Download 11.6 Dot Product and the Angle between Two Vectors

1454
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
y
v
w
1
2
That is, the vectors v and w are on the lines with slopes
satisfy
b d
! D "1
a c
b
a
and
)
x
d
c , respectively.
bd D "ac
)
The lines are perpendicular if and only if their slopes
ac C bd D 0
We now consider the case where one of the vectors, say v, is perpendicular to the x-axis. In this case a D 0, and the vectors are
perpendicular if and only if w is parallel to the x-axis, that is, d D 0. So ac C bd D 0 ! c C b ! 0 D 0.
11.6 Dot Product and the Angle between Two Vectors
Preliminary Questions
1. Is the dot product of two vectors a scalar or a vector?
SOLUTION
The dot product of two vectors is the sum of products of scalars, hence it is a scalar.
2. What can you say about the angle between a and b if a ! b < 0?
a!b
Since the cosine of the angle between a and b satisfies cos ! D kakkbk
, also cos ! < 0. By definition 0 # ! # ", but
since cos ! < 0 then ! is in ."=2; "#. In other words, the angle between a and b is obtuse.
SOLUTION
3. Which property of dot products allows us to conclude that if v is orthogonal to both u and w, then v is orthogonal to u C w?
SOLUTION One property is that two vectors are orthogonal if and only if the dot product of the two vectors is zero. The second
property is the Distributive Law. Since v is orthogonal to u and w, we have v ! u D 0 and v ! w D 0. Therefore,
v ! .u C w/ D v ! u C v ! w D 0 C 0 D 0
We conclude that v is orthogonal to u C w.
4. Which is the projection of v along v: (a) v or (b) ev ?
SOLUTION
The projection of v along itself is v, since
vjj D
!v ! v"
v!v
vDv
Also, the projection of v along ev is the same answer, v, because
#
$
v ! ev
vjj D
ev D kvkev D v
ev ! ev
5. Let ujj be the projection of u along v. Which of the following is the projection u along the vector 2v and which is the projection
of 2u along v?
(a) 12 ujj
(b) ujj
(c) 2ujj
SOLUTION
Since ujj is the projection of u along v, we have,
ujj D
!u ! v"
v!v
v
The projection of u along the vector 2v is
#
$
#
$
#
$
!u ! v"
u ! 2v
2u ! v
4u ! v
2v D
2v D
vD
v D ujj
2v ! 2v
4v ! v
4v ! v
v!v
That is, ujj is the projection of u along 2v, so our answer is (b) for the first part. Notice that the projection of u along v is the
projection of u along the unit vector ev , hence it depends on the direction of v rather than on the length of v. Therefore, the
projection of u along v and along 2v is the same vector.
On the other hand, the projection of 2u along v is as follows:
#
$
!u ! v"
2u ! v
vD2
v D 2ujj
v!v
v!v
giving us answer (c) for the second part.
S E C T I O N 11.6
Dot Product and the Angle between Two Vectors
6. Which of the following is equal to cos !, where ! is the angle between u and v?
(a) u ! v
(b) u ! ev
SOLUTION
1455
(c) eu ! ev
By the Theorems on the Dot Product and the Angle Between Vectors, we have
cos ! D
u!v
u
v
D
!
D eu ! ev
kukkvk
kuk kvk
The correct answer is (c).
Exercises
In Exercises 1–4, compute the dot product.
1. h3; 1i ! h4; "7i
SOLUTION
2.
˝1
1
6; 2
The dot product of the two vectors is the following scalar:
h3; 1i ! h4; "7i D 3 ! 4 C 1 ! ."7/ D 5
˛ ˝ 1˛
! 3; 2
SOLUTION
The dot product is
%
3. i ! j
SOLUTION
& %
&
1 1
1
1
1 1
1
1
3
;
! 3;
D !3C ! D C D
6 2
2
6
2 2
2
4
4
By the orthogonality of i and j, we have i ! j D 0
4. j ! j
SOLUTION
Since j has length 1, we have j ! j D 1
In Exercises 5–6, determine whether the two vectors are orthogonal and, if not, whether the angle between them is acute or obtuse.
˝
˛ ˝1 7˛
4
5. 12
5 ; "5 ,
2;"4
SOLUTION
We find the dot product of the two vectors:
%
& %
&
# $ # $
12 4
1 7
12 1
4
7
12
28
13
;" ! ;" D
! C "
! "
D
C
D
5
5
2 4
5 2
5
4
10
20
5
The dot product is positive, hence the angle between the vectors is acute.
6. h12; 6i,
SOLUTION
h2; "4i
Since h12; 6i ! h2; "4i D 12 ! 2 C 6 ! ."4/ D 0, the vectors are orthogonal.
In Exercises 7–8, find the angle between the vectors. Use a calculator if necessary.
p
p ˛
˝ p ˛ ˝
7. 2; 2 , 1 C 2; 1 " 2
D p E
D
p
p E
SOLUTION We write v D 2; 2 and w D 1 C 2; 1 " 2 . To use the formula for the cosine of the angle ! between two
vectors we need to compute the following values:
p
p
kvk D 4 C 2 D 6
q
p
p
p
kwk D .1 C 2/2 C .1 " 2/2 D 6
p
p
p
v!wD2C2 2C 2"2D3 2
Hence,
p
p
v!w
3 2
2
cos ! D
D p p D
kvkkwk
2
6 6
and so,
! D cos"1
p
2
D "=4
2
1456
C H A P T E R 11
˝ p ˛
8. 5; 3 ,
˝p
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
˛
3; 2
D p E
Dp E
We denote v D 5; 3 and w D
3; 2 . To use the formula for the cosine of the angle ! between two vectors we
need to compute the following values:
p
p
kvk D 25 C 3 D 28
p
p
kwk D 3 C 4 D 7
p
p
p
v!wD 5! 3C 3!2 D 7 3
SOLUTION
Hence,
cos ! D
p
p
v!w
7 3
3
D p p D
kvkkwk
2
28 7
and so,
"1
! D cos
p
3
D "=6
2
In Exercises 9–12, simplify the expression.
9. .v " w/ ! v C v ! w
SOLUTION
By properties of the dot product we obtain
.v " w/ ! v C v ! w D v ! v " w ! v C v ! w D kvk2 " v ! w C v ! w D kvk2
10. .v C w/ ! .v C w/ " 2v ! w
SOLUTION
Using properties of the dot product we obtain
.v C w/ ! .v C w/ " 2v ! w D v ! .v C w/ C w ! .v C w/ " 2v ! w D v ! v C v ! w C w ! v C w ! w " 2v ! w
D kvk2 C v ! w C v ! w C kwk2 " 2v ! w D kvk2 C kwk2
11. .v C w/ ! v " .v C w/ ! w
SOLUTION
We use properties of the dot product to write
.v C w/ ! v " .v C w/ ! w D v ! v C w ! v " v ! w " w ! w
D kvk2 C w ! v " w ! v " kwk2 D kvk2 " kwk2
12. .v C w/ ! v " .v " w/ ! w
SOLUTION
By properties of the dot product we get
.v C w/ ! v " .v " w/ ! w D .v C w/ ! v " w ! .v " w/
Dv!vCw!v"w!vCw!w
D v ! v C w ! w D kvk2 C kwk2
In Exercises 13–16, use the properties of the dot product to evaluate the expression, assuming that u ! v D 2, kuk D 1, and kvk D 3.
13. u ! .4v/
SOLUTION
Using properties of the dot product we get
u ! .4v/ D 4.u ! v/ D 4 ! 2 D 8:
14. .u C v/ ! v
SOLUTION
Using the distributive law and the dot product relation with length we obtain
.u C v/ ! v D u ! v C v ! v D u ! v C kvk2 D 2 C 32 D 11:
15. 2u ! .3u " v/
SOLUTION
By properties of the dot product we obtain
2u ! .3u " v/ D .2u/ ! .3u/ " .2u/ ! v D 6.u ! u/ " 2.u ! v/
D 6kuk2 " 2.u ! v/ D 6 ! 12 " 2 ! 2 D 2
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
16. .u C v/ ! .u " v/
SOLUTION
We use the distributive law, commutativity and the relation with length to write
.u C v/ ! .u " v/ D u ! .u " v/ C v ! .u " v/ D kuk2 " u ! v C u ! v " kvk2
D kuk2 " kvk2 D 12 " 32 D "8
17. Find the angle between v and w if v ! w D "kvk kwk.
SOLUTION
Using the formula for dot product, and the given equation v ! w D "kvk kwk, we get:
kvk kwk cos ! D "kvk kwk;
which implies cos ! D "1, and so the angle between the two vectors is ! D ".
18. Find the angle between v and w if v ! w D 12 kvk kwk.
SOLUTION
Using the formula for dot product, and the given equation v ! w D 12 kvk kwk, we get:
kvk kwk cos ! D
which implies cos ! D
1
2,
1
kvk kwk;
2
and so the angle between the two vectors is ! D "=3.
!
3.
32 C 52
19. Assume that kvk D 3, kwk D 5 and that the angle between v and w is ! D
2
(a) Use the relation kv C wk D .v C w/ ! .v C w/ to show that kv C
(b) Find kv C wk.
SOLUTION
wk2
D
C 2v ! w.
For part (a), we use the distributive property to get:
kv C wk2 D .v C w/ ! .v C w/
D v!vCv!wCw!vCw!w
D kvk2 C 2v ! w C kwk2
D 32 C 52 C 2v ! w
For part (b), we use the definition of dot product on the previous equation to get:
kv C wk2 D 32 C 52 C 2v ! w
D 34 C 2 ! 3 ! 5 ! cos "=3
Thus, kv C wk D
p
D 34 C 15 D 49
49 D 7:
20. Assume that kvk D 2, kwk D 3, and the angle between v and w is 120ı . Determine:
(a) v ! w
(b) k2v C wk
(c) k2v " 3wk
SOLUTION
(a) We use the relation between the dot product and the angle between two vectors to write
# $
1
v ! w D kvkkwk cos ! D 2 ! 3 cos 120ı D 6 ! "
D "3
2
(b) By the relation of the dot product with length and by properties of the dot product we have
k2v C wk2 D .2v C w/ ! .2v C w/ D 4v ! v C 2v ! w C 2w ! v C w ! w
D 4kvk2 C 4v ! w C kwk2
We now substitute v ! w D "3 from part (a) and the given information, obtaining
k2v C wk2 D 4 ! 22 C 4."3/ C 32 D 13
)
k2v C wk D
p
13 $ 3:61
(c) We express the length in terms of a dot product and use properties of the dot product. This gives
k2v " 3wk2 D .2v " 3w/ ! .2v " 3w/ D 4v ! v " 6v ! w " 6w ! v C 9w ! w
D 4kvk2 " 12v ! w C 9kwk2
Substituting v ! w D "3 from part (a) and the given values yields
k2v " 3wk2 D 4 ! 22 " 12."3/ C 9 ! 32 D 133
)
k2v " 3wk D
p
133 $ 11:53
1457
1458
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
21. Show that if e and f are unit vectors such that ke C fk D 32 , then ke " fk D
SOLUTION
p
7
2 .
Hint: Show that e ! f D 18 .
We use the relation of the dot product with length and properties of the dot product to write
9=4 D ke C fk2 D .e C f/ ! .e C f/ D e ! e C e ! f C f ! e C f ! f
D kek2 C 2e ! f C kfk2 D 12 C 2e ! f C 12 D 2 C 2e ! f
We now find e ! f:
9=4 D 2 C 2e ! f
e ! f D 1=8
)
Hence, using the same method as above, we have:
ke " fk2 D .e " f/ ! .e " f/ D e ! e " e ! f " f ! e C f ! f
D kek2 " 2e ! f C kfk2 D 12 " 2e ! f C 12 D 2 " 2e ! f D 2 " 2=8 D 7=4:
Taking square roots, we get:
ke " fk D
p
7
2
22. Find k2e " 3fk assuming that e and f are unit vectors such that ke C fk D
SOLUTION
p
3=2.
We use the relation of the dot product with length and properties of the dot product to write
3=2 D ke C fk2 D .e C f/ ! .e C f/ D e ! e C e ! f C f ! e C f ! f
D kek2 C 2e ! f C kfk2 D 12 C 2e ! f C 12 D 2 C 2e ! f
We now find e ! f:
3=2 D 2 C 2e ! f
e ! f D "1=4
)
Hence, using the same method as above, we have:
k2e " 3fk2 D .2e " 3f/ ! .2e " 3f/
D k2ek2 " 2 ! 2e ! 3f C k3fk2 D 22 " 12e ! f C 32 D 13 C 3 D 16:
Taking square roots, we get:
k2e " 3fk D 4
23. Find the angle ! in the triangle in Figure 1.
y
(0, 10)
(10, 8)
(3, 2)
x
FIGURE 1
SOLUTION
We denote by u and v the vectors in the figure.
y
(0, 10)
v
(10, 8)
u
(3, 2)
x
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1459
Hence,
cos ! D
v!u
kvkkuk
(1)
We find the vectors v and u, and then compute their length and the dot product v ! u. This gives
v D h0 " 10; 10 " 8i D h"10; 2i
u D h3 " 10; 2 " 8i D h"7; "6i
q
p
kvk D ."10/2 C22 D 104
q
p
kuk D ."7/2 C ."6/2 D 85
v ! u D h"10; 2i ! h"7; "6i D ."10/ ! ."7/ C 2 ! ."6/ D 58
Substituting these values in (1) yields
cos ! D p
58
p $ 0:617
104 85
Hence the angle of the triangle is 51:91ı .
24. Find all three angles in the triangle in Figure 2.
y
(2, 7)
(6, 3)
x
(0, 0)
FIGURE 2
SOLUTION
We denote by u, v and w the vectors and by !1 , !2 , and !3 the angles shown in the figure. We compute the vectors:
u D h2; 7i
v D h6; 3i
w D h6 " 2; 3 " 7i D h4; "4i
y
(2, 7)
3
w
u
2
1
(6, 3)
v
x
(0, 0)
Since the angles are acute the cosines are positive, so we have
cos !1 D
ju ! vj
;
kukkvk
cos !2 D
jv ! wj
;
kvkkwk
cos !3 D 180 " .!1 C !2 /
We compute the lengths and the dot products in (1):
u ! v D h2; 7i ! h6; 3i D 2 ! 6 C 7 ! 3 D 33
v ! w D h6; 3i ! h4; "4i D 6 ! 4 C 3 ! ."4/ D 12
(1)
1460
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
p
p
22 C 72 D 53
p
p
kvk D 62 C 32 D 45
q
p
kwk D 42 C ."4/2 D 32
kuk D
Substituting in (1) and solving for acute angles yields
33
cos !1 D p p $ 0:676
53 45
)
!1 $ 47:47ı
12
cos !2 D p p $ 0:316
45 32
)
!2 $ 71:58ı
The sum of the angles in a triangle is 180ı , hence
!3 D 180ı " .47:47 C 71:58/ $ 60:95ı :
In Exercises 25–26, find the projection of u along v.
25. u D h2; 5i,
SOLUTION
v D h1; 1i
We first compute the following dot products:
u ! v D h2; 5i ! h1; 1i D 7
v ! v D kvk2 D 12 C 12 D 2
The projection of u along v is the following vector:
ujj D
26. u D h2; "3i,
SOLUTION
v D h1; 2i
!u ! v"
v!v
vD
%
&
7
7 7
vD ;
2
2 2
We first compute the following dot products:
u ! v D h2; "3i ! h1; 2i D "4
v ! v D kvk2 D 12 C 22 D 5
The projection of u along v is the following vector:
ujj D
27. Find the length of OP in Figure 3.
!u ! v"
v!v
vD
%
&
"4
"4 "8
vD
;
5
5 5
y
u = ⟨3, 5⟩
u⊥
v = ⟨8, 2⟩
P
O
x
FIGURE 3
SOLUTION
This is just the component of u D h3; 5i along v D h8; 2i. We first compute the following dot products:
u ! v D h3; 5i ! h8; 2i D 34
v ! v D kvk2 D 82 C 22 D 68
The component of u along v is the length of the projection of u along v
'! u ! v " '
34 p
34
'
'
v' D
kvk D
68
'
v!v
68
68
28. Find ku? k in Figure 3.
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1461
SOLUTION From the previous problem (see solution above) we know that the component of u along v is 1=2, and thus the
projection is uk D h4; 1i. Using the standard formula for u? , we obtain
u? D u " uk D h3; 5i " h4; 1i D h"1; 4i
In Exercises 29–30, find the decomposition a D ajj C a? with respect to b.
29. a D h1; 0i,
b D h1; 1i
SOLUTION
Step 1. We compute a ! b and b ! b
a ! b D h1; 0i ! h1; 1i D 1 ! 1 C 0 ! 1 D 1
b ! b D kbk2 D 12 C 12 D 2
Step 2. We find the projection of a along b:
ajj D
#
$
%
&
a!b
1
1 1
b D h1; 1i D ;
b!b
2
2 2
Step 3. We find the orthogonal part as the difference:
a? D a " ajj D h1; 0i "
%
& %
&
1 1
1 1
;
D ;"
2 2
2 2
Hence,
a D ajj C a? D
30. a D h2; "3i,
SOLUTION
%
& %
&
1 1
1 1
;
C ;" :
2 2
2 2
b D h5; 0i
We first compute a ! b and b ! b to find the projection of a along b:
a ! b D h2; "3i ! h5; 0i D 10
b ! b D kbk2 D 52 C 02 D 25
Hence,
ajj D
#
$
a!b
10
bD
h5; 0i D h2; 0i
b!b
25
We now find the vector a? orthogonal to b by computing the difference:
a " ajj D h2; "3i " h2; 0i D h0; "3i
Thus, we have
31. Let e" D hcos !; sin !i. Show that e" ! e
SOLUTION
a D ajj C a? D h2; 0i C h0; "3i
D cos.! "
/ for any two angles ! and
.
First, e" is a unit vector since by a trigonometric identity we have
ke" k D
p
p
cos2 ! C sin2 ! D 1 D 1
The cosine of the angle ˛ between e" and the vector i in the direction of the positive x-axis is
cos ˛ D
e" ! i
D e" ! i D ..cos !/i C .sin !/j/ ! i D cos !
ke" k ! kik
The solution of cos ˛ D cos ! for angles between 0 and " is ˛ D !. That is, the vector e" makes an angle ! with the x-axis. We
now use the trigonometric identity
cos ! cos
C sin ! sin
D cos.! "
/
to obtain the following equality:
e" ! e
D hcos !; sin !i ! hcos ; sin i D cos ! cos
C sin ! sin
D cos.! "
/
1462
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
32.
Let v and w be vectors in the plane.
(a) Use Theorem 2 to explain why the dot product v ! w does not change if both v and w are rotated by the same angle !.
Dp p E
(b) Sketch the vectors e1 D h1; 0i and e2 D 22 ; 22 , and determine the vectors e01 ; e02 obtained by rotating e1 ; e2 through an
angle
!
4.
Verify that e1 ! e2 D e01 ! e02 .
SOLUTION
(a) By Theorem 2,
v ! w D kvkkwk cos ˛
where ˛ is the angle between v and w. Since rotation by an angle ! does not change the angle between the vectors, nor the norms
of the vectors, the dot product v ! w remains unchanged.
y
e2 =
〈
2
2
,
2
2
〉
x
e1 = 〈1, 0〉
(b) Notice from the picture that if weD p
rotate
e E by "=4, we get e2 , and when we rotate e2p by the same
amount
we get a unit
p 1
p
p
2
2
2
2
2
0
0
vector along the y axis. Thus, e1 D 2 ; 2 and e2 D h0; 1i. Note that e1 ! e2 D 1 ! 2 C 0 ! 2 D 2 and e01 ! e02 D
0!
p
2
2
C1!
p
2
2
D
p
2
2 .
Thus, e1 ! e2 D e01 ! e02 .
33.
Let v and w be nonzero vectors and set u D ev C ew . Use the dot product to show that the angle between u and v is
equal to the angle between u and w. Explain this result geometrically with a diagram.
SOLUTION We denote by ˛ the angle between u and v and by ˇ the angle between u and w. Since ev and ew are vectors in the
directions of v and w respectively, ˛ is the angle between u and ev and ˇ is the angle between u and ew . The cosines of these angles
are thus
u ! ev
u ! ev
u ! ew
u ! ew
cos ˛ D
D
I cos ˇ D
D
kukkev k
kuk
kukkew k
kuk
To show that cos ˛ D cos ˇ (which implies that ˛ D ˇ) we must show that
u ! ev D u ! ew :
We compute the two dot products:
u ! ev D .ev C ew / ! ev D ev ! ev C ew ! ev D 1 C ew ! ev
u ! ew D .ev C ew / ! ew D ev ! ew C ew ! ew D ev ! ew C 1
We see that u ! ev D u ! ew . We conclude that cos ˛ D cos ˇ, hence ˛ D ˇ. Geometrically, u is a diagonal in the rhombus OABC
(see figure), hence it bisects the angle ^AOC of the rhombus.
v
A
B
ev
O
34.
ample.
SOLUTION
u
ew
C
w
Let v, w, and a be nonzero vectors such that v ! a D w ! a. Is it true that v D w? Either prove this or give a counterexThe equality v ! a D w ! a is equivalent to the following equality:
v!aD w!a
v!a"w!aD 0
.v " w/ ! a D 0
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1463
That is, v " w is orthogonal to a rather than v D w. Consider the following counterexample:
a D h1; 1iI
v D h1; 0iI
w D h0; 1i
Obviously, v ¤ w, but v ! a D w ! a since
v ! a D h1; 0i ! h1; 1i D 1 ! 1 C 1 ! 0 D 1
w ! a D h0; 1i ! h1; 1i D 0 ! 1 C 1 ! 1 D 1
35. Calculate the force (in newtons) required to push a 40-kg wagon up a 10ı incline (Figure 4).
40 kg
10°
FIGURE 4
SOLUTION Gravity exerts a force Fg of magnitude 40g newtons where g D 9:8. The magnitude of the force required to push
the wagon equals the component of the force Fg along the ramp. Resolving Fg into a sum Fg D Fjj C F? , where Fjj is the force
along the ramp and F? is the force orthogonal to the ramp, we need to find the magnitude of Fjj . The angle between Fg and the
ramp is 90ı " 10ı D 80ı . Hence,
Fjj D kFg k cos 80ı D 40 ! 9:8 ! cos 80ı $ 68:07 N:
F||
80°
10°
Fg
F
Therefore the minimum force required to push the wagon is 68.07 N. (Actually, this is the force required to keep the wagon from
sliding down the hill; any slight amount greater than this force will serve to push it up the hill.)
36. A force F is applied to each of two ropes (of negligible weight) attached to opposite ends of a 40-kg wagon and making an
angle of 35ı with the horizontal (Figure 5). What is the maximum magnitude of F (in newtons) that can be applied without lifting
the wagon off the ground?
F
F
35°
35°
40 kg
FIGURE 5
SOLUTION With two ropes at either end, both at the same angle with the horizontal and both with the same force, pulling on the
40-kg wagon, each rope will need to lift 20 kg. Let’s look at the situation on the right-hand side of the wagon. We resolve the force
F on the right-hand rope into a sum F D Fjj C F? where Fjj is the horizontal force and F? is the force orthogonal to the ground.
The wagon will not be lifted off the ground if the magnitude of F? , that is the component of F along the direction orthogonal to the
ground, is equal to (but not more than) the magnitude of the force due to gravity from 20 kg (remember, each rope needs to only
lift half of the wagon, and remember also that the acceleration due to gravity is 9.8 meters per second squared). That is,
(1)
20.9:8/ D kF? k
The angle between F and a vector orthogonal to the ground is
90ı
" 35ı
D
55ı ,
hence, 20.9:8/ D 196 D
F
F⊥
55°
35°
F||
Fg
This gives us
196 D kFk cos 55ı
)
kFk D
196
$ 341 Newtons
cos 55ı
The maximum force that can be applied is of magnitude 341 newtons on each rope.
kFk cos 55ı
1464
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
37. A light beam travels along the ray determined by a unit vector L, strikes a flat surface at point P , and is reflected along the ray
determined by a unit vector R, where !1 D !2 (Figure 6). Show that if N is the unit vector orthogonal to the surface, then
R D 2.L ! N/N " L
Incoming light
Reflected light
N
L
R
1
2
P
FIGURE 6
SOLUTION
We denote by W a unit vector orthogonal to N in the direction shown in the figure, and let !1 D !2 D !.
Incoming light
Reflected light
N
L
R
W
We resolve the unit vectors R and L into a sum of forces along N and W. This gives
R D cos.90 " !/W C cos !N D sin !W C cos !N
L D " cos.90 " !/W C cos !N D " sin !W C cos !N
(1)
N
W
Now, since
L ! N D kLkkNk cos ! D 1 ! 1 cos ! D cos !
N
N
L
R
90 +
90 −
W
W
we have by (1):
2.L ! N/N " L D .2 cos !/N " L D .2 cos !/N " .." sin !/W C .cos !/N/
D .2 cos !/N C .sin !/W " .cos !/N D .sin !/W C .cos !/N D R
38. Verify the Distributive Law:
u ! .v C w/ D u ! v C u ! w
SOLUTION
We denote the components of the vectors u, v, and w by
u D ha1 ; a2 iI
v D hb1 ; b2 iI
w D hc1 ; c2 i
We compute the left-hand side:
u ! .v C w/ D ha1 ; a2 i .hb1 ; b2 i C hc1 ; c2 i/
D ha1 ; a2 i ! hb1 C c1 ; b2 C c2 i
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1465
D ha1 .b1 C c1 /; a2 .b2 C c2 /i
Using the distributive law for scalars and the definitions of vector sum and the dot product we get
u ! .v C w/ D ha1 b1 C a1 c1 ; a2 b2 C a2 c2 i
D ha1 b1 ; a2 b2 i C ha1 c1 ; a2 c2 i
D ha1 ; a2 i ! hb1 ; b2 i C ha1 ; a2 i ! hc1 ; c2 i
Du!vCu!w
39. Verify that .$v/ ! w D $.v ! w/ for any scalar $.
SOLUTION
We denote the components of the vectors v and w by
v D ha1 ; a2 i w D hb1 ; b2 i
Thus,
.$v/ ! w D .$ha1 ; a2 i/ ! hb1 ; b2 i D h$a1 ; $a2 i ! hb1 ; b2 i
D $a1 b1 C $a2 b2
Recalling that $, ai , and bi are scalars and using the definitions of scalar multiples of vectors and the dot product, we get
.$v/ ! w D $.a1 b1 C a2 b2 / D $ .ha1 ; a2 i ! hb1 ; b2 i/ D $.v ! w/
Further Insights and Challenges
40. Prove the Law of Cosines, c 2 D a2 C b 2 " 2ab cos !, by referring to Figure 7. Hint: Consider the right triangle 4PQR.
R
a
c
a sin
P
b
Q
b − a cos
FIGURE 7
SOLUTION
""! "! "!
We denote the vertices of the triangle by S, Q, and R. Since RQ D RS C SQ, we have
""! 2 ""! ""! !"! "!" !"! "!"
c 2 D kRQk D RQ ! RQ D RS C SQ ! RS C SQ
"! "! "! "! "! "! "! "!
D RS ! RS C RS ! SQ C SQ ! RS C SQ ! SQ
"! 2
"! "!
"! 2
D kRSk C 2RS ! SQ C kSQk
"! "!
c 2 D a2 C 2RS ! SQ C b 2
(1)
R
a
c
P
S
Q
b
"! "!
"!
"!
We find the dot product RS ! SQ. The angle between the vectors RS and SQ is !, hence,
"! "!
"!
"!
SR ! SQ D kSRk ! kSQk cos ! D ab cos !:
Therefore,
"! "!
"! "!
RS ! SQ D "SR ! SQ D "ab cos !
Substituting (2) in (1) yields
c 2 D a2 " 2ab cos ! C b 2 D a2 C b 2 " 2ab cos !:
(Note that we did not need to use the point P .)
(2)
1466
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
41. In this exercise, we prove the Cauchy–Schwarz inequality: If v and w are any two vectors, then
jv ! wj # kvk kwk
6
(a) Let f .x/ D kxv C wk2 for x a scalar. Show that f .x/ D ax 2 C bx C c, where a D kvk2 , b D 2v ! w, and c D kwk2 .
(b) Conclude that b 2 " 4ac # 0. Hint: Observe that f .x/ % 0 for all x.
SOLUTION
(a) We express the norm as a dot product and compute it:
f .x/ D kxv C wk2 D .xv C w/ ! .xv C w/
D x 2 v ! v C xv!w C xw ! v C w ! w D kvk2 x 2 C 2.v ! w/x C kwk2
Hence, f .x/ D ax 2 C bx C c, where a D kvk2 , b D 2v ! w, and c D kwk2 .
(b) If f has distinct real roots x1 and x2 , then f .x/ is negative for x between x1 and x2 , but this is impossible since f is the
square of a length.
y
x1
x2
x
f (x) = ax2 + bx + c, a > 0
Using properties of quadratic functions, it follows that f has a nonpositive discriminant. That is, b 2 " 4ac # 0. Substituting the
values for a, b, and c, we get
4.v ! w/2 " 4kvk2 kwk2 # 0
.v ! w/2 # kvk2 kwk2
Taking the square root of both sides we obtain
jv ! wj # kvkkwk
42. Use (6) to prove the Triangle Inequality
kv C wk # kvk C kwk
Hint: First use the Triangle Inequality for numbers to prove
j.v C w/ ! .v C w/j # j.v C w/ ! vj C j.v C w/ ! wj
SOLUTION
Using the relation between the length and dot product we have
kv C wk2 D .v C w/ ! .v C w/ D v ! v C v ! w C w ! v C w ! w
D kvk2 C 2v ! w C kwk2
(1)
Obviously, v ! w # jv ! wj. Also, by the Cauchy–Schwarz inequality jv ! wj # kvkkwk. Therefore, v ! w # kvkkwk, and combining
this with (1) we get
kvCwk2 D kvk2 C 2v ! w C kwk2 # kvk2 C 2kvkkwk C kwk2 D .kvk C kwk/2
That is,
kvCwk2 # .kvk C kwk/2
Taking the square roots of both sides and recalling that the length is nonnegative, we get
kvCwk # kvk C kwk
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1467
43. This exercise gives another proof of the relation between the dot product and the angle ! between two vectors v D ha1 ; b1 i
and w D ha2 ; b2 i in the plane. Observe that v D kvk hcos !1 ; sin !1 i and w D kwk hcos !2 ; sin !2 i, with !1 and !2 as in Figure 8.
Then use the addition formula for the cosine to show that
v ! w D kvk kwk cos !
y
y
a2
y
w
w
a1
b2
2
v
1
x
v
b1
x
=
2
−
x
1
FIGURE 8
SOLUTION
Using the trigonometric function for angles in right triangles, we have
a2 D kvk sin !1 ;
a1 D kvk cos !1
b2 D kwk sin !2 ;
b1 D kwk cos !2
Hence, using the given identity we obtain
v ! w D ha1 ; a2 i ! hb1 ; b2 i D a1 b1 C a2 b2 D kvk cos !1 kwk cos !2 C kvk sin !1 kwk sin !2
D kvkkwk.cos !1 cos !2 C sin !1 sin !2 / D kvkkwk cos.!1 " !2 /
That is,
v ! w D kvkkwk cos.!/
44. Let v D hx; yi and
v" D hx cos ! C y sin !; "x sin ! C y cos !i
Prove that the angle between v and v" is !.
SOLUTION The dot product of the vectors v and v" is
v ! v" D hx; yi ! hx cos ! C y sin !; "x sin ! C y cos !i
D x.x cos ! C y sin !/ C y."x sin ! C y cos !/
D x 2 cos ! C xy sin ! " xy sin ! C y 2 cos !
D .x 2 C y 2 / cos !
That is,
v ! v" D .x 2 C y 2 / cos !
(1)
On the other hand, if ˛ denotes the angle between v and v" , we have
v ! v" D kvkkv" k cos ˛
(2)
We compute the lengths. Using the identities cos2 ! C sin2 ! D 1 and 2 sin ! cos ! D sin 2!, we obtain
q
p
kvk D hx; yi D x 2 C y 2
q
kv" k D .x cos ! C y sin !/2 C ."x sin ! C y cos !/2
q
D x 2 cos2 ! C xy sin 2! C y 2 sin2 ! C x 2 sin2 ! " xy sin 2! C y 2 cos2 !
q
q
q
D x 2 .cos2 ! C sin2 !/ C y 2 .sin2 ! C cos2 !/ D x 2 ! 1 C y 2 ! 1 D x 2 C y 2
Substituting the lengths in (2) yields
v ! v" D
We now equate (1) and (3) to obtain
q
q
x 2 C y 2 ! x 2 C y 2 cos ˛ D .x 2 C y 2 / cos ˛
.x 2 C y 2 / cos ! D .x 2 C y 2 / cos ˛
cos ! D cos ˛
The solution for angles between 0ı and 180ı is ˛ D 0. That is, the angle between v and v" is !.
(3)