6.3 vectors in the plane

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6.3 Vectors in the Plane uantities in geometry and physics, such as area, ti mperature, can be represented by a single real numb uantities, such as force and velocity, involve both de and direction and cannot be completely erized by a single real number. esent such a quantity, we use a directed line segme ected line segment PQ has initial point P and te and we denote its magnitude (length) by PQ P Initial Point Q Terminal Point PQ

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6.3 Vectors in the Plane. Many quantities in geometry and physics, such as area, time, and temperature, can be represented by a single real number. Other quantities, such as force and velocity, involve both magnitude and direction and cannot be completely - PowerPoint PPT Presentation

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Page 1: 6.3 Vectors in the Plane

6.3 Vectors in the Plane

Many quantities in geometry and physics, such as area, time,and temperature, can be represented by a single real number.

Other quantities, such as force and velocity, involve both magnitude and direction and cannot be completely characterized by a single real number.

To represent such a quantity, we use a directed line segment.The directed line segment PQ has initial point P and terminalpoint Q and we denote its magnitude (length) by .PQ

P

Initial Point

QTerminal Point

PQ

Page 2: 6.3 Vectors in the Plane

Vector Representation by Directed Line Segments

Let u be represented by the directed line segment from P = (0,0) to Q = (3,2), and let v be represented by the directed line segment from R = (1,2) to S = (4,4). Show that u = v.

1 2 3 4

4

3

2

1

P

QR

S

u

v

Using the distance formula, show that u and v have the same length.Show that their slopes are equal.

( ) ( )( ) ( ) 132414

130203

22

22

=−+−=

=−+−=

v

u

Slopes of u and v are both3

2

Page 3: 6.3 Vectors in the Plane

Component Form of a Vector

The component form of the vector with initial point P = (p1, p2)and terminal point Q = (q1, q2) is

PQ vvvpqpq ==−−= 212211 ,,

The magnitude (or length) of v is given by

( ) ( ) =−+−= 222

211 pqpqv 2

22

1 vv +

Page 4: 6.3 Vectors in the Plane

Find the component form and length of the vector v that hasinitial point (4,-7) and terminal point (-1,5)

-2 2 4

6

4

2

-2

-4

-6

-8

Let P = (4, -7) = (p1, p2) and Q = (-1, 5) = (q1, q2).

Then, the components of v = are given by

21,vv

v1 = q1 – p1 =

v2 = q2 – p2 =

-1 – 4 = -5

5 – (-7) = 12

Thus, v = 12,5−and the length of v is

1316912)5( 22 ==+−=v

Page 5: 6.3 Vectors in the Plane

Vector Operations

The two basic operations are scalar multiplication and vectoraddition. Geometrically, the product of a vector v and a scalark is the vector that is times as long as v. If k is positive, then kv has the same direction as v, and if k is negative, then kv has the opposite direction of v.

k

v ½ v 2v -vv

2

3−

Page 6: 6.3 Vectors in the Plane

Definition of Vector Addition & Scalar Multiplication

Let u = and v = be vectors and let k be a scalar (real number). Then the sum of u and v is

21,uu 21,vv

u + v = 2211 , vuvu ++

and scalar multiplication of k times u is the vector

2121 ,, kukuuukku ==

Page 7: 6.3 Vectors in the Plane

Vector Operations

Ex. Let v = and w = . Find the following vectors. a. 2v b. w – v

5,2− 4,3

-2 2

6

10

8

4

2

-2-4

10,42 −=v

v

2v

1 2 3 4

4

3

2

1

5-1

w

1,554),2(3 −=−−−=−vw

-v

w - v

Page 8: 6.3 Vectors in the Plane

Writing a Linear Combination of Unit Vectors

Let u be the vector with initial point (2, -5) and terminal point (-1, 3). Write u as a linear combination of the standard unitvectors of i and j.

-2 2 4

6

4

2

-2

-4

-6

-8

(2, -5)

u

(-1, 3)

Solution

53,21 +−−=u

8,3−=

ji 83 +−=

-2 2

6

10

8

4

2

-2-4

Graphically,it looks like…

-3i

8j

Page 9: 6.3 Vectors in the Plane

Writing a Linear Combination of Unit Vectors

Let u be the vector with initial point (2, -5) and terminal point(-1, 3).Write u as a linear combination of the standard unitvectors i and j.

Begin by writing the component form of the vector u.

u = −1− 2,3− −5( )

u = −3,8

u = −3i + 8 j

Page 10: 6.3 Vectors in the Plane

Unit Vectors

=v

v=

1

v

⎝ ⎜

⎠ ⎟vu = unit vector

Find a unit vector in the direction of v =

v

v=

−2,5

−2( )2

+ 5( )2

−2,5

=1

29−2,5

=−2

29,

5

29

Page 11: 6.3 Vectors in the Plane

Vector Operations

Let u = -3i + 8j and let v = 2i - j. Find 2u - 3v.

2u - 3v = 2(-3i + 8j) - 3(2i - j)

= -6i + 16j - 6i + 3j

= -12i + 19 j