Lecture Slides on Scalar Multiplication | MATH 114, Study notes of Mathematics

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Math 114 – Rimmer

13.2 Vectors

Quantities such as area, volume, mass, and time can be

characterized by a single real variable

Other quantities such as displacement, velocity, and

force involve both magnitude and direction.

To represent these quantities we use a vector represented by a

directed line segment (arrow)

initial

point

terminal

point

v =

The magnitude of a vector

is represented by v or v.

Any other vector that has the same

magnitude and direction as

is called an equivalent or equal vector ⇒ =

u

v

u v

u

Math 114 – Rimmer

13.2 Vectors

We can multiply a vector by a real number.

This is called.

c

scalar multiplication

c v has a magnitude that is c times the magnitude of v.

c v has has the same direction as v if c >0.

c v has has the opposite direction as v if c < 0.

v

2 v

3

2

v

− v

− v

Math 114 – Rimmer

13.2 Vectors We can add a vector to another vector.

This is called , +

v u

vector addition v u

v

u

v

u

v + u

Connect the terminal point of the first vector

to the initial point of the second vector

When connected this way, the sum is the vector

from the initial point of the first vector

to the terminal point of the second vector

( )

Vector subtraction is just vector

addition in disguise

−

−

v u

v + u

− u

v

− u v − u

v

u

v

u

v + u

v − u

This can all be summed up using

the parallelogram determined by v and u.

Math 114 – Rimmer

13.2 Vectors So far we have studied vectors geometrically.

We now want to look at vectors algebraically.

1 2

v = a , a

( ) 1 2

a , a

v

1

a

2

a

Standard unit vectors

( ) ( )

2 2

1 2

v = a + a

The magnitude of v is found by:

v = 1 ⇒the vector is called a unit vector

i = 1, 0 j = 0, Now v can be written as :

v = a i + a j

v

Math 114 – Rimmer

13.2 Vectors

Now for 3 dimensions we have:

1, 0, 0

0,1, 0

0,0,

=

=

=

i

j

k

, ,

a a a

a a a

= + +

=

v i j k

v

v

More specifically,

the component of

the component of

the component of

a

a

a

=

=

=

i v

j v

k v

, , and are called

the components of

a a a

v

Math 114 – Rimmer

13.2 Vectors

Scalar Multiplication:

v = a , a , a scaled by a factor c

c v = ca , ca , ca

( )

multiply each component by c

Vector Addition:

v = a , a , a added to u = b b , , b

v + u = a + b a , + b , a + b

( )

add componentwise

1 2 y + y

2 y

1 y

1 2 x + x

1 x

2 x

Math 114 – Rimmer

13.2 Vectors

A

B

( )

1 2

a , a

( )

1 2

b b ,

Vector from a point A to another point B :

v = b − a , b − a

A unit vector u in the same direction as another vector v :

Scale the vector v by the reciprocal of v

a , a , a

 

= =

 

 

 

v

u

v v

In 3-dimensions

v = b − a , b − a , b − a

1 2 3

, ,

a a a

u =

v v v

3

4 v =3, 4

v = 5

3 4

5 5

u = ,

Math 114 – Rimmer

13.2 Vectors

( ) ( )

Find the component form and magnitude of the vector with the

initial point 3, 2, 0 and terminal point 4,1, 5.

Find 3.

Find a unit vector in the direction of.

v

v

v

v = 4 − 3,1 − 2,5 − 0 = 1, −1,5 or i − j + 5 k

( )

3 v = 3 1,3⋅ ⋅ −1 ,3 5 ⋅

= 3, −3,15 or 3 i − 3 j + 15 k

( )

v = 1 + − 1 + 5 = 1 + 1 + 25

= 27

= 3 3

, ,

−

u =

, ,

−

=

or

i − j + k

Math 114 – Rimmer

13.2 Vectors

1 2

A 200 lb. traffic light supported by two cables hangs in equilibrium.

As shown in figure (b), let the weight of the light be represented by and

the forces in the two cables by and.

w

F F

1 2 1 2

As shown in figure (c), the forces can be arranged to form a triangle.

Equilibrium implies that the sum of the forces is.

Find and , and find the magnitudes of and.

0

F F F F

( ) ( )

= cos 20 , sin 20 1 1 1

F F F

 

( ) ( ) 2 2 2

F = − F cos 15 , F sin 15

 

F



2

F



w = 0, − 200

( ) ( ) ( ) ( )

2 2 2

• • = cos 20 − cos 15 , sin 20 + sin 15 − 200 =0, 0 1 1 1

F F w F F F F

   

( )

( )

2

cos 15

cos 20

= 1

F

F





( )

( )

( ) ( )

2

2

cos 15

sin 20 sin 15 200

cos 20

• =

F

F



 



( )

( )

( ) ( ) 2

cos 15

sin 20 sin 15 200

cos 20

 

 + =

   

F



 



( )

( )

( ) ( )

200

2 cos 15

sin 20 sin 15 cos 20

F =    

2

⇒ F ≈ 327.66 lbs. 1

F ≈ 336.81 lbs.

Math 114 – Rimmer

13.2 Vectors

If we were interested in having these forces in Newtons, we can just convert

at the end using the coversion factor 1 lb. (force) ≈4.448 Newtons

Where does this conversion factor come from?

2

1ft.

1 lb. (force) 1 slug

sec.

≈ ⋅

2

1 m.

1 Newton 1 kg

sec.

≈ ⋅

So we need to convert slugs into kg. and ft. into meters.

1 lb. (force) ≈1 slug

1ft.

⋅

2

14.5939029 kg.

sec. 1 slug

⋅

0.3048 m.

1 ft.

⋅

2

m.

4.448221604 kg

sec.

= ⋅

⇒ 1 lb. (force) ≈4.448 Newtons