Vectors & Scalars. Vectors & Scalars. Vectors are measurements which have both magnitude (size) and a directional component. EXAMPLES OF VECTOR VALUES: Displacement Velocity Acceleration Force Direction counts in all of these measurements. Scalars are measurements which have only
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Vectors are measurements which have both
magnitude (size) and a directional component.
EXAMPLES OF VECTOR VALUES:
Displacement
Velocity
Acceleration
Force
Direction counts in all of these measurements.
Scalars are measurements which have only
magnitude (size) and no directional component
EXAMPLES OF SCALAR VALUES:
Distance
Speed
Temperature
A
B
Comparing Vector & Scalar ValuesDisplacement (a vector) versus distance (a scalar)
We want to get from point A to point B. If we follow the road
around the lake our direction is always changing. There is no specific
direction. The distance traveled on the road is a scalar quantity.
A straight line between A and B is the displacement. It has a specific
direction and is therefore a vector.
60
N
40
70
30
80
W
E
20
90
S
10
SPEEDOMETER
COMPASS
Speed & VelocitySpeed and velocity are not the same.
Velocity requires a directional component and is
therefore a vector quantity.
Speed tells us how fast we are going but not which way.
Speed is a scalar (direction doesn’t count!)
VELOCITY
+
Left = 
Up = +
Down = 
Right = +
Rectangular Coordinates
90 o North
y
Quadrant II
+
Quadrant I

+
0 o East
x
West 180 o
360 o
Quadrant III
Quadrant IV

270 o South
+y
0O East
West 180O
+x
 x
360O
 y
270O South
MEASURING THESAME DIRECTIONIN DIFFERENT WAYS
120O
240O
30O West of North
30O Left of +y
60O North of West
60O Above  x
RADIANS = ARC LENGTH / RADIUS LENGTH
CIRCUMFERENCE OF A CIRCLE = 2 x RADIUS
RADIANS IN A CIRCLE = 2 R / R
1 CIRCLE = 2 RADIANS = 360O
1 RADIAN = 360O / 2 = 57.3O
/2 radians
y
Quadrant II
+
Quadrant I

+
0 radians
radians
x
2 radians
Quadrant III
Quadrant IV

3/2 radians
+y
West 180O
+x
+10
 x
360O
10
 y
270O South
VECTOR NOTATIONS
315O or
(7/4) RADIANS
45O or
45O SOUTH OF EAST
0O East
10, 10 (X = +10, Y = 10)
14.1 METERS @ 315O,
14.1 METERS @ (7/4) RADIANS,
14.1 FEET AT 45O SOUTH OF EAST
GRAPHIC REPRESENTATION OF VECTORS
30 METERS @ 45O
30 METERS
@ 90O
50 METERS @ 0O
VECTOR ARROWS MAY BE DRAWN
ANYWHERE ON THE PAGE AS
LONG AS THE PROPER LENGTH AND
DIRECTION ARE MAINTAINED
= 10 METERS
SCALE
GRAPHIC ADDITION
BE DRAWN TO
SCALE & POINTED IN
THE PROPER
DIRECTION
Adding Vectors
A
D
R
B
C
B
C
A
D
+ + + =
A
B
C
D
R
30 METERS @ 45O
C
Vector B
50 METERS @ 0O
B
Vector C
30 METERS
@ 90O
A
= 10 METERS
SCALE
Drawing Vectors to Scale
To add the vectors
Place them head to tail
Angle is measured at 40o
Resultant = 9 x 10 = 90 meters
30 METERS @ 45O
Vector  A
30 METERS @ 225O
WORKING WITH VECTORS GRAPHIC SUBTRACTION
+

+
0 radians
radians
x
2 radians

3/2 radians
Trig Function Signs
Sin Cos Tan
Quadrant I
+ + +
Quadrant II
+ 
/2 radians
90 o
  +
Quadrant III
 + 
Quadrant IV
0 o
180 o
360 o
270 o
Ax
A
= SIN
Ay
A
VECTOR COMPONENTS
30 METERS @ 45O
Vector B
50 METERS @ 0O
= SIN
Vy
V
Vector C
30 METERS
@ 90O
= COS
Vx
V
ADDING & SUBTRACTING VECTORS USING COMPONENTSADD THE FOLLOWING
THREE VECTORS USING
COMPONENTS
ADDING & SUBTRACTING VECTORS USING COMPONENTS
(2) ADD THE X COMPONENTS OF EACH VECTOR
ADD THE Y COMPONENTS OF EACH VECTOR
X =SUM OF THE Xs = 21.2 + 50 + 0 = +71.2
Y =SUM OF THE Ys = 21.2 + 0 + 30 = +51.2
(3) CONSTUCT A NEW RIGHT TRIANGLE USING THE
X AS THE BASE AND Y AS THE OPPOSITE SIDE
Y = +51.2
X = +71.2
THE HYPOTENUSE IS THE RESULTANT VECTOR
X = +71.2
(+71.2)2 + (+51.2)2 = 87.7
(4) USE THE PYTHAGOREAN THEOREM TO THE LENGTH
(MAGNITUDE) OF THE RESULTANT VECTOR
ANGLE
TAN1 (51.2/71.2)
ANGLE = 35.7 O
QUADRANT I
(5) FIND THE ANGLE (DIRECTION) USING INVERSE
TANGENT OF THE OPPOSITE SIDE OVER THE
ADJACENT SIDE
RESULTANT = 87.7 METERS @ 35.7 O
A
B
C
R
Vector A
30 METERS @ 45O
Vector B
50 METERS @ 0O
+ ( ) + =
A
B
C
R
Vector C
30 METERS
@ 90O
SUBTRACTING VECTORS USING COMPONENTS
Vector A
30 METERS @ 45O
 Vector B
50 METERS @ 180O
Vector C
30 METERS @ 90O
(2) ADD THE X COMPONENTS OF EACH VECTOR
ADD THE Y COMPONENTS OF EACH VECTOR
X = SUM OF THE Xs = 21.2 + (50) + 0 = 28.8
Y =SUM OF THE Ys = 21.2 + 0 + 30 = +51.2
(3) CONSTUCT A NEW RIGHT TRIANGLE USING THE
X AS THE BASE AND Y AS THE OPPOSITE SIDE
Y = +51.2
X = 28.8
THE HYPOTENUSE IS THE RESULTANT VECTOR
X = 28.8
(28.8)2 + (+51.2)2 = 58.7
(4) USE THE PYTHAGOREAN THEOREM TO THE LENGTH
(MAGNITUDE) OF THE RESULTANT VECTOR
ANGLE
TAN1 (51.2/28.8)
ANGLE = 60.6 0
(1800 –60.60 ) = 119.40
QUADRANT II
(5) FIND THE ANGLE (DIRECTION) USING INVERSE
TANGENT OF THE OPPOSITE SIDE OVER THE
ADJACENT SIDE
RESULTANT = 58.7 METERS @ 119.4O