Multivariate Calculus - Bruce E. Shapiro

LECTURE 3. THE CROSS PRODUCT 23
Proof of equation 3.3.
To show that Uv is perpendicular to both v and u we compute the dot products:
⎛ ⎞
u · (Uv) = u T Uv = ( α β γ ) cβ − bγ
⎝aγ − cα⎠
bα − aβ
= α(cβ − bγ) + βaγ − cα) + γ(bα − aβ) = 0
⎛ ⎞
v · (Uv) = v T Uv = ( a b c ) cβ − bγ
⎝aγ − cα⎠
bα − aβ
= a(cβ − bγ) + b(aγ − cα) + c(bα − aβ) = 0
To show that the product given by equation (3.3) has length ‖u‖‖v‖ sin θ, observe
that
‖u‖ 2 = (‖u‖ sin θ) 2 + (‖u‖ cos θ) 2
‖u‖ 2 ‖v‖ 2 = (‖u‖‖v‖ sin θ) 2 + (‖u‖‖v‖ cos θ) 2
= (‖u‖‖v‖ sin θ) 2 + (u · v) 2
(‖u‖‖v‖ sin θ) 2 = ‖u‖ 2 ‖v‖ 2 − (u · v) 2
= ( α 2 + β 2 + γ 2) ( a 2 + b 2 + c 2) − (αa + βb + γc) 2
= α 2 ( a 2 + b 2 + c 2) + β 2 ( a 2 + b 2 + c 2) + γ 2 ( a 2 + b 2 + c 2)
−αa (αa + βb + γc) − βb (αa + βb + γc) − γc (αa + βb + γc)
= α 2 b 2 + α 2 c 2 + β 2 a 2 + β 2 c 2 + γ 2 a 2 + γ 2 b 2 − 2αβab − 2αγac − 2βγbc
Now consider the magnitude ‖Uv‖,
‖Uv‖ 2 =
and therefore
⎛ ⎞
cβ − bγ
⎝
aγ − cα⎠
∥ bα − aβ ∥
2
= (cβ − bγ) 2 + (aγ − cα ) 2 + (bα − aβ) 2
= c 2 β 2 − 2bcβγ + b 2 γ 2 + a 2 γ 2 − 2acαγ + c 2 α 2 + b 2 α 2 − 2abαβ + a 2 β 2
= (‖u‖‖v‖ sin θ) 2
‖Uv‖ = ‖u‖‖v‖ sin θ = ‖u × v‖
Hence the cross product’s components are given by equation (3.3).
Definition 3.2 (Determinant of a Square Matrix.) Let
( ) a b
M =
c d
Math 250, Fall 2006 Revised December 6, 2006.