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MATH 550, VECTOR ANALYSIS, EXTRA
HOMEWORK 3 AND SOLUTIONS
Problem 0. Suppose a, b, c ∈ R
3
. Show that (a × b) × c = −a(b · c) + b(a · c).
Solution: Let a = 〈a
1
, a
2
, a
3
〉, b = 〈b
1
, b
2
, b
3
〉, c = 〈c
1
, c
2
, c
3
〉. Then
a × b = 〈a 2
b
3
− a
3
b
2
, a
3
b
1
− a
1
b
3
, a
1
b
2
− a
2
b
1
〉, and
(a × b) × c
(a
3
b
1
− a
1
b
3
)c
3
− (a
1
b
2
− a
2
b
1
)c
2
(a
1
b
2
− a
2
b
1
)c
1
− (a
2
b
3
− a
3
b
2
)c
3
(a
2
b
3
− a
3
b
2
)c
2
− (a
3
b
1
− a
1
b
3
)c
1
−a
1
(b
2
c
2
3
c
3
) + b
1
(a
2
c
2
3
c
3
−a
2
(b
1
c
1
3
c
3
) + b
2
(a
1
c
1
3
c
3
−a
3
(b
1
c
1
2
c
2
) + b
3
(a
1
c
1
2
c
2
−a
1
(b
1
c
1
2
c
2
3
c
3
) + b
1
(a
1
c
1
2
c
2
3
c
3
−a
2
(b
1
c
1
2
c
2
3
c
3
) + b
2
(a
1
c
1
2
c
2
3
c
3
−a 3
(b 1
c 1
c 2
c 3
) + b 3
(a 1
c 1
c 2
c 3
a 1
a
2
a
3
(b
1
c
1
2
c
2
3
c
3
b 1
b
2
b
3
(a
1
c
1
2
c
2
3
c
3
= −a(b · c) + b(a · c).
Suppose ˆu is a unit vector in the abstract vector space V , and θ ∈ R. For every vector
r in V define:
R(θ,
u)r = r cos θ + ˆu(ˆu · r)(1 − cos θ) + ˆu × r sin θ.
Suppose v, w ∈ V.
Problem 1. Show that [R(θ, uˆ)v] · [R(θ, uˆ)w] = v · w.
Solution: We use direct calculation:
R(θ, uˆ)v · R(θ, uˆ)w
= [v cos θ + ˆu(ˆu · v)(1− cos θ) + ˆu × v sin θ] · [w cos θ + ˆu(ˆu · w)(1 − cos θ) + ˆu × w sin θ]
= v · w cos
2
θ + (ˆu · v)(ˆu · w) cos θ(1 − cos θ) + v · uˆ × w cos θ sin θ
- (ˆu · w)(ˆu · v) cos θ(1 − cos θ) + (ˆu · v)(ˆu · w)(1 − cos θ)
2
2
θ
Because ˆu is perpendicular to ˆu × v and ˆu × w, we have ˆu · uˆ × w = 0 and ˆu × v · uˆ = 0.
Also by the triple product identity v ·
u × w = ˆu × w · v = ˆu · w × v = −
u · v × w, and
uˆ × v · w = ˆu · v × w. Hence the terms involving sin θ cos θ cancel out. Collecting terms a
bit we get:
R(θ, uˆ)v · R(θ, uˆ)w = v · w cos
2
θ + (ˆu · v)(ˆu · w) sin
2
θ + (ˆu × v) · (ˆu × w) sin
2
θ
Using the triple product identity and problem 0 we get
(ˆu × v) · (ˆu × w) = [(ˆu × v) ×
u] · w = [−
u(v ·
u) + v(ˆu ·
u)] · w = v · w − (ˆu · v)(ˆu · w).
Substituting this into the previous expression we get
R(θ, uˆ)v · R(θ, uˆ)w = v · w(cos
2
θ + sin
2
θ) = v · w.
Using problem 0 we can show the following Jacobi identity holds:
(a × b) × c + (c × a) × b + (b × c) × a = 0. To see this we compute:
(a × b) × c = −a(b · c) + b(a · c)
(c × a) × b = −c(a · b) + a(c · b)
(b × c) × a = −b(c · a) + c(b · a).
Adding both sides we clearly get a zero vector on the right-hand-side.
Problem 2. Show that R(θ, uˆ)(v × w) = [R(θ, uˆ)v] × [R(θ, uˆ)w].
Solution: As before we directly calculate:
R(θ, uˆ)v × R(θ, uˆ)w
= [v cos θ + ˆu(ˆu · v)(1− cos θ) + ˆu × v sin θ]×[w cos θ + ˆu(ˆu · w)(1− cos θ) + ˆu × w sin θ]
= v × w cos
2
θ + (v × uˆ)(ˆu · w) cos θ(1 − cos θ) + v × (ˆu × w) cos θ sin θ
- (ˆu × w)(ˆu · v) cos θ(1 − cos θ) + ˆu × ˆu(ˆu · v)(ˆu · w)(1 − cos θ)
2
2
θ
= [v × w − (v × uˆ)(ˆu · w) − (ˆu × w)(ˆu · v)] cos
2
θ + (ˆu × v) × (ˆu × w) sin
2
θ
[v × (ˆu × w) − uˆ × (ˆu × w)(ˆu · v) + (ˆu × v) × w − (ˆu × v) × uˆ(ˆu · w)] cos θ sin θ
[(v × uˆ)(ˆu · w) + (ˆu × w)(ˆu · v)] cos θ
[ˆu × (ˆu × w)(ˆu · v) + (ˆu × v) × uˆ(ˆu · w)] sin θ
We must work on each of these terms. The first term is (using problem 0):
v × w − (v × uˆ)(ˆu · w) − (ˆu × w)(ˆu · v) = v × w + [−v(w · ˆu) + w(v · uˆ)] × ˆu
= v × w + [(v × w) × ˆu] × uˆ
= v × w − (v × w)(ˆu · ˆu) + ˆu[(v × w) · uˆ]
= ˆu[ˆu · (v × w)].
The second term is (using problem 0 and the triple product identity):
(ˆu × v) × (ˆu × w) = −uˆ[v · (ˆu × w)] + v[ˆu · (ˆu × w)] = ˆu[v · (w × uˆ)] = ˆu[(v × w) · ˆu]
= ˆu[ˆu · (v × w)].
