product
Definition of a Dot Product
If v=a1i+b1j and w = a2i+b2j are
vectors, the dot product is defined as
v  w  a1a2 + b1b2
The dot product of two vectors is the
sum of the products of their
horizontal and vertical components.
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) = 25 + 4 = 29
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
v  w  v w cos
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
vw
-1 v  w
cos 
and   cos
v w
v w
Example
• Find the angle  between v=2i-4j and
w=3i+2j.
-1 v  w
  cos
Solution:
 cos
-1
 cos -1
 cos -1
v w
2 * 3 + -4 * 2
2 2 + (-4) 2 * 32 + 2 2
6-8
-2
 cos -1
20 13
2 65
-1
 97.1
65
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
• Are the vectors v=3i-2j and w=3i+2j
orthogonal?
Solution:
v  w  3 * 3 + 2 * -2
 9-4  5  0
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
projwv 
vw
w
2
w
Example
• If v=3i+4j and w=2i-5j, find the projection
of v onto w
Solution:
proj wv 
vw
w
2
w
3 * 2 + 4 * -5
 2
(2i - 5 j )
2
[2 + (-5) ]
- 14
- 28 70

(2i - 5 j ) 
i+
j
29
29
29
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.
v1  proj wv 
vw
w
2
w, v2  v - v1
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.
v1 
Solution:
vw
w
2
w, v2  v - v1
3 * 2 + 1* -3
(2i - 3 j )
2
2
2 + (-3)
3
6
9
 (2i - 3 j )  i j
13
13 13
6
9
v2  (3 - )i + (-3 + ) j
13
13
33 30
 ij
13 13
v1 
Definition of Work
• The work W done by a force F in moving an
object from A to B is
• W = F · AB.