Solutions Manual
Introduction to Electrodynamics 4th Edition
By
David J. Griffiths
( All Chapters Included - 100% Verified Solutions )
1
,CHAPTER 1. VECTOR ANALYSIS 3
CHAPTER 1. VECTOR ANALYSIS 3
4 CHAPTER 1. VECTOR ANALYSIS
CHAPTER 1. VECTOR ANALYSIS 3
Chapter 1
Chapter 1
Chapter 1
VectorChapter
Analysis
1
Vector Analysis
Vector Analysis
Vector Analysis
Problem 1.1
✒
✣
}} }
Problem 1.1
(a) From the diagram,
Problem |B1.1
+ C| cos θ3 = |B| cos θ1 + |C| cos θ2 . Multiply by |A|. ✒
✣
Problem 1.1
|A||B + C| cos θ(a)3 =From
|A||B|
thecos θ1 + |A||C| cos θ2θ.3 = |B| cos θ1 + |C| cos θ2 . ✒
✣
C
diagram, |B + C| cos |C| sin θ2
So:(a)A·(B
From the diagram, + C| cos 3 = |B| cos ✓1θ ++|C| cos ✓2cos
. Multiply by |A|.
C
+
|B ✓=
+ C) = A·B
|A||B++ A·C.
cos (Dot
3 + C| cos θ3 = is
product
|A||B| cos 1 distributive)
cos|A||C| θ2θ.2 .
C
C| θ
(a) From the diagram, |B |B| θ1 + |C| cos |C| sin
B
|A||B + C| cos =A·(B θ3cos + |A||C| cos 2 . cosproduct
C
✓3 + |A||B| ✓=1 A·B ✓(Dot
+
So:
|A||B cos+ C)= |A||B| cos+θ1A·C.
+ |A||C| θ2 . is distributive)
C
C| |C| sin θ2
B
θ2
So: A·(B|B ++C| C) = A·B + A·C. (Dot product 2 . distributive)
is
C
Similarly: sin = |B| sin + |C| sin Mulitply by |A| n̂.
+
θ θ
So: A·(B3 + C) = A·B 1+ A·C. (Dot product θ is distributive) θ3 ✯
B
θ2
|A||B + C| sin θ3 n̂Similarly:
Similarly: |B
If n̂ is|A||B
the unit vector
= |A||B||B
+|A||B
C| sin+
Similarly: ✓C|
|B +=
sin+θ1C|n̂sin
C||B|
sinsin
θ ✓
=
θ3 = |B|sin
+ |A||C|
+
|B| |C|
sin sin
θ +
sinθθ1n̂.
✓ .
|C| 2
+ |C| sin θ2 . Mulitply by |A| n̂.B
Mulitply
sin θ
3 sin θ3 n̂ =1 |A||B| sin θ21 n̂ + |A||C| sin θ2 n̂.
pointing out
3
of the θpage,
1
it✓ follows
2 . by
Mulitply
that
|A| n̂.
by |A| n̂. θ1 θ3 ✯
θ2
B
}
θ3 ✯ |B| sin θ1
✲ }
|B| sin
A×(B If +
+ C| sin
C)the
n̂ is
|A||B
= unit
(A×B)
✓3 n̂
If =
n̂+is
C|the
|A||B|
is the+unit
If n̂ vector
3 n̂sin
sin θunit
(A×C).
✓1 n̂ +pointing
=vector
|A||B|
vector out
pointing (Cross
pointing
of
sin
the
1 n̂ +
|A||C|
product
out
sin
|A||C|
out n̂.sin
2of
is
of theit page,
page,
theθ2page,
n̂.
distributive)
follows it follows
that
it follows that! "#θ1B $ !! "#
that |B| cos
θ1
θ1 $ ! |C|"#
cos
}
"# θ2 $ !✲ "# $ ✲
$|B| sinAθ1
A
A×(B + C) = (A×B) + (A×C). (Cross product is distributive) ! "# |B| cos θ1
$ A cos θ2
|C|
A×(B + C) = (A×B) + (A×C). (Cross product is distributive)
(b) For the A⇥(B + C)case,
general = (A⇥B) see G. + E.
(A⇥C).
Hay’s (Cross
Vector product
and is distributive)
Tensor Analysis, Chapter
|B| cos θ1 |C| cos θ2
1, Section 71,(dot product) and
(b)(b)
ForFor
the the general
general case,case,
see G.see E. G.
Hay’sE. Hay’s
Vector Vector
and Tensor and Analysis,
Tensor Chapter 1,Chapter
Analysis, Section 7 (dot Section 7 (dot
product) and product)
Section
(b) For8 the(cross product)
general case,
Section
Section see G.
8 (cross
8 (cross E. product)
Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and
product)
Section
Problem 1.2 Problem 8 (cross
Problem product)
1.2 1.2
Problem 1.2 CC ✻ C
TheThe
The triple cross-product triple cross-product
triple cross-product
is not is not isinassociative.
in general general
not associative.
in general For example,
associative.
For example, For example, ✻ ✻
The triple cross-product
suppose
suppose A = is
B not
and in
C general
is
= B and C istoperpendicularassociative.
perpendicular to A, For
as example,
in the
to diagram. diagram.
A, as in the diagram.
suppose A = B and C isAperpendicular A, as in the ✲
suppose A Then
Then (B×C) points
= Then and(B×C)
B (B×C) is perpendicular
C points out-of-the-page,
points to A,
out-of-the-page, andasA×(B×C)
in
and theA×(B×C)
diagram.
points down,points down, ✲AA==BB ✲ A = B
Then (B⇥C) hasout-of-the-page,
andpoints magnitude ABC. and
out-of-the-page, Butand
A×(B×C)
A⇥(B⇥C)
(A×B) = 0, sopoints
points
(A×B)×C down,
down, = 0 ̸= ❂
and has and has magnitude ABC. But (A×B) = 0, so=(A×B)×C =B×C
❂
0 ̸= ❂
and magnitude
has magnitude ABC.
A×(B×C).
A×(B×C). ABC. But But(A×B)
(A⇥B)==0,0,so so (A×B)×C
(A⇥B)⇥C = 00 6≠= ❄A×(B×C)
B×C ❄ A×(B×C)
A×(B×C).A⇥(B⇥C).
B×C ❄ A×(B×C)
Problem 1.3 z
Problem 1.3 1.3 Problem 1.3
Problem z✻✻ z✻
√
A = +1 x̂ + 1√ ŷ −p1 ẑ; A = 3; B√= 1 x̂ + 1 ŷ + 1 ẑ; p
√
A =x̂ +1
A = +1 + 1x̂ŷ+−1 ŷ1Aẑ;=1A+1
ẑ;=A =3;1 ŷB3; B=1 =1 x̂1Ax̂
+=+1 1ŷ3; +
B131√
ŷ+√ ẑ;ẑ;1BB = 1 ŷ3.
x̂ = 3.+ 1 ẑ; ✣B
A·B = +1 +x̂1+ − 1 =− 1= ẑ;
ABpcos pθ = = 3 cos +
θ ⇒ cos θ.
√ √ √ √ 11 ✣ θB ✣
✲By
A·B =A·B+1=++1 1− + 11A·B
=
θ = cos
11−1
==
= 1%+1
=
& AB
1 +cos
AB 1 −cos
θ 1=✓=◦=1 3=3AB3 3cos
cos
cos
θ ✓θ⇒)
= cos
cos 3 θ =cos
✓ 3= 3 ..θ ⇒ cos θ.
3 ≈ 70.5288
θ θ
% &
3
❲ ✲y ✲y
✓ =−1cos1 1 3 ⇡
% & 1 θ= cos◦−1 31 ≈ 70.5288◦
70.5288 ✰ A
θ = cos 3 ≈ 70.5288
x
❲ ❲
✰ A✰ A
Problem 1.4 x x
Problem 1.4 The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
Problem 1.4 Problem 1.4
we might pick the base (A) and the left side (B):
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
The cross-product The
of any two vectorsofin
cross-product anythe
twoplane willingive
vectors a vector
the plane willperpendicular to the plane.
give a vector perpendicular to For example,
the plane. For exam
we might pick the base (A) and the left side (B):
we might pick the we Pearson
c base
⃝2005 might pick
and the
(A)Education, base
theInc.,
left (A)
side
Upper and
(B):
Saddle the NJ.
River, leftAllside (B):
rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
A= 1 x̂reproduced,
+ 2 ŷ +in0 any
ẑ; B =or by
form 1 x̂any+means,
0 ŷ +without
3 ẑ. permission in writing from the publisher.
A = −1 x̂ + 2 ŷ⃝2005
+ 0 ẑ;Pearson
c
B = Education,
−1 x̂ + 0Inc.,
ŷ +Upper
3 ẑ. Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
c 2012 Pearson Education, Inc., Upper Saddle River, NJ.
reproduced, in any form or by any means, without permission in writing from the All rights reserved. This material is
publisher.
⃝2005
c Pearson Education, Inc., Upper Saddleprotected under
River, NJ.all
Allcopyright laws as they
rights reserved. currently
This exist.
material is No portion of this material may be
protected under all copyright laws as theyreproduced, in anyNo
currently exist. form or by any
portion means,
of this without
material permission
may be in writing from the publisher.
reproduced, in any form or by any means, without permission in writing from the publisher.
2
,CHAPTER 1. VECTOR ANALYSIS 5
x̂ ŷ ẑ
A⇥B = 1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ.
1 0 3
This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its
length:
p A⇥B
|A⇥B| = 36 + 9 + 4 = 7. n̂ = |A⇥B| = 7 x̂ + 7 ŷ + 7 ẑ .
6 3 2
Problem 1.5
x̂ ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(By Cz Bz Cy ) (Bz Cx Bx Cz ) (Bx Cy By Cx )
= x̂[Ay (Bx Cy By Cx ) Az (Bz Cx Bx Cz )] + ŷ() + ẑ()
(I’ll just check the x-component; the others go the same way)
= x̂(Ay Bx Cy Ay By Cx Az Bz Cx + Az Bx Cz ) + ŷ() + ẑ().
B(A·C) C(A·B) = [Bx (Ax Cx + Ay Cy + Az Cz ) Cx (Ax Bx + Ay By + Az Bz )] x̂ + () ŷ + () ẑ
= x̂(Ay Bx Cy + Az Bx Cz Ay By Cx Az Bz Cx ) + ŷ() + ẑ(). They agree.
Problem 1.6
A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B) = B(A·C) C(A·B)+C(A·B) A(C·B)+A(B·C) B(C·A) = 0.
So: A⇥(B⇥C) (A⇥B)⇥C = B⇥(C⇥A) = A(B·C) C(A·B).
If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or
one is zero), or else B·C = B·A = 0, in which case B is perpendicular to A and C (including the case B = 0.)
Conclusion: A⇥(B⇥C) = (A⇥B)⇥C () either A is parallel to C, or B is perpendicular to A and C.
Problem 1.7
r = (4 x̂ + 6 ŷ + 8 ẑ) (2 x̂ + 8 ŷ + 7 ẑ) = 2 x̂ 2 ŷ + ẑ
p
r = 4+4+1= 3
r̂ = r = + 13 ẑ
3 x̂ 3 ŷ
2 2
r
Problem 1.8
(a) Āy B̄y + Āz B̄z = (cos Ay + sin Az )(cos By + sin Bz ) + ( sin Ay + cos Az )( sin By + cos Bz )
= cos2 Ay By + sin cos (Ay Bz + Az By ) + sin2 Az Bz + sin2 Ay By sin cos (Ay Bz + Az By ) +
cos Az Bz
2
= (cos2 + sin2 )Ay By + (sin2 + cos2 )Az Bz = Ay By + Az Bz . X
(b) (Ax )2 + (Ay )2 + (Az )2 = ⌃3i=1 Ai Ai = ⌃3i=1 ⌃3j=1 Rij Aj ⌃3k=1 Rik Ak = ⌃j,k (⌃i Rij Rik ) Aj Ak .
⇢
1 if j = k
This equals A2x + A2y + A2z provided ⌃3i=1 Rij Rik =
0 if j 6= k
Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but also
necessary. For suppose A = (1, 0, 0). Then ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 , and this must equal 1 (since we
2 2 2
want Ax +Ay +Az = 1). Likewise, ⌃3i=1 Ri2 Ri2 = ⌃3i=1 Ri3 Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0).
Then we want 2 = ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 + ⌃i Ri2 Ri2 + ⌃i Ri1 Ri2 + ⌃i Ri2 Ri1 . But we already
know that the first two sums are both 1; the third and fourth are equal, so ⌃i Ri1 Ri2 = ⌃i Ri2 Ri1 = 0, and so
on for other unequal combinations of j, k. X In matrix notation: R̃R = 1, where R̃ is the transpose of R.
c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
3
, 6 CHAPTER 1. VECTOR ANALYSIS
CHAPTER 1. VECTOR ANALYSIS 5
CHAPTER 1. VECTOR ANALYSIS 5
Problem 1.9
y✻ z′ ✻
✻ y
y✻
✿ ❃ z′ ✻
✻y
✿ ❃
✲ x LookingLooking
down the axis:
down the axis: ✒ ❄
✲x Looking down the axis: ✒ ❄
■ &y
′
✠ ■z ✰✰ & y ′ &
✠ z z ✰
✰′ x &x
′ x
z x
A 120 rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax = Az ,
Ay = Ax , Az = Ay .
0 1
0 01
R = @1 0 0A
0 10
Problem 1.10
(a) No change. (Ax = Ax , Ay = Ay , Az = Az )
(b) A ! A, in the sense (Ax = Ax , Ay = Ay , Az = Az )
(c) (A⇥B) ! ( A)⇥( B) = (A⇥B). That is, if C = A⇥B, C ! C . No minus sign, in contrast to
behavior of an “ordinary” vector, as given by (b). If A and B are pseudovectors, then (A⇥B) ! (A)⇥(B) =
(A⇥B). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vector
and a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself a vector.
Angular momentum (L = r⇥p) and torque (N = r⇥F) are pseudovectors.
(d) A·(B⇥C) ! ( A)·(( B)⇥( C)) = A·(B⇥C). So, if a = A·(B⇥C), then a ! a; a pseudoscalar
changes sign under inversion of coordinates.
Problem 1.11
(a)rf = 2x x̂ + 3y 2 ŷ + 4z 3 ẑ
(b)rf = 2xy 3 z 4 x̂ + 3x2 y 2 z 4 ŷ + 4x2 y 3 z 3 ẑ
(c)rf = ex sin y ln z x̂ + ex cos y ln z ŷ + ex sin y(1/z) ẑ
Problem 1.12
(a) rh = 10[(2y 6x 18) x̂ + (2x 8y + 28) ŷ]. rh = 0 at summit, so
2y 6x 18 = 0
2y 18 24y + 84 = 0.
2x 8y + 28 = 0 =) 6x 24y + 84 = 0
22y = 66 =) y = 3 =) 2x 24 + 28 = 0 =) x = 2.
Top is 3 miles north, 2 miles west, of South Hadley.
(b) Putting in x = 2, y = 3:
h = 10( 12 12 36 + 36 + 84 + 12) = 720 ft.
(c) Putting in x = 1, y = 1: rh = 10[(2 6 18) x̂ + (2 8 + 28) ŷ] = 10( 22 x̂ + 22 ŷ) = 220( x̂ + ŷ).
p
|rh| = 220 2 ⇡ 311 ft/mile; direction: northwest.
c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
⃝2005
c Pearson Education, Inc.,reproduced, in any
Upper Saddle form
River, NJ.or All
by rights
any means, without
reserved. Thispermission
material isin writing from the publisher.
⃝2005
c Pearson Education,
protected underInc., Upper Saddle
all copyright laws River, NJ.
as they All rights
currently reserved.
exist. This material
No portion is
of this material may be
protected under all copyright
reproduced, laws
in any as they
form or bycurrently
any means,exist. No portion
without of this
permission in material may the
writing from be publisher.
reproduced, in any form or by any means, without permission in writing from the publisher.
4
Introduction to Electrodynamics 4th Edition
By
David J. Griffiths
( All Chapters Included - 100% Verified Solutions )
1
,CHAPTER 1. VECTOR ANALYSIS 3
CHAPTER 1. VECTOR ANALYSIS 3
4 CHAPTER 1. VECTOR ANALYSIS
CHAPTER 1. VECTOR ANALYSIS 3
Chapter 1
Chapter 1
Chapter 1
VectorChapter
Analysis
1
Vector Analysis
Vector Analysis
Vector Analysis
Problem 1.1
✒
✣
}} }
Problem 1.1
(a) From the diagram,
Problem |B1.1
+ C| cos θ3 = |B| cos θ1 + |C| cos θ2 . Multiply by |A|. ✒
✣
Problem 1.1
|A||B + C| cos θ(a)3 =From
|A||B|
thecos θ1 + |A||C| cos θ2θ.3 = |B| cos θ1 + |C| cos θ2 . ✒
✣
C
diagram, |B + C| cos |C| sin θ2
So:(a)A·(B
From the diagram, + C| cos 3 = |B| cos ✓1θ ++|C| cos ✓2cos
. Multiply by |A|.
C
+
|B ✓=
+ C) = A·B
|A||B++ A·C.
cos (Dot
3 + C| cos θ3 = is
product
|A||B| cos 1 distributive)
cos|A||C| θ2θ.2 .
C
C| θ
(a) From the diagram, |B |B| θ1 + |C| cos |C| sin
B
|A||B + C| cos =A·(B θ3cos + |A||C| cos 2 . cosproduct
C
✓3 + |A||B| ✓=1 A·B ✓(Dot
+
So:
|A||B cos+ C)= |A||B| cos+θ1A·C.
+ |A||C| θ2 . is distributive)
C
C| |C| sin θ2
B
θ2
So: A·(B|B ++C| C) = A·B + A·C. (Dot product 2 . distributive)
is
C
Similarly: sin = |B| sin + |C| sin Mulitply by |A| n̂.
+
θ θ
So: A·(B3 + C) = A·B 1+ A·C. (Dot product θ is distributive) θ3 ✯
B
θ2
|A||B + C| sin θ3 n̂Similarly:
Similarly: |B
If n̂ is|A||B
the unit vector
= |A||B||B
+|A||B
C| sin+
Similarly: ✓C|
|B +=
sin+θ1C|n̂sin
C||B|
sinsin
θ ✓
=
θ3 = |B|sin
+ |A||C|
+
|B| |C|
sin sin
θ +
sinθθ1n̂.
✓ .
|C| 2
+ |C| sin θ2 . Mulitply by |A| n̂.B
Mulitply
sin θ
3 sin θ3 n̂ =1 |A||B| sin θ21 n̂ + |A||C| sin θ2 n̂.
pointing out
3
of the θpage,
1
it✓ follows
2 . by
Mulitply
that
|A| n̂.
by |A| n̂. θ1 θ3 ✯
θ2
B
}
θ3 ✯ |B| sin θ1
✲ }
|B| sin
A×(B If +
+ C| sin
C)the
n̂ is
|A||B
= unit
(A×B)
✓3 n̂
If =
n̂+is
C|the
|A||B|
is the+unit
If n̂ vector
3 n̂sin
sin θunit
(A×C).
✓1 n̂ +pointing
=vector
|A||B|
vector out
pointing (Cross
pointing
of
sin
the
1 n̂ +
|A||C|
product
out
sin
|A||C|
out n̂.sin
2of
is
of theit page,
page,
theθ2page,
n̂.
distributive)
follows it follows
that
it follows that! "#θ1B $ !! "#
that |B| cos
θ1
θ1 $ ! |C|"#
cos
}
"# θ2 $ !✲ "# $ ✲
$|B| sinAθ1
A
A×(B + C) = (A×B) + (A×C). (Cross product is distributive) ! "# |B| cos θ1
$ A cos θ2
|C|
A×(B + C) = (A×B) + (A×C). (Cross product is distributive)
(b) For the A⇥(B + C)case,
general = (A⇥B) see G. + E.
(A⇥C).
Hay’s (Cross
Vector product
and is distributive)
Tensor Analysis, Chapter
|B| cos θ1 |C| cos θ2
1, Section 71,(dot product) and
(b)(b)
ForFor
the the general
general case,case,
see G.see E. G.
Hay’sE. Hay’s
Vector Vector
and Tensor and Analysis,
Tensor Chapter 1,Chapter
Analysis, Section 7 (dot Section 7 (dot
product) and product)
Section
(b) For8 the(cross product)
general case,
Section
Section see G.
8 (cross
8 (cross E. product)
Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and
product)
Section
Problem 1.2 Problem 8 (cross
Problem product)
1.2 1.2
Problem 1.2 CC ✻ C
TheThe
The triple cross-product triple cross-product
triple cross-product
is not is not isinassociative.
in general general
not associative.
in general For example,
associative.
For example, For example, ✻ ✻
The triple cross-product
suppose
suppose A = is
B not
and in
C general
is
= B and C istoperpendicularassociative.
perpendicular to A, For
as example,
in the
to diagram. diagram.
A, as in the diagram.
suppose A = B and C isAperpendicular A, as in the ✲
suppose A Then
Then (B×C) points
= Then and(B×C)
B (B×C) is perpendicular
C points out-of-the-page,
points to A,
out-of-the-page, andasA×(B×C)
in
and theA×(B×C)
diagram.
points down,points down, ✲AA==BB ✲ A = B
Then (B⇥C) hasout-of-the-page,
andpoints magnitude ABC. and
out-of-the-page, Butand
A×(B×C)
A⇥(B⇥C)
(A×B) = 0, sopoints
points
(A×B)×C down,
down, = 0 ̸= ❂
and has and has magnitude ABC. But (A×B) = 0, so=(A×B)×C =B×C
❂
0 ̸= ❂
and magnitude
has magnitude ABC.
A×(B×C).
A×(B×C). ABC. But But(A×B)
(A⇥B)==0,0,so so (A×B)×C
(A⇥B)⇥C = 00 6≠= ❄A×(B×C)
B×C ❄ A×(B×C)
A×(B×C).A⇥(B⇥C).
B×C ❄ A×(B×C)
Problem 1.3 z
Problem 1.3 1.3 Problem 1.3
Problem z✻✻ z✻
√
A = +1 x̂ + 1√ ŷ −p1 ẑ; A = 3; B√= 1 x̂ + 1 ŷ + 1 ẑ; p
√
A =x̂ +1
A = +1 + 1x̂ŷ+−1 ŷ1Aẑ;=1A+1
ẑ;=A =3;1 ŷB3; B=1 =1 x̂1Ax̂
+=+1 1ŷ3; +
B131√
ŷ+√ ẑ;ẑ;1BB = 1 ŷ3.
x̂ = 3.+ 1 ẑ; ✣B
A·B = +1 +x̂1+ − 1 =− 1= ẑ;
ABpcos pθ = = 3 cos +
θ ⇒ cos θ.
√ √ √ √ 11 ✣ θB ✣
✲By
A·B =A·B+1=++1 1− + 11A·B
=
θ = cos
11−1
==
= 1%+1
=
& AB
1 +cos
AB 1 −cos
θ 1=✓=◦=1 3=3AB3 3cos
cos
cos
θ ✓θ⇒)
= cos
cos 3 θ =cos
✓ 3= 3 ..θ ⇒ cos θ.
3 ≈ 70.5288
θ θ
% &
3
❲ ✲y ✲y
✓ =−1cos1 1 3 ⇡
% & 1 θ= cos◦−1 31 ≈ 70.5288◦
70.5288 ✰ A
θ = cos 3 ≈ 70.5288
x
❲ ❲
✰ A✰ A
Problem 1.4 x x
Problem 1.4 The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
Problem 1.4 Problem 1.4
we might pick the base (A) and the left side (B):
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
The cross-product The
of any two vectorsofin
cross-product anythe
twoplane willingive
vectors a vector
the plane willperpendicular to the plane.
give a vector perpendicular to For example,
the plane. For exam
we might pick the base (A) and the left side (B):
we might pick the we Pearson
c base
⃝2005 might pick
and the
(A)Education, base
theInc.,
left (A)
side
Upper and
(B):
Saddle the NJ.
River, leftAllside (B):
rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
A= 1 x̂reproduced,
+ 2 ŷ +in0 any
ẑ; B =or by
form 1 x̂any+means,
0 ŷ +without
3 ẑ. permission in writing from the publisher.
A = −1 x̂ + 2 ŷ⃝2005
+ 0 ẑ;Pearson
c
B = Education,
−1 x̂ + 0Inc.,
ŷ +Upper
3 ẑ. Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
c 2012 Pearson Education, Inc., Upper Saddle River, NJ.
reproduced, in any form or by any means, without permission in writing from the All rights reserved. This material is
publisher.
⃝2005
c Pearson Education, Inc., Upper Saddleprotected under
River, NJ.all
Allcopyright laws as they
rights reserved. currently
This exist.
material is No portion of this material may be
protected under all copyright laws as theyreproduced, in anyNo
currently exist. form or by any
portion means,
of this without
material permission
may be in writing from the publisher.
reproduced, in any form or by any means, without permission in writing from the publisher.
2
,CHAPTER 1. VECTOR ANALYSIS 5
x̂ ŷ ẑ
A⇥B = 1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ.
1 0 3
This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its
length:
p A⇥B
|A⇥B| = 36 + 9 + 4 = 7. n̂ = |A⇥B| = 7 x̂ + 7 ŷ + 7 ẑ .
6 3 2
Problem 1.5
x̂ ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(By Cz Bz Cy ) (Bz Cx Bx Cz ) (Bx Cy By Cx )
= x̂[Ay (Bx Cy By Cx ) Az (Bz Cx Bx Cz )] + ŷ() + ẑ()
(I’ll just check the x-component; the others go the same way)
= x̂(Ay Bx Cy Ay By Cx Az Bz Cx + Az Bx Cz ) + ŷ() + ẑ().
B(A·C) C(A·B) = [Bx (Ax Cx + Ay Cy + Az Cz ) Cx (Ax Bx + Ay By + Az Bz )] x̂ + () ŷ + () ẑ
= x̂(Ay Bx Cy + Az Bx Cz Ay By Cx Az Bz Cx ) + ŷ() + ẑ(). They agree.
Problem 1.6
A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B) = B(A·C) C(A·B)+C(A·B) A(C·B)+A(B·C) B(C·A) = 0.
So: A⇥(B⇥C) (A⇥B)⇥C = B⇥(C⇥A) = A(B·C) C(A·B).
If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or
one is zero), or else B·C = B·A = 0, in which case B is perpendicular to A and C (including the case B = 0.)
Conclusion: A⇥(B⇥C) = (A⇥B)⇥C () either A is parallel to C, or B is perpendicular to A and C.
Problem 1.7
r = (4 x̂ + 6 ŷ + 8 ẑ) (2 x̂ + 8 ŷ + 7 ẑ) = 2 x̂ 2 ŷ + ẑ
p
r = 4+4+1= 3
r̂ = r = + 13 ẑ
3 x̂ 3 ŷ
2 2
r
Problem 1.8
(a) Āy B̄y + Āz B̄z = (cos Ay + sin Az )(cos By + sin Bz ) + ( sin Ay + cos Az )( sin By + cos Bz )
= cos2 Ay By + sin cos (Ay Bz + Az By ) + sin2 Az Bz + sin2 Ay By sin cos (Ay Bz + Az By ) +
cos Az Bz
2
= (cos2 + sin2 )Ay By + (sin2 + cos2 )Az Bz = Ay By + Az Bz . X
(b) (Ax )2 + (Ay )2 + (Az )2 = ⌃3i=1 Ai Ai = ⌃3i=1 ⌃3j=1 Rij Aj ⌃3k=1 Rik Ak = ⌃j,k (⌃i Rij Rik ) Aj Ak .
⇢
1 if j = k
This equals A2x + A2y + A2z provided ⌃3i=1 Rij Rik =
0 if j 6= k
Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but also
necessary. For suppose A = (1, 0, 0). Then ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 , and this must equal 1 (since we
2 2 2
want Ax +Ay +Az = 1). Likewise, ⌃3i=1 Ri2 Ri2 = ⌃3i=1 Ri3 Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0).
Then we want 2 = ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 + ⌃i Ri2 Ri2 + ⌃i Ri1 Ri2 + ⌃i Ri2 Ri1 . But we already
know that the first two sums are both 1; the third and fourth are equal, so ⌃i Ri1 Ri2 = ⌃i Ri2 Ri1 = 0, and so
on for other unequal combinations of j, k. X In matrix notation: R̃R = 1, where R̃ is the transpose of R.
c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
3
, 6 CHAPTER 1. VECTOR ANALYSIS
CHAPTER 1. VECTOR ANALYSIS 5
CHAPTER 1. VECTOR ANALYSIS 5
Problem 1.9
y✻ z′ ✻
✻ y
y✻
✿ ❃ z′ ✻
✻y
✿ ❃
✲ x LookingLooking
down the axis:
down the axis: ✒ ❄
✲x Looking down the axis: ✒ ❄
■ &y
′
✠ ■z ✰✰ & y ′ &
✠ z z ✰
✰′ x &x
′ x
z x
A 120 rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax = Az ,
Ay = Ax , Az = Ay .
0 1
0 01
R = @1 0 0A
0 10
Problem 1.10
(a) No change. (Ax = Ax , Ay = Ay , Az = Az )
(b) A ! A, in the sense (Ax = Ax , Ay = Ay , Az = Az )
(c) (A⇥B) ! ( A)⇥( B) = (A⇥B). That is, if C = A⇥B, C ! C . No minus sign, in contrast to
behavior of an “ordinary” vector, as given by (b). If A and B are pseudovectors, then (A⇥B) ! (A)⇥(B) =
(A⇥B). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vector
and a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself a vector.
Angular momentum (L = r⇥p) and torque (N = r⇥F) are pseudovectors.
(d) A·(B⇥C) ! ( A)·(( B)⇥( C)) = A·(B⇥C). So, if a = A·(B⇥C), then a ! a; a pseudoscalar
changes sign under inversion of coordinates.
Problem 1.11
(a)rf = 2x x̂ + 3y 2 ŷ + 4z 3 ẑ
(b)rf = 2xy 3 z 4 x̂ + 3x2 y 2 z 4 ŷ + 4x2 y 3 z 3 ẑ
(c)rf = ex sin y ln z x̂ + ex cos y ln z ŷ + ex sin y(1/z) ẑ
Problem 1.12
(a) rh = 10[(2y 6x 18) x̂ + (2x 8y + 28) ŷ]. rh = 0 at summit, so
2y 6x 18 = 0
2y 18 24y + 84 = 0.
2x 8y + 28 = 0 =) 6x 24y + 84 = 0
22y = 66 =) y = 3 =) 2x 24 + 28 = 0 =) x = 2.
Top is 3 miles north, 2 miles west, of South Hadley.
(b) Putting in x = 2, y = 3:
h = 10( 12 12 36 + 36 + 84 + 12) = 720 ft.
(c) Putting in x = 1, y = 1: rh = 10[(2 6 18) x̂ + (2 8 + 28) ŷ] = 10( 22 x̂ + 22 ŷ) = 220( x̂ + ŷ).
p
|rh| = 220 2 ⇡ 311 ft/mile; direction: northwest.
c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
⃝2005
c Pearson Education, Inc.,reproduced, in any
Upper Saddle form
River, NJ.or All
by rights
any means, without
reserved. Thispermission
material isin writing from the publisher.
⃝2005
c Pearson Education,
protected underInc., Upper Saddle
all copyright laws River, NJ.
as they All rights
currently reserved.
exist. This material
No portion is
of this material may be
protected under all copyright
reproduced, laws
in any as they
form or bycurrently
any means,exist. No portion
without of this
permission in material may the
writing from be publisher.
reproduced, in any form or by any means, without permission in writing from the publisher.
4
