Linear Algebra, Theory And Applications, 2012a

2.6. MATRICES AND CALCULUS 65
where the u i are the components of the vector u (t) in terms of the fixed vectors i ∗ , j ∗ , k ∗ .
Therefore,
u ′ (t) =Ω (t) × u (t) =Q ′ (t) Q (t) T u (t) (2.27)
where
Ω (t) =ω 1 (t) i ∗ +ω 2 (t) j ∗ +ω 3 (t) k ∗ .
because
Ω (t) × u (t) ≡
∣
i ∗ j ∗ k ∗
w 1 w 2 w 3
u 1 u 2 u 3 ∣ ∣∣∣∣∣
≡
i ∗ ( w 2 u 3 − w 3 u 2) + j ∗ ( w 3 u 1 − w1) 3 + k
∗ ( w 1 u 2 − w 2 u 1) .
This proves the lemma and yields the existence part of the following theorem.
Theorem 2.6.4 Let i (t) , j (t) , k (t) be as described. Then there exists a unique vector Ω (t)
such that if u (t) is a vector whose components are constant with respect to i (t) , j (t) , k (t) ,
then
u ′ (t) =Ω (t) × u (t) .
Proof: It only remains to prove uniqueness. Suppose Ω 1 also works. Then u (t) =Q (t) u
and so u ′ (t) =Q ′ (t) u and
Q ′ (t) u = Ω × Q (t) u = Ω 1 × Q (t) u
for all u. Therefore,
(Ω − Ω 1 ) × Q (t) u = 0
for all u and since Q (t) is one to one and onto, this implies (Ω − Ω 1 ) ×w = 0 for all w and
thus Ω − Ω 1 = 0.
Now let R (t) be a position vector and let
r (t) =R (t)+r B (t)
where
r B (t) ≡ x (t) i (t)+y (t) j (t)+z (t) k (t) .
R(t)
✍
r B (t)
❘
✒
r(t)
In the example of the earth, R (t) is the position vector of a point p (t) on the earth’s
surface and r B (t) is the position vector of another point from p (t) , thus regarding p (t)
as the origin. r B (t) is the position vector of a point as perceived by the observer on the
earth with respect to the vectors he thinks of as fixed. Similarly, v B (t) anda B (t) will be
the velocity and acceleration relative to i (t) , j (t) , k (t), and so v B = x ′ i + y ′ j + z ′ k and
a B = x ′′ i + y ′′ j + z ′′ k. Then
v ≡ r ′ = R ′ + x ′ i + y ′ j + z ′ k+xi ′ + yj ′ + zk ′ .
By , (2.27), if e ∈{i, j, k} , e ′ = Ω × e because the components of these vectors with respect
to i, j, k are constant. Therefore,
xi ′ + yj ′ + zk ′ = xΩ × i + yΩ × j + zΩ × k
= Ω× (xi + yj + zk)