Vectors
A vector quantity has both magnitude (size) and direction
A scalar quantity only has size (i.e. temperature, time, energy, etc.)
Parts of a vector:
length – represents the
magnitude
tail
head
We can perform math operations with vectors!
Vector Addition
A motor boat is moving 15 km/hr relative to the water.
The river current is 10 km/hr downstream.
How fast does the boat go (relative to the shore)
upstream and downstream?
Boat Upstream Vector
Vector Addition
A motor boat is moving 15 km/hr relative to the water.
The river current is 10 km/hr downstream.
How fast does the boat go (relative to the shore)
upstream and downstream?
Boat Upstream Vector
Boat Downstream Vector
Vector Addition
A motor boat is moving 15 km/hr relative to the water.
The river current is 10 km/hr downstream.
How fast does the boat go (relative to the shore)
upstream and downstream?
Boat Upstream Vector
Boat Downstream Vector
Current Vector = 10 km/hr downstream
Boat Velocity Upstream
Upstream: Place vectors head to tail,
net result, 5 km/hr upstream
  
s  a b
Boat Velocity Upstream
Upstream: Place vectors head to tail,
Boat Velocity
Upstream: Place vectors head to tail,
net result, 5 km/hr upstream
Start
Finish
Difference
Boat Velocity
Downstream: Place vectors head to tail,
Boat Velocity
Downstream: Place vectors head to tail,
net result,
Boat Velocity
Downstream: Place vectors head to tail,
net result, 25 km/hr downstream
  
s  a b
Commutative law
   
a b  b a
Forces On An Airplane
When will it fly?
Gravity
Propulsion
Net Force?
Forces On An Airplane
When will it fly?
Gravity
Propulsion
Net Force
Plane Dives to the Ground
Forces On An Airplane
When will it fly?
Gravity
Propulsion
Lift
Net Force?
Friction
When will it fly?
Gravity
Propulsion
Lift
Net Force = 0 up or down
Plane rolls along the runway like a car because of propulsion.
Forces On An Airplane
When will it fly?
Gravity
Propulsion
Lift
Net Force
Plane Flies as long as Lift > Gravity
Friction
When will it fly?
Gravity
Propulsion
Lift
Air Resistance
Net Force = 0
Equilibrium
Flight
When will it fly?
Gravity
Propulsion
Lift
Air Resistance
Net Force
Plane Flies as long as Lift > Gravity
AND Propulsion > Air Resistance
Vector Components
A component of a vector is the projection of the vector on an
axis
y
Magnitude, size is:

A

A 6
We can write the vector A as
the sum of an x-component and
y-component:
Ay
x



A  Ax x  Ay y
Ax
Ax , Ay = the x and y components of the vector A
x hat and and y hat are the unit vectors
y



A  Ax x  Ay y

A
Ay

Ax
opp
x
If we only know the mag. of
A, and the angle,  it makes
with the x-axis, how do we
find the x, and y components?
adj
adj Ax
cos  
 
hyp A
opp Ay
sin  
 
hyp A


Ax  A cos 


Ax  A sin 
y



A  Ax x  Ay y

A
Ay

opp
If we only know the x and
y components, how can we
find the magnitude of A?
x
Ax
adj

A
2
x
2
y
A A
This comes from Pythagorean’s
theorem
GO TO HITT
Adding (and subtracting) vectors by components
Let’s say I have two vectors:
I want to calculate the vector
sum of these vectors:


A  Ax x  Ay


B  Bx x  B y
 


A  B   Ax  Bx  x  Ay  By  y
Let’s say the vectors have the following values:


A  Ax x  Ay


B  Bx x  B y



y 3x  4 y



y  5 x  8 y

y

y





A  Ax x  Ay y  3 x  4 y





B  Bx x  B y y  5 x  8 y
 


A  B   Ax  Bx  x  Ay  By  y


 3  5x  4  8 y



 2 x  12 y
y
Our result is consistent
with the graphical method!

y
B
A

x
B
A

x
What’s the magnitude
of our new vector?
 
A B 
 22  122
 4  144  148  12.2

y
How would you find the angle, , the
vector makes with the y-axis?
 


A  B  2 x  12 y
A
+

B
opp = 2

x

y

adj = 12

x
opp 2 1
1 1
0
tan  


   tan
 9.5
adj 12 6
6
GO TO HITT
Multiplying vectors by scalars:



A  Ax x  Ay y



a A  a Ax x  a Ay y



A  3 x  4 y and
 it was multiplied by
So if the vector A was:
the scalar, a = 5 then the new vector:
Scalar Product: (aka dot product):


a A  5  3x  5  4  y


 15 x  20 y
   
a  b  a b cos 
mag. of a
mag. of b
angle between
the vectors
Scalar Product: (aka dot product):
   
a  b  a b cos 
vectors
scalars
The dot product is the product
of two quantities:
(1) mag. of one vector
(2) scalar component of the
second vector along the
direction of the first
 






a  b  (a x x  a y y  a z z )  (bx x  by y  bz z )
 a x bx  a y by  a z bz
Go To HITT
Vector Product (aka cross product)
The vector product produces a new vector who’s magnitude
is given by:
    
c  a  b  a b sin 
The direction of the new vector is given by the,
“right hand rule”
Mathematically, we can find the direction using
matrix operations.




a  ax x  a y y  az z




b  bx x  by y  bz z

x
 
a  b  ax
bx

y
ay
by

z
az
bz
The cross product
is determined from
three determinants

x
 
a  b  ax
bx

y
ay
by

z
az
bz
The determinants are used to find the
components of the vector
1st : Strike out the first column and first row!
2nd : Cross multiply the four components – and subtract:
a y bz  a z by
x - component
3rd : Strike out the 2nd column and first row

x
 
a  b  ax
bx

y
ay
by

z
az
bz
4th : Cross multiply the four components,subtract, and
multiply by -1:
 axbz  az bx 
y - component

x
 
a  b  ax
bx

y
ay
by

z
az
bz
5th: Cross out the last column and first row
6th : Cross multiply and subtract four elements
a x by  a y bx
z-component
So then the new vector will be:
  



c  a  b  a y bz  az by  x  axbz  az bx  y  axby  a y bx  z
We’ll look more at the scalar product when we talk about angular
momentum.
Example:

x
 
a b  2



a  2x  3 y



b  4 x  2 y
 
y z



3 0  2  2  3  4 x  2  0   0  4 y  2  2  3  4 z
4 2 0



 4   12x  0y  4   12z


 16 x  16 z
Example:

x


a b  2



a  2x  3 y



b  4 x  2 y
 
y z



3 0  3  0   0  2x  2  0   0  4 y  2  2  3  4 z
4 2 0



 0x  0y  4   12z

 16 z
Notice the resultant vector is in the z – direction!