Solution Manual for Introduction to Electrodynamics (5th Edition) by David J. Griffiths

All 12 Chapters Covered




SOLUTIONS

,Contents
1 Vector Analysis 4
2 Electrostatics 26
3 Potential 53
4 Electric Fields in Matter 92
5 Magnetostatics 110
6 Magnetic Fields in Matter 133
7 Electrodynamics 145
8 Conservation Laws 168
9 Electromagnetic Waves 185
10 Potentials and Fields 210
11 Radiation 231
12 Electrodynamics and Relativity 262

,Chapter 1

Vector Analysis

Problem 1.1
✒✣
(a) From the diagram, |B + C| cos ✓3 = |B|cos ✓1 + |C| cos ✓ 2. Multiply by |A|.


}
in θ2
|A||B + C| cos ✓3 = |A||B| cos ✓1 + |A||C| cos ✓ 2.
So: A·(B + C) = A·B + A·C. (Dot product is distributive) |C| s
Similarly: |B + C|sin ✓3 = |B|sin ✓1 + |C|sin ✓2 . Mulitply by |A|n̂. θ2
|A||B + C|sin ✓3n̂ = |A||B|sin ✓1n̂ + |A||C|sin ✓2n̂.
If n̂ is the unit vector pointing out of the page, it follows that θ1
θ3 ✯
B|✲
}
| in θ1
s
A
! "# $ ! |C| cos "#θ2 $
A⇥(B + C) = (A⇥B) + (A⇥C). (Cross product is distributive) |B| cos θ1

(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and Section 8
(cross product)
Problem 1.2 C
The triple cross-product is not in general associative. For example, ✻
suppose A = B and C is perpendicular to A, as in the diagram. ✲A = B
Then (B⇥C) points out-of-the-page, and ⇥B)A⇥(B⇥C) points down,
= 0, so (A⇥B)⇥C = 0 6= ❂
and has magnitude ABC. But (A B×C
A⇥(B⇥C). ❄
A×(B×C)
Problem 1.3 z ✻
p p
A = +1 x̂ + 1 ŷ 1 ẑ; A = 3; B = 1 x̂ + 1 ŷ + 1 ẑ; B = 3.
p p 1 ✣B
A·B = +1 + 1 1 = 1 = AB cos ✓ = 3 3 cos ✓ )cos ✓= 3. θ
✓ = cos 1 1 ⇡ 70.5288 ✲y
3 ❲
✰ A
x
Problem 1.4
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example, we
might pick the base (A) and the left side (B):
A = 1 x̂ + 2 ŷ + 0 ẑ; B = 1 x̂ + 0 ŷ + 3 ẑ.

, 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
n̂ = A⇥B = x̂ + ŷ + ẑ .
6 3 2
|A⇥B| = 36 + 9 + 4 = 7.
|A⇥B| 7 7 7

Problem 1.5 ŷ ẑ
x̂
A⇥(B⇥C) = Ax Ay Az
(By Cz Bz Cy ) (Bz Cx BxCz ) (BxCy 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 BxCy + Az BxCz 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
= (4 x̂ + 6 ŷ + 8 ẑ) (2 x̂ + 8 ŷ + 7 ẑ) =
2 x̂ 2 ŷ + ẑ
p
= 4+4+1 = 3
ˆ = = 2x̂ 2ŷ + 1ẑ
3 3 3
Problem 1.8
(a) Āy B̄y + Ā z 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 AyBy sin cos (Ay Bz + Az By ) + cos2 AzBz
= (cos2 + sin2 )AyBy + (sin2 + cos2 )Az Bz = Ay By + Az Bz . X
(b) (Ax)2 + (Ay)2 + (Az)2 = Σ3 AiAi = Σ3 i=1 3 j=1 Rij Aj Σ3 k=1 RikAk = Σj,k (ΣiRij Rik) Aj Ak .
Σ ⇢
This equals A2 + A2 + A2 provided Σi=1 3
Rij Rik = 1 if j = k
x y z i=1
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.2
For2
suppose
2
A = (1, 0, 0). Then Σj,k (Σi Rij Rik) Aj Ak = Σi Ri1Ri1, and this must equal 1 (since we
want Ax +Ay +Az = 1). Likewise, Σ3 i=1 Ri2Ri2 = Σ
3
i=1 Ri3Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0).
Then we want 2 = Σj,k (Σi Rij Rik) Aj Ak = Σi Ri1Ri1 + Σi Ri2Ri2 + Σi Ri1Ri2 + Σi Ri2Ri1. But we already
know that the first two sums are both 1; the third and fourth are equal, so Σi Ri1Ri2 = Σi Ri2Ri1 = 0, and so on for other
unequal combinations of j, k. X In matrix notation: R̃R = 1, where R̃ is the transpose of R.

,Problem 1.9 z ′✻
✻y
y✻
✿ ❃
Looking down the axis: ✒ ❄
✲x ■
✠
z &&y ′
z✰✰′ x
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
001
R = @1 0 0A
0 1 0

Problem 1.10
(a)
No change. (Ax = Ax, Ay = Ay, Az = Az)
(b) A ! A, in the sense (A = A , A = A , A = A )
x x y y z z
(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 pseudoscalar
changes sign under inversion of coordinates.
Problem 1.11 a ! a;
(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 18 24y + 84 = 0.
2y
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.

,Problem 1.13
p
= (x x0 ) x̂ + (y y 0 ) ŷ + (z z 0 ) ẑ; = (x x0 )2 + (y y 0 )2 + (z z 0 )2 .
(a) r( 2 ) = @ [(x x0 )2+(y y 0 )2+(z z 0 )2 ] x̂+ @
() ŷ+ @ () ẑ = 2(x x0 ) x̂+2(y y 0 ) ŷ+2(z z 0 ) ẑ = 2 .
@x @y @z
ẑ
1 1 1
1 @ @ @
(b) r( )= [(x x0 )2 + (y y 0 )2 + (z z 0 )2 ] 2 x̂ + () 2 ŷ + () 2
@x @y @z
z 0 ) ẑ
1 3 1 3 1 3
= () 2 2(x x0 ) x̂ () 2 2(y y 0 ) ŷ () 2 2(z
2 3 2 2
= () 2[(x x0 ) x̂ + (y y 0 ) ŷ + (z z 0 ) ẑ] = (1/ 3
) = (1/ 2
)ˆ.
(c) @ ( n) = n n 1@
=n n 1( 1
1
2 x) = n
n 1 ˆ x, so
@x @x 2
r( n) = n n 1 ˆ
Problem 1.14
y = +y cos + z sin ; multiply by sin : y sin = +y sin cos + z sin2 . z = y sin
+ z cos ; multiply by cos : z cos = y sin cos + z cos2 . Add: y sin + z cos =
2
z(sin
@y
+ cos2 ) =@yz. Likewise, y cos
@z
z sin @z= y.
So )
@y = cos ; @z = sin ; @y = sin ; @z = cos . Therefore
(rf)y = @y = @y @y + @z @y = + cos (rf)y + sin (rf)z
@ ff
@ @ ff @y
@ @y @ff @z
@ @z So rf transforms as a vector. qed
(rf)z = @z = @y @z + @z @z = sin (rf)y + cos (rf)z
Problem 1.15
(a)r·va = @ (x2) + @
(3xz2) + @
( 2xz) = 2x + 0 2x = 0.
@x @y @z
@ @ @
(b)r·vb = (xy) + (2yz) + (3xz) = y + 2z + 3x.
@x @y @z
@
(c)r·vc = (y 2 ) + @ (2xy + z 2 ) + @
(2yz) = 0 + (2x) + (2y) = 2(x + y)
@x @y @z

Problem 1.16 h i
3
r·v = @
( x)+ @
( y )+ @
(z)= @ x(x2 + y2 + z 2)
2
@x r3 @y r3 @z r3 @x
h i h 3
i
+ @
@y
y(x2 + y2 + z 2 ) 32
+ @
@z
z(x2 + y2 + z 2 ) 2
3 5 3 5 3
= () 2 + x( 3/2)() 2 2x + () 2 + y( 3/2)() 2 2y + () 2
5
+ z( 3/2)() 2 2z = 3r 3 3r 5(x2 + y2 + z 2) = 3r 3 3r 3 = 0.
This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the origin.
How, then, can r·v = 0? The answer is that r·v = 0 everywhere except at the origin, but at the origin our
calculation is no good, since r = 0, and the expression for v blows up. In fact, r·v is infinite at that one point, and
zero elsewhere, as we shall see in Sect. 1.5.
Problem 1.17
v@vy = cos@ v vy + sin vz@; vvz z = sin ⇣vy + cos vz@v . ⌘ ⇣ ⌘
@vz @z
y y @v y @y y @z
cos + @v z @y sin . Use result in Prob. 1.14:
@y = cos + @y sin = @y @y + @z @y
@y @y +
@z @y
⇣
@y ⌘ ⇣ ⌘
@vy cos +
@vy
sin cos + @vz cos + @vz sin sin .
@vz = @y @z
@ vz ⇣ @y ⌘ @z ⇣ ⌘
@ vy @v y @y @ vy @ z @v z @y @vz @z
cos
@z = @y @z + @z @z sin + @y @z + @z @z
⇣ sin +
@z @z cos =
@v
⌘ ⇣ @vz
⌘
sin + @vz @y sin +
@vy sin + y cos
= @y @z
cos cos . So




@z

,@v y @vz @vy
@y cos 2 + @vy
@z sin cos + @v
@yz sin cos + @v
@zz sin 2 + @vy
@y sin 2 @vy
@z sin cos
@y + @z =
— @v
@y sin cos +
z @vz
cos 2
@z
@vy 2 @vz 2 @vy @vz
= cos2 + sin + sin + cos2 = + . X
@y @z @y @z
Problem 1.18
x̂
ŷ ẑ@
(a) r⇥va = @@x @
= x̂(0 6xz) + ŷ(0 + 2z) + ẑ(3z 2 0) =
@y @z 6xz x̂ + 2z ŷ + 3z 2 ẑ.
x2 3xz 2 2xz
x̂
@ ŷ
@ ẑ@
(b) r⇥vb = @x @y @z
= x̂(0 2y) + ŷ(0 3z) + ẑ(0 x) =
2y x̂ 3z ŷ x ẑ.
xy 2yz 3xz
@x̂ ŷ
@ ẑ@
(c) r⇥vc = @x @y @z
= x̂(2z 2z) + ŷ(0 0) + ẑ(2y 2y) =
0.
y2 (2xy + z 2 ) 2yz
Problem 1.19

y As we go from point A to point B (9 o’clock to 10 o’clock), x
increases, y increases, vx increases, and vy decreases, so @vx/@y > 0,
v while @vy /@y < 0. On the circle, vz = 0, and there is no dependence on
z, so Eq. 1.41 says
v B v ✓ ◆
@v y @vx
r⇥ v = ẑ
@x @y
A x
points in the negative z direction (into the page), as the right hand
z rule would suggest. (Pick any other nearby points on the circle and
v you will come to the same conclusion.) [I’m sorry, but I cannot
remember who suggested this cute illustration.]

Problem 1.20
v = y x̂ + x ŷ; or v = yz x̂ + xz ŷ + xy ẑ; or v = (3x2 z z 3 ) x̂ + 3 ŷ + (x3 3xz 2 ) ẑ;
or v = (sin x)(cosh y) x̂ (cos x)(sinh y) ŷ ; etc.
Problem 1.21 ⇣ ⌘ ⇣ ⌘ ⇣ ⌘
@(fg) @(fg) @(fg)
(i) r(f g) = x̂ + ŷ + ẑ = f @g + g @f x̂ + f @g + g @f ŷ + f @g + g @f ẑ
⇣@x @y ⌘@z ⇣ @x @x ⌘ @y @y @z @z
= f @g @g @g @f @f @f

@x x̂ + @y ŷ + @z ẑ + g @x x̂ + @y ŷ + @z ẑ = f (rg) + g(rf ). qed
@
(iv) r·(A⇥B) = ( @ @
@ Ay Bz Az By ) + @y(Az Bx AxB z ) + (A B Ay B x)
@z x y
= A@xy
⇣ @Bz @Ay @By @A⌘
z @ x⌘ Az @x By @x + Az @y ⇣
@Az @Bx z ⇣z
@B
Bz @A x
= Bx @Axz + B@B @A @A @A
+ Bx @A
@yx
Ax @y @B z @ y⌘
@B
+Ax @z y y+ B+ yBy@zx Ay @z x zBx @ z
@A x @B@A @A y y y

@y @z @z @x @y
—Ax @y @z
@x
⇣ + B z
⌘
Az @By @Bx
@z @x @x @y
= B· (r⇥A) A· (r⇥B). qed
⇣ ⌘ ⇣ ⌘ ⇣ ⌘
(v) r⇥ (fA) = @ (fAz) @ (fAy) x̂ + @ (fAx) @ (fAz) ŷ + @ (fAy) @(fAx)
ẑ
@y @z @z @x @x @y

, ⇣ ⌘ ⇣ @Az
⌘
= f @A z + Az
@f
@y f@@A
z
y
Ay @f x̂ + f @Ax + Ax @ z
@f f@x Az @f ŷ
@y ⇣ @z ⌘@z @x

h⇣
@A y
+ f @x⌘+ Ay @x f @y@f @A x A @f
x ⇣ ẑ ⌘i
@Ay @Az @y
= f @ Az x̂ + @Ax @Ay
⌘
@ Ax
⇣ẑ
@y @z @z ŷ + @x @y
h⇣ ⌘ ⇣@x ŷ + Ax ⌘i
—Ay
x
x̂ + A z ẑ
Ay @f @ z Az @y @f @f A
@x @z
@f @f
@y
@f
@x
= f (r⇥A) A⇥ (rf). qed
Problem 1.22 ⇣ ⌘ ⇣ ⌘
@B @By @B
(a) (A·r) B = Ax @Bx + Ay @Bx + Az @Bx x̂ + Ax y + Ay ⌘ + Az y ŷ
@x ⇣ @y @z @Bz
@x @y @z
@B ẑ.
+ Ax @Bz z

@x
+ Ay @y + Az @z
x x̂+y ŷ+z ẑ
(b) r̂ = r =
r
p ⇣ . Let’s just do the x⌘component.
x12+y2+z2
[(r̂·r)r̂] =
p
x @
+ y @ +z @ p x
x @x @y @z x2 +y2 +z2
n h i h i h io
= 11r xx 1
+ x(3 2) (1p 2)3 2x1 2 + yx1 x 2 1(p x)3 2y
p1
1
2
+ zx 2 1(p )3 2z
2 2
1
1 x x
=
r r r3
x + xy + xz = r r r3 x + y + z =r r r = 0.
Same goes for the other components. Hence: (r̂·r) r̂ = 0 .
⇣ + 3xz2 @ ⌘
(c) (va·r) vb = x2 @ 2xz @ (xy x̂ + 2yz ŷ + 3xz ẑ)
@x @y @z
= x2 (y x̂ + 0 ŷ + 3z ẑ) + 3xz 2 (x x̂ + 2z ŷ + 0 ẑ) 2xz (0 x̂ + 2y ŷ + 3x ẑ)
= x2 y + 3x2 z 2 x̂ + 6xz 3 4xyz ŷ + 3x2 z 6x2 z ẑ
= x2 y + 3z 2 x̂ + 2xz 3z 2 2y ŷ 3x2 z ẑProblem
1.23
@
(ii) [r(A·B)] = x (A@x
xBx + Ay By + Az Bz) = @Ax @B y @Bxx
@B @Ay @Bx @B
@B zy @Az @Bz
@x Bx + Ax @x + @x By + Ay @x + @x Bz + Az
— Az
@x
[A⇥(r⇥B)]x = Ay (r⇥B)z@AyAz (r⇥B)
@Ax y = Ay @Ax
@x @Az
@y @z @x

[B⇥(r⇥A)]
[(A·r)B] = xA= By + A B =—ABz + Az
@y @z @x
@x +Ax @x y z @z x x @x + Ay @y @z
x @ @y @ @ @Bx @Bx @Bx
@Ax @Ax
[(B·r)A]x = Bx @x +
@ABxy @y + Bz @z
So [A⇥(r⇥B) + @B
B⇥(r⇥A)@B+ (A·r)B @B + (B·r)A]x@A
= Ay @By x x z y @Ax @Ax
+ Bz
@Az
Ay Az + Az + By B y @y@AxBz @z @x
+Ax @B
@x x @y
@Bx @z
@Bx @x @Ax @x @Ax
+
@x + Ay @y + Az @z + Bx@
+
y @x +@ yBy @y Bz @z /@
y / / + /
@
y
@x + Ax @x + By @x
+ Ay @x
= Bx @Ax /
@
z
@ Bx
/ @A + Az @A@/z
@
z
y x @Ax @By
/
@
z
@ Bx @Bx

@x + + @x +
@Ax
+Bz + @Az @Ax @Bx @Bz @Bx

= [r(A·B)]x (same
(vi) [r⇥(A⇥B)] for y(A⇥B)
and z)
x = y z
@ @z (A⇥B)
@
y = @y (A xBy
@
Ay B x ) @z (Az Bx AxBz )
@
@@A @B y @A y @ Bx @ Bx
@ Az @ Ax @ Bz
= x
@y + Bz + Ax
By + Ax @y @y @y Bx Ay @z Bx Az @z @z @z
[(B·r)A (A·r)B@A
+ xA(r·B) @AB(r·A)] x @Bx
= Bx @Ax x @Bx @Bx @Bx @By @Bz @Ax @Ay @Az
@x + By @y + Bz @z Ax @x Ay @y Az @z
+ Ax @x
+ @y
+ @z Bx @x
+ @y
+ @z

, @y
+ Ax /
@
x
+ /
@
x
+ + + Bx
@x /
@y
@y @z
= By @Ax @Bx @Bx @ By @Bz
@z
@Ax
@x/ @Ax @Ay @Az
@B x @B x
+ Ay + Az
@y
+ Bz@z @Ax @z
= [r⇥( A⇥B)]x (same for y and z)
Problem 1.24
r(f/g) = @ (f/g) x̂ + @
(f/g) ŷ + @
(f/g) ẑ
@x@f @g
@y @f @g @z @f @g
g @y f @y g @z f @z
= h g⇣f2@x
g @x
x̂ @f g2 ŷ⌘ ⇣g2 ˆ ⌘i
1 @f @f + @g grf frg
= g+ z @g @g

g2 @x x̂ + @y ŷ + @z ẑ f @x x̂ + @y ŷ + @z ẑ = g2 . qed
r·(A/g) = @ (Ax/g) + @ (Ay/g) + @ (Az/g)
@x@ Ax @y @Ay
@z @ Az
@g @g @g
g Ay @y g Az @x
=
g h
@x ⇣A2x @x + @Ay
@y
g2 ⌘ +⇣
@z
g2 ⌘i ·A A·rg
1 g@A x
= g @Az @g @g
= gr
+ @y + @z — A x@g + Ay @y + Az @z . qed
[r⇥(A/g)]x = y (Az /g) @z (Ay /g)
g2 @x @x g2
@ @
@ @Az @Ay
@g
g Az @ g g A
h@y⇣ 2 @y @z y @z
= ⌘ g2 ⇣
= 1 g
g @Az @Ay @g ⌘i
@g
g2 @y @z — Az @y
Ay @z
⇥ ⇥
g(r A)x +(A
= g2 (same for y and z). qed
Problem 1.25
x̂ ŷ ẑ
(a) A⇥B = x 2y 3z = x̂(6xz) + ŷ(9zy) + ẑ( 2x2 6y 2 )
3y 2x 0
r·(A⇥B) = @ (6xz) + @ (9zy) + @ ( 2x2 6y 2 ) = 6z + 9z + 0 = 15z
⇣ @x ⌘ @y @z ⇣ ⌘
r⇥A = x̂ @ (3z) @ (2y) + ŷ @ (x) @ (3z) + ẑ @ (2y) @ (x) = 0; B·(r⇥A) = 0
⇣ @y ⌘
@z @z @x⇣ @x ⌘ @y
r⇥B = x̂ @ (0) @ ( 2x) + ŷ @ (3y) @ (0) + ẑ @ ( 2x) @ (3y) = 5 ẑ; A·(r⇥B) = 15z
@y @z @z @x @x @y
?
r·(A⇥B) = B·(r⇥A) A·(r⇥B) = 0 ( 15z) = 15z. X
(b) A·B = 3xy 4xy = xy ; r(A·B) = r( xy) = x̂ @ ( xy) + ŷ @ ( xy) = y x̂ x ŷ
@x @y

x̂ ŷ ẑ
A⇥(r⇥B) = x 2y 3z = x̂( 10y) + ŷ(5x); B⇥(r⇥A) = 0
0 0 5
⇣ ⌘
(A·r)B = x @ + 2y @ + 3z @ (3y x̂ 2x ŷ) = x̂(6y) + ŷ( 2x)
⇣ @x ⌘
@y @z
@ @
(B·r)A = 3y @x
2x @y
(x x̂ + 2y ŷ + 3z ẑ) = x̂(3y) + ŷ( 4x)
A⇥(r⇥B) + B⇥(r⇥A) + (A·r)B + (B·r)A
= 10y x̂ + 5x ŷ + 6y x̂ 2x ŷ + 3y x̂ 4x ŷ = y x̂ x ŷ = r·(A·B). X
⇣ ⌘ ⇣ ⌘
(c) r⇥(A⇥B) = x̂ @ ( 2x2 6y 2 ) @ (9zy) + ŷ @ (6xz) @ ( 2x2 6y 2 ) + ẑ @ (9zy) @
(6xz)
@y @z @z @x @x @y
= x̂( 12y 9y) + ŷ(6x + 4x) + ẑ(0) = 21y x̂ + 10x ŷ
r·A = @ (x) + @ (2y) + @ (3z) = 1 + 2 + 3 = 6; r·B = @
(3y) + @
( 2x) = 0
@x @y @z @x @y

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