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Vectors and the Geometry of Space
The Dot Product of Two Vectors
Written by Richard Gill
Associate Professor of Mathematics
Tidewater Community College, Norfolk Campus, Norfolk, VA
With Assistance from a VCCS LearningWare Grant
In our first two lessons on vectors, you have studied:
• Properties of vectors;
• Notation associated with vectors;
• Vector Addition;
• Multiplication by a Scalar.
In this lesson you will study the dot product of two vectors. The dot product of two vectors generates a scalar as described below.
The Definition of Dot ProductThe dot product of two vectors
.
,,
2211
2121
vuvuvu
isvvvanduuu
The dot product of two vectors
.
,,,,
332211
321321
vuvuvuvu
isvvvvanduuuu
Properties of the Dot Product
2vvv 5.
0v0 4.
vcuvuc)vuc( 3.
wuvu)wv(u :holdsproperty vedistributi The 2.
.uvvu :holdsproperty ecommutativ The 1.
scalar. a be clet and
spacein or plane in the vectorsbe w and,v,uLet
The proof of property 5.
2
23
22
21
22
32
22
1
2
23
22
21321321
321
Therefore
,,v,,v
.,,vv Suppose
vvv
vvvvvvv
vvvvvvvvv
vv
Proofs of other properties are similar.
Consider two non-zero vectors. We can use their dot product and their magnitudes to calculate the angle between the two vectors. We begin with the sketch.
u
v
uv
From the Law of Cosines where c is the side opposite the angle theta:
cos2
cos2222
222
vuvuuv
abbac
22
2
2
)()(
)()(
uvuv
uuvuuvvv
uvuuvv
uvuvuv
: thatfollowsIt .www that have we
productdot for developedjust that weproperties theFrom2
cos2222
vuvuuv
From the previous slide
Substituting from above
vu
vu
vuvu
vuvu
cos
cos
cos22
cos222222
vuvuuvuv
Simplifying
You have just witnessed the proof of the following theorem:
.vu
vucos then v and u
vectorsnonzero obetween tw angle theis If
Example 1
Find the angle between the vectors:
.14,v and 5,3
u
Solution
106.9or 87.1
......2911.0cos
1734
7
116259
512
1,45,3
1,45,3cos
vu
vu
Example 2
.6-3,-3,z and 2-1,3,-w
:if z and w ctorsbetween ve angle theFind
Solution
2
03699419
1239
6,3,32,1,3
6,3,32,1,3cos
vu
vu
True or False? Whenever two non-zero vectors are perpendicular, their dot product is 0.
Think before you click.Congratulate yourself if you chose True!
002
cos and 02
cos
vuvu
vu
vu
vu
True or False? Whenever two non-zero vectors are perpendicular, their dot product is 0.
This is true. Since the two vectors are perpendicular,
the angle between them will be .
2or 90
True or False? Whenever you find the angle between
two non-zero vectors the formula
will generate angles in the interval
vu
vu
cos
.2
0
True or False? Whenever you find the angle between
two non-zero vectors the formula
will generate angles in the interval
vu
vu
cos
.2
0
This is False. For example, consider the vectors:
.2,1-v and 1,4
u
vu
vucos
Finish this on your own then click for the answer.
x
y
True or False? Whenever you find the angle between
two non-zero vectors the formula
will generate angles in the interval
vu
vu
cos
.2
0
This is False. For example, consider the vectors:
.2,1-v and 1,4
u
4.139or 43.2
...7592.085
7
517
18
1,21,4
1,21,4
vu
vucos
When the cosine is negative the angle between the two vectors is obtuse.
An Application of the Dot Product: Projection
The Tractor Problem: Consider the familiar example of a heavy box being dragged across the floor by a rope. If the box weighs 250 pounds and the angle between the rope and the horizontal is 25 degrees, how much force does the tractor have to exert to move the box?
Discussion: the force being exerted by the tractor can be interpreted as a vector with direction of 25 degrees. Our job is to find the magnitude.
25
v
The Tractor Problem, Slide 2: we are going to look at the force vector as the sum of its vertical component vector and its horizontal component. The work of moving the box across the floor is done by the horizontal component.
x
v
1v
2v
y
25
lbsv
v
vv
v
v
8.275......9063.0
250
25cos
25cos
25cos
1
1
1
Hint: for maximum accuracy, don’t round off until the end of the problem. In this case, we left the value of the cosine in the calculator and did not round off until the end.
Conclusion: the tractor has to exert a force of 275.8 lbs before the box will move.
The problem gets more complicated when the direction in which the object moves is not horizontal or vertical.
u
v
1w
2w
x
y
.w toorthogonal is and
v toparallel is w that so
wu sketch, In the
12
1
21
w
w
.wcomponent the
by done being is work the
of all then v ofdirection
in theorigin at the
object an moving is uby
drepresente force a If
1
.v toorthogonal u
ofcomponent vector thecalled is
.proj as denoted is and v
onto u of projection thecalled is
2
v1
1
w
uw
w
A Vocabulary Tip:
When two vectors are perpendicular we say that they are orthogonal.
When a vector is perpendicular to a line or a plane we say that the vector is normal to the line or plane.
u
v
1w
2w
x
y
The following theorem will prove very useful in the remainder of this course.
vv
vuuprojv
2
:follows as calculated
becan v onto u of projection the
then vectors,nonzero are v and u If
.v vector of multiple a is v
onto u of projection thesoscalar a is
sparenthesi theinsidequantity The
.v toorthogonal u ofcomponent
vector thefind and uproj find 6,3 is v and 5,9 is u if :Example v
1
2
22
8.3,6.7
3,645
573,6
936
2730
3,63,6
3,69,5
w
vv
vuuprojv
2.5,6.28.3,6.79,5
9,58.3,6.7
w
that notice sketch, theFrom
2
2
21
w
w
uw
(5,9)
(6,3)
Work is traditionally defined as follows: W = FD where F is the constant force acting along the line of motion and D is the distance traveled along the line of motion.
Example: an object is pulled 12 feet across the floor using a force of 100 pounds. Find the work done if the force is applied at an angle of 50 degrees above the horizontal.
50
100 lb
12 ft
Solution A: using W = FD we use the projection of F in the x direction.
pounds-foot 35.771)12)(50cos100( FDW
pounds.-foot 35.771)12(50cos10012,050sin100,100cos50
thatalso Note .12,0 isvector direction theand
50sin100,100cos50 is form coordinatein vector force that thenote :BSolution
Two Ways to Calculate Work
v
vprojv
u W:B Method
u W:A Method
:B Methodby or A Methodby calculated becan v vector along
n applicatio ofpoint its moving u forceconstant aby doneWork W
Example: Find the work done by a force of 20 lb acting in the direction N50W while moving an object 4 ft due west.
lbsft 28.61W
)4)(40)(cos20(proj W:SolutionA v
ftlbvu
lbsft 28.61)0(140sin20)4(140cos20
0,4 and 140sin20,140cos20u :SolutionB
vuW
v
u
v
50
4v and 20
u
axes.-z and y-, x-,positive with themakes v that ][0, interval in the
and , , angles theare v vector nonzero a of anglesdirection The
v.v vector of cosinesdirection thecalled are
,cos and ,cos ,cos angles,direction theseof cosines The
We can calculate the direction cosines by using the unit vectors along each positive axis and the dot product.
i j
k
The last topic of this lesson concerns Direction Cosines.
v
v
v
vvv
kv
kv
v
v
v
vvv
jv
jv
v
v
v
vvv
iv
iv
3321
2321
1321
1
1,0,0,,cos
1
0,1,0,,cos
1
0,0,1,,cos
For comments on this presentation you may email the author Professor Richard Gill [email protected] or the publisher of the VML Dr. Julia Arnold [email protected]