Copyright © 2011 Pearson, Inc. 6.1 Vectors in the Plane.
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Transcript of Copyright © 2011 Pearson, Inc. 6.1 Vectors in the Plane.
Copyright © 2011 Pearson, Inc.
6.1Vectors in the
Plane
Slide 6.1 - 2 Copyright © 2011 Pearson, Inc.
What you’ll learn about
Two-Dimensional Vectors Vector Operations Unit Vectors Direction Angles Applications of Vectors
… and whyThese topics are important in many real-world applications, such as calculating the effect of the wind on an airplane’s path.
Slide 6.1 - 3 Copyright © 2011 Pearson, Inc.
Directed Line Segment
Slide 6.1 - 4 Copyright © 2011 Pearson, Inc.
Two-Dimensional Vector
A two - dimensional vector v is an ordered pair of real
numbers, denoted in component form as a,b . The
numbers a and b are the components of the vector v.
The standard representation of the vector a,b is the
arrow from the origin to the point (a,b). The magnitude
of v is the length of the arrow and the direction of v is the
direction in which the arrow is pointing. The vector
0 = 0,0 , called the zero vector, has zero length and
no direction.
Slide 6.1 - 5 Copyright © 2011 Pearson, Inc.
Initial Point, Terminal Point, Equivalent
Slide 6.1 - 6 Copyright © 2011 Pearson, Inc.
Head Minus Tail (HMT) Rule
If an arrow has initial point x1, y
1( ) and terminal point
x2 , y2( ), it represents the vector x2 −x1, y2 −y1 .
Slide 6.1 - 7 Copyright © 2011 Pearson, Inc.
Magnitude
If v is represented by the arrow from x1, y
1( ) to x2 , y2( ),
then
v = x2 −x1( )2+ y2 −y1( )
2.
If v= a,b , then v = a2 +b2 .
Slide 6.1 - 8 Copyright © 2011 Pearson, Inc.
Example Finding Magnitude of a Vector
Find the magnitude of v represented by PQ,
where P =(3,−4) and Q=(5,2).
Slide 6.1 - 9 Copyright © 2011 Pearson, Inc.
Example Finding Magnitude of a Vector
v = x2 −x1( )2+ y2 −y1( )
2
= 5−3( )2+ 2−(−4)( )
2
=2 10
Find the magnitude of v represented by PQ,
where P =(3,−4) and Q=(5,2).
Slide 6.1 - 10 Copyright © 2011 Pearson, Inc.
Vector Addition and Scalar Multiplication
Let u = u1,u2 and v= v1,v2 be vectors and let k be
a real number (scalar). The sum (or resultant) of the
vectors u and v is the vector
u+ v= u1 +v1,u2 +v2 .
The product of the scalar k and the vector u is
ku=k u1,u2 = ku1,ku2 .
Slide 6.1 - 11 Copyright © 2011 Pearson, Inc.
Example Performing Vector Operations
Let u = 2, −1 and v= 5,3 . Find 3u+ v.
Slide 6.1 - 12 Copyright © 2011 Pearson, Inc.
Example Performing Vector Operations
3u = 3 2( ), 3 −1( ) = 6,−3
3u+ v = 6,−3 + 5,3 = 6+5,−3+3 = 11,0
Let u = 2, −1 and v= 5,3 . Find 3u+ v.
Slide 6.1 - 13 Copyright © 2011 Pearson, Inc.
Unit Vectors
A vector u with | u | =1 is a unit vector. If v is not
the zero vector 0,0 , then the vector u=v|v |
=1|v |
v
is a unit vector in the direction of v.
Slide 6.1 - 14 Copyright © 2011 Pearson, Inc.
Example Finding a Unit Vector
Find a unit vector in the direction of v = 2, −3 .
Slide 6.1 - 15 Copyright © 2011 Pearson, Inc.
Example Finding a Unit Vector
v = 2, −3 = 22 + −3( )2= 13, so
v
v=
1
132, −3 =
2
13, −
3
13
Find a unit vector in the direction of v = 2, −3 .
Slide 6.1 - 16 Copyright © 2011 Pearson, Inc.
Standard Unit Vectors
The two vectors i = 1,0 and j = 0,1 are the standard
unit vectors. Any vector v can be written as an expression
in terms of the standard unit vector:
v= a,b
= a,0 + 0,b
=a 1,0 +b 0,1
=ai +bj
Slide 6.1 - 17 Copyright © 2011 Pearson, Inc.
Resolving the Vector
If v has direction angle θ, the components of v canbe computed using the formula
v = vcosθ, vsinθ .
From the formula above, it follows that the unit vector
in the direction of v is u =vv= cosθ,sinθ .
Slide 6.1 - 18 Copyright © 2011 Pearson, Inc.
Example Finding the Components of a Vector
Find the components of the vector v with
direction angle 120o and magnitude 8.
Slide 6.1 - 19 Copyright © 2011 Pearson, Inc.
Example Finding the Components of a Vector
v = a,b = 8cos120o,8sin120o
= 8 −12
⎛
⎝⎜⎞
⎠⎟,8
32
⎛
⎝⎜
⎞
⎠⎟
= −4,4 3
So a=−4 and b=4 3.
Find the components of the vector v with
direction angle 120o and magnitude 8.
Slide 6.1 - 20 Copyright © 2011 Pearson, Inc.
Example Finding the Direction Angle of a Vector
Let u = −2,3 and v= −4,−1 .
Find the magnitude and direction angle of each vector.
Slide 6.1 - 21 Copyright © 2011 Pearson, Inc.
Example Finding the Direction Angle of a Vector
Let α be the direction angle of u, then
|u|= −2( )2+32 = 13
−2 =u cosα
−2 = 13cosα
cosα =−2
13
Let u = −2,3 and v= −4,−1 .
Find the magnitude and direction angle of each vector.
α =cos−1−2
13
⎛
⎝⎜⎞
⎠⎟
90º<α <180º
α ≈123.69º
Slide 6.1 - 22 Copyright © 2011 Pearson, Inc.
Example Finding the Direction Angle of a Vector
Let β be the direction angle of v, then
|v |= −4( )2+ −1( )
2= 17
−4 = v cosβ
−4 = 17 cosβ
cosβ =−4
17
Let u = −2,3 and v= −4,−1 .
Find the magnitude and direction angle of each vector.
β =360º−cos−1−4
17
⎛
⎝⎜⎞
⎠⎟
180º< β < 2700º
β ≈194.04º
Slide 6.1 - 23 Copyright © 2011 Pearson, Inc.
Example Finding the Direction Angle of a Vector
Interpret
The direction angle for u
is α ≈123.69º.
The direction angle for vis α ≈194.04º.
Let u = −2,3 and v= −4,−1 .
Find the magnitude and direction angle of each vector.
Slide 6.1 - 24 Copyright © 2011 Pearson, Inc.
Velocity and Speed
The velocity of a moving object is a vector
because velocity has both magnitude and
direction. The magnitude of velocity is speed.
Slide 6.1 - 25 Copyright © 2011 Pearson, Inc.
Quick Review
1. Find the values of x and y.
2. Solve for θ in degrees. θ =sin-1 3
11
⎛
⎝⎜⎞
⎠⎟
Slide 6.1 - 26 Copyright © 2011 Pearson, Inc.
Quick Review
3. A naval ship leaves Port Northfolk and averages
43 knots (nautical mph) traveling for 3 hr on a bearing
of 35o and then 4 hr on a course of 120o.
What is the boat's bearing and distance from
Port Norfolk after 7 hr.
The point P is on the terminal side of the angle θ. Find the measure of θ if 0o <θ < 360o.4. P(5,7)5. P(-5,7)
Slide 6.1 - 27 Copyright © 2011 Pearson, Inc.
Quick Review Solutions
1. Find the values of x and y.
x =−7.5, y=7.5 3
2. Solve for θ in degrees. θ =sin-1 3
11
⎛
⎝⎜⎞
⎠⎟64.8º
Slide 6.1 - 28 Copyright © 2011 Pearson, Inc.
Quick Review Solutions
3. A naval ship leaves Port Northfolk and averages
43 knots (nautical mph) traveling for 3 hr on a bearing
of 35o and then 4 hr on a course of 120o.
What is the boat's bearing and distance from
Port Norfolk after 7 hr.
distance = 224.2; bearing = 84.9º
The point P is on the terminal side of the angle θ. Find the measure of θ if 0o <θ < 360o.4. P(5,7) 54.5º5. P(−5,7) 125.5º