CHAPTER 1. VECTOR ANALYSI ̓S
,CHAPTER 1. VECTOR ANALYSI ̓S 5
Chapter 1
Vector Analysis
Problem 1.1
(a) From the di̓agram, |B + C| cos θ3 = |B| cos θ1 + |C| cos θ2. Multi̓ply by |A|.
|A||B + C| cos θ3 = |A||B| cos θ1 + |A||C| cos θ2.
So: A·(B + C) = A·B + A·C. (Dot product i̓s di̓stri̓buti̓ve) si ̓n θ2
Si̓mi̓larly: |B + C| si̓n θ3 = |B| si̓n θ1 + |C| si̓n θ2. Muli̓tply by |A| n̂ .
|A||B + C| si̓n θ3 n̂ = |A||B| si̓n θ1 n̂ + |A||C| si̓n θ2 n̂ .
I̓f n̂ i̓s the uni̓t vector poi̓nti̓ng out of the page, i̓t follows that
A⇥(B + C) = (A⇥B) + (A⇥C). (Cross product i̓s di̓stri̓buti̓ve)
(b) For the general case, see G. E. Hay’s Vector and Tensor Analysi̓s, Chapter 1, Secti̓on 7 (dot product) and
Secti̓on 8 (cross product)
Problem 1.2
The tri̓ple cross-product i̓s not i̓n general associ̓ati̓ve. For example,
suppose A = B and C i̓s perpendi̓cular to A, as i̓n the di̓agram.
= B
Then (B⇥C) poi̓nts out-of-the-page, and A⇥(B⇥C) poi̓nts down,
and has magni̓tude ABC. But (A⇥B) = 0, so (A⇥B)⇥C = 0 =/
A⇥(B⇥C).
Problem 1.3
√ √
A = +1 x̂ + 1 ŷ — 1 ˆz ; A = 3; B = 1 x̂ + 1 ŷ + 1 ẑ ; B = 3.
√ √ 1
A·B = +1 + 1 — 1 = 1 = AB cos θ = 3 3 cos θ ⇒ cos θ = 3.
θ
y
70.5288○
x
Problem 1.4
The cross-product of any two vectors i̓n the plane wi̓ll gi̓ve a vector perpendi̓cular to the plane. For example,
we mi̓ght pi̓ck the base (A) and the left si̓de (B):
A = —1 x̂ + 2 ŷ + 0 ẑ ; B = —1 x̂ + 0 ŷ + 3 ẑ .
O
c 2012 Pearson Educati ̓on, I ̓nc., Upper Saddle Ri ̓ver, NJ. All ri ̓ghts reserved. Thi ̓s materi ̓al i̓s
protected under all copyri ̓ght laws as they currently exi ̓st. No porti ̓on of thi ̓s materi ̓al may be
reproduced, i ̓n any form or by any means, wi ̓thout permi ̓ssi ̓on i ̓n wri ̓ti ̓ng from the publi ̓sher.
, CHAPTER 1. VECTOR ANALYSI ̓S
x̂ ŷ ẑ
A⇥B = —1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ .
—1 0 3
Thi̓s has the ri̓ght di̓recti̓on, but the wrong magni̓tude. To make a uni̓t vector out of i̓t, si̓mply di̓vi̓de by i̓ts
length:
√
|A⇥B| = 36 + 9 + 4 = 7. n̂ = A⇥B = .
|A⇥B|
Problem 1.5
x̂ ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(ByCz — BzCy) (BzCx — BxCz) (BxCy — ByCx)
= x̂[A y (B x C y —ByCx) —Az(BzCx —BxCz)] + ŷ ( ) + ẑ ( )
(I̓’ll just check the x-component; the others go the same way)
= x̂(A y B x C y — AyByCx — AzBzCx + AzBxCz) + ŷ ( ) + ẑ( ) .
B(A·C) — C(A·B) = [Bx(AxCx + AyCy + AzCz) — Cx(AxBx + AyBy + AzBz)] x̂ + () ŷ + () ẑ
= x̂(A y B x C y + AzBxCz — AyByCx — AzBzCx) + 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).
I̓f thi̓s i̓s zero, then ei̓ther A i̓s parallel to C (i̓ncludi̓ng the case i̓n whi̓ch they poi̓nt i̓n opposi̓te di̓recti̓ons, or
one i̓s zero), or else B·C = B·A = 0, i̓n whi̓ch case B i̓s perpendi̓cular to A and C (i̓ncludi̓ng the case B = 0.)
Problem 1.7
= (4 x̂ + 6 ŷ + 8 ẑ ) — (2 x̂ + 8 ŷ + 7 ẑ ) =2 —2
√
= 4+4+1= 3
—
Problem 1.8
(a) A¯y B̄ y + A¯z B̄ z = (cos Ay + si̓n Az)(cos By + si̓n Bz) + (— si̓n Ay + cos Az)( —si̓n By + cos Bz)
2 2
= cos2 AyBy + si̓n cos (AyBz + AzBy) + si̓n AzBz + si̓n AyBy — si̓n cos (AyBz + AzBy) +
cos AzBz
2
= (cos2 + si̓n )AyBy + (si̓n + cos2 )AzBz = AyBy + AzBz. X
2 2
(b) (A )2 + (A )2 + (A )2 = ⌃3 A A = ⌃3i ̓=1 ⌃3j=1 Ri ̓jAj ⌃3k=1 Ri ̓kAk = ⌃j,k (⌃iR
̓ i ̓j Ri ̓k) Aj Ak.
x y z i ̓=1 i ̓ i ̓
⇢
2 2 2 1 i ̓f j = k
Thi̓s equals A + A + A provi̓ded ⌃3 Ri ̓jRi ̓k =
x y z i ̓=1 0 i ̓f j /= k
Moreover, i̓f R i̓s to preserve lengths for all vectors A, then thi̓s condi̓ti̓on i̓s not only suffi̓ci̓ent but also
necessary.
2 For
2 suppose
2 A = (1, 0, 0). Then ⌃j,k (⌃i ̓ Ri ̓jRi ̓k) AjAk = ⌃i ̓ Ri ̓1Ri ̓1, and thi̓s must equal 1 (si̓nce we
wantAx +Ay +Az = 1). Li̓kewi̓se, ⌃i3̓=1 Ri̓2R i̓2 = ⌃3i ̓=1 R i̓3 R i̓3 = 1. To check the case j /= k, choose A = (1, 1, 0).
Then we want 2 = ⌃j,k (⌃i ̓ Ri ̓jRi ̓k) AjAk = ⌃i ̓ Ri ̓1Ri ̓1 + ⌃i ̓ Ri ̓2Ri ̓2 + ⌃i ̓ Ri ̓1Ri ̓2 + ⌃i ̓ Ri ̓2Ri ̓1. But we already
know that the fi̓rst two sums are both 1; the thi̓rd and fourth are equal, so ⌃i ̓ Ri ̓1Ri ̓2 = ⌃i ̓ Ri ̓2Ri ̓1 = 0, and so
on for other unequal combi̓nati̓ons of j, k. X I̓n matri̓x notati̓on: R̃ R = 1, where R̃ i̓s the transpose of R.
,
,CHAPTER 1. VECTOR ANALYSI ̓S 5
Chapter 1
Vector Analysis
Problem 1.1
(a) From the di̓agram, |B + C| cos θ3 = |B| cos θ1 + |C| cos θ2. Multi̓ply by |A|.
|A||B + C| cos θ3 = |A||B| cos θ1 + |A||C| cos θ2.
So: A·(B + C) = A·B + A·C. (Dot product i̓s di̓stri̓buti̓ve) si ̓n θ2
Si̓mi̓larly: |B + C| si̓n θ3 = |B| si̓n θ1 + |C| si̓n θ2. Muli̓tply by |A| n̂ .
|A||B + C| si̓n θ3 n̂ = |A||B| si̓n θ1 n̂ + |A||C| si̓n θ2 n̂ .
I̓f n̂ i̓s the uni̓t vector poi̓nti̓ng out of the page, i̓t follows that
A⇥(B + C) = (A⇥B) + (A⇥C). (Cross product i̓s di̓stri̓buti̓ve)
(b) For the general case, see G. E. Hay’s Vector and Tensor Analysi̓s, Chapter 1, Secti̓on 7 (dot product) and
Secti̓on 8 (cross product)
Problem 1.2
The tri̓ple cross-product i̓s not i̓n general associ̓ati̓ve. For example,
suppose A = B and C i̓s perpendi̓cular to A, as i̓n the di̓agram.
= B
Then (B⇥C) poi̓nts out-of-the-page, and A⇥(B⇥C) poi̓nts down,
and has magni̓tude ABC. But (A⇥B) = 0, so (A⇥B)⇥C = 0 =/
A⇥(B⇥C).
Problem 1.3
√ √
A = +1 x̂ + 1 ŷ — 1 ˆz ; A = 3; B = 1 x̂ + 1 ŷ + 1 ẑ ; B = 3.
√ √ 1
A·B = +1 + 1 — 1 = 1 = AB cos θ = 3 3 cos θ ⇒ cos θ = 3.
θ
y
70.5288○
x
Problem 1.4
The cross-product of any two vectors i̓n the plane wi̓ll gi̓ve a vector perpendi̓cular to the plane. For example,
we mi̓ght pi̓ck the base (A) and the left si̓de (B):
A = —1 x̂ + 2 ŷ + 0 ẑ ; B = —1 x̂ + 0 ŷ + 3 ẑ .
O
c 2012 Pearson Educati ̓on, I ̓nc., Upper Saddle Ri ̓ver, NJ. All ri ̓ghts reserved. Thi ̓s materi ̓al i̓s
protected under all copyri ̓ght laws as they currently exi ̓st. No porti ̓on of thi ̓s materi ̓al may be
reproduced, i ̓n any form or by any means, wi ̓thout permi ̓ssi ̓on i ̓n wri ̓ti ̓ng from the publi ̓sher.
, CHAPTER 1. VECTOR ANALYSI ̓S
x̂ ŷ ẑ
A⇥B = —1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ .
—1 0 3
Thi̓s has the ri̓ght di̓recti̓on, but the wrong magni̓tude. To make a uni̓t vector out of i̓t, si̓mply di̓vi̓de by i̓ts
length:
√
|A⇥B| = 36 + 9 + 4 = 7. n̂ = A⇥B = .
|A⇥B|
Problem 1.5
x̂ ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(ByCz — BzCy) (BzCx — BxCz) (BxCy — ByCx)
= x̂[A y (B x C y —ByCx) —Az(BzCx —BxCz)] + ŷ ( ) + ẑ ( )
(I̓’ll just check the x-component; the others go the same way)
= x̂(A y B x C y — AyByCx — AzBzCx + AzBxCz) + ŷ ( ) + ẑ( ) .
B(A·C) — C(A·B) = [Bx(AxCx + AyCy + AzCz) — Cx(AxBx + AyBy + AzBz)] x̂ + () ŷ + () ẑ
= x̂(A y B x C y + AzBxCz — AyByCx — AzBzCx) + 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).
I̓f thi̓s i̓s zero, then ei̓ther A i̓s parallel to C (i̓ncludi̓ng the case i̓n whi̓ch they poi̓nt i̓n opposi̓te di̓recti̓ons, or
one i̓s zero), or else B·C = B·A = 0, i̓n whi̓ch case B i̓s perpendi̓cular to A and C (i̓ncludi̓ng the case B = 0.)
Problem 1.7
= (4 x̂ + 6 ŷ + 8 ẑ ) — (2 x̂ + 8 ŷ + 7 ẑ ) =2 —2
√
= 4+4+1= 3
—
Problem 1.8
(a) A¯y B̄ y + A¯z B̄ z = (cos Ay + si̓n Az)(cos By + si̓n Bz) + (— si̓n Ay + cos Az)( —si̓n By + cos Bz)
2 2
= cos2 AyBy + si̓n cos (AyBz + AzBy) + si̓n AzBz + si̓n AyBy — si̓n cos (AyBz + AzBy) +
cos AzBz
2
= (cos2 + si̓n )AyBy + (si̓n + cos2 )AzBz = AyBy + AzBz. X
2 2
(b) (A )2 + (A )2 + (A )2 = ⌃3 A A = ⌃3i ̓=1 ⌃3j=1 Ri ̓jAj ⌃3k=1 Ri ̓kAk = ⌃j,k (⌃iR
̓ i ̓j Ri ̓k) Aj Ak.
x y z i ̓=1 i ̓ i ̓
⇢
2 2 2 1 i ̓f j = k
Thi̓s equals A + A + A provi̓ded ⌃3 Ri ̓jRi ̓k =
x y z i ̓=1 0 i ̓f j /= k
Moreover, i̓f R i̓s to preserve lengths for all vectors A, then thi̓s condi̓ti̓on i̓s not only suffi̓ci̓ent but also
necessary.
2 For
2 suppose
2 A = (1, 0, 0). Then ⌃j,k (⌃i ̓ Ri ̓jRi ̓k) AjAk = ⌃i ̓ Ri ̓1Ri ̓1, and thi̓s must equal 1 (si̓nce we
wantAx +Ay +Az = 1). Li̓kewi̓se, ⌃i3̓=1 Ri̓2R i̓2 = ⌃3i ̓=1 R i̓3 R i̓3 = 1. To check the case j /= k, choose A = (1, 1, 0).
Then we want 2 = ⌃j,k (⌃i ̓ Ri ̓jRi ̓k) AjAk = ⌃i ̓ Ri ̓1Ri ̓1 + ⌃i ̓ Ri ̓2Ri ̓2 + ⌃i ̓ Ri ̓1Ri ̓2 + ⌃i ̓ Ri ̓2Ri ̓1. But we already
know that the fi̓rst two sums are both 1; the thi̓rd and fourth are equal, so ⌃i ̓ Ri ̓1Ri ̓2 = ⌃i ̓ Ri ̓2Ri ̓1 = 0, and so
on for other unequal combi̓nati̓ons of j, k. X I̓n matri̓x notati̓on: R̃ R = 1, where R̃ i̓s the transpose of R.
,
