Practice Problems
Scalar Dot Products of Two Vectors
Directions:
On this worksheet we will be investigating the properties of the dot products of two vectors. There are two principle ways to calculate the
scalar dot product
,
A
B
, of two vectors. As the name implies, it is important to notice that the dot product of two vectors does NOT produce a new vector; instead it results in a scalar - that is, a value that only has magnitude or size, not direction. These methods are:
A
B
= |A| |B| cos
q
|A| and |B| represent the magnitudes of vectors
A
and
B
, while
q
is the
size
of the angle between them when placed tail-to-tail
A
B
= A
_{x}
B
_{x}
+ A
_{y}
B
_{y}
A
_{x}
and B
_{x}
represent the horizontal components of vectors
A
and
B
, while A
_{y}
and B
_{y}
represent their vertical components
The example vectors displayed in the table below are not drawn to scale; however, they do indicate correct relative directions.
A
B
C
D
omit
Question 1
Would the dot product
A
C
be positive, negative, or zero?
positive
negative
zero
omit
Question 2
Would the dot product
A
B
be positive, negative, or zero?
positive
negative
zero
omit
Question 3
Would the dot product
A
D
be positive, negative, or zero?
zero
positive
negative
omit
Question 4
Given the vectors:
C
= (6 newtons, 52º) and
B
= (14 meters, 90º)
What is the dot product of W = 9
C
2
B
?
[NOTE: work is defined as the dot product of a force vector,
F
, with its displacement vector,
s
.]
1191 Joules
1512 Joules
1181 Joules
931 Joules
omit
Question 5
Given the vectors:
D
= (12 meters, 142º) and
A
= (15 meters, 0º)
What is the value of G = 6
D
9
A
-7659 m
^{2}
-2380 m
^{2}
6036 m
^{2}
-7615 m
^{2}
omit
Question 6
Another application for the scalar dot product is
B
A
which determines the number of
magnetic flux lines
, ϕ, measured in webers, that pass through a given cross-sectional area. Suppose the magnetic field is given by the vector
B
= (15 Teslas, 0º) and the cross-sectional area vector is given by
A
= (28 m
^{2}
, 180º).
What is the magnitude of the flux (or field lines) passing through the specified area?
[NOTE: a positive sign means that the flux lines are exiting surface,
A
, while a negative sign means that the flux lines are entering surface
A
.]
0 webers
-13 webers
43 webers
-420 webers
omit
Question 7
As stated earlier, the work done on an object by a constant force is defined by the formula W =
F
s
. In this formula,
F
is an applied force which does NOT change in either magnitude nor in direction, and
s
is the length of the path along which this force is exerted. The work-kinetic energy theorem states that the net work done by one or more forces acting on an object as it moves between two positions is equal to the change in the object's KE.
How much would work would be done on a 6-kg mass if
F
and
s
are defined as:
F
= 9
A
and
s
= 2
B
where
A
= (15 newtons, 0º) and
B
= (14 meters, 90º)?
-3780 Joules
163 Joules
0 Joules
3780 Joules
omit
Question 8
How much would the kinetic energy of a 6-kg mass change if
F
and
s
are defined as:
F
= 9
C
and
s
= 6
E
where
C
= (6 newtons, 52º) and
E
= (14 newtons, 232º) ?
[NOTE: when work is positive the object's KE is increasing; while negative work means that the object's KE is decreasing.]
0 Joules
4536 Joules
-3574 Joules
-2793 Joules
omit
Question 9
How much would the kinetic energy of a 6-kg mass change if
F
and
s
are defined as:
F
= 9
C
and
s
= -6
E
?
2793 Joules
this value cannot be determined
3574 Joules
0 Joules
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