A First Course in Linear Algebra, 2017a

4.9. The Cross Product 181
There is another version of 4.14 which may be easier to remember. We can express the cross product
as the determinant of a matrix, as follows.
∣ ⃗i ⃗j ⃗ k ∣∣∣∣∣
⃗u ×⃗v =
u 1 u 2 u 3
(4.16)
∣ v 1 v 2 v 3
Expanding the determinant along the top row yields
∣ ∣ ∣ ∣ ∣ ∣ ∣∣∣
⃗i(−1) 1+1 u 2 u 3 ∣∣∣ ∣∣∣
+⃗j (−1) 2+1 u 1 u 3 ∣∣∣ ∣∣∣
+
v 2 v 3 v 1 v ⃗ k (−1) 3+1 u 1 u 2 ∣∣∣
3 v 1 v 2 =⃗i
∣ u ∣ 2 u 3 ∣∣∣ −⃗j
v 2 v 3
∣ u ∣ 1 u 3 ∣∣∣
+
v 1 v ⃗ k
3
∣ u ∣
1 u 2 ∣∣∣
v 1 v 2
Expanding these determinants leads to
(u 2 v 3 − u 3 v 2 )⃗i − (u 1 v 3 − u 3 v 1 )⃗j +(u 1 v 2 − u 2 v 1 )⃗k
which is the same as 4.15.
The cross product satisfies the following properties.
Proposition 4.50: Properties of the Cross Product
Let ⃗u,⃗v,⃗w be vectors in R 3 ,andk a scalar. Then, the following properties of the cross product hold.
1. ⃗u ×⃗v = −(⃗v ×⃗u),and⃗u ×⃗u =⃗0
2. (k⃗u) ×⃗v = k (⃗u ×⃗v)=⃗u × (k⃗v)
3. ⃗u × (⃗v + ⃗w)=⃗u ×⃗v +⃗u × ⃗w
4. (⃗v +⃗w) ×⃗u =⃗v ×⃗u + ⃗w ×⃗u
Proof. Formula 1. follows immediately from the definition. The vectors ⃗u ×⃗v and ⃗v ×⃗u have the same
magnitude, |⃗u||⃗v|sinθ, and an application of the right hand rule shows they have opposite direction.
Formula 2. is proven as follows. If k is a non-negative scalar, the direction of (k⃗u) ×⃗v is the same as
the direction of ⃗u ×⃗v,k (⃗u ×⃗v) and ⃗u × (k⃗v). The magnitude is k times the magnitude of ⃗u ×⃗v which is the
same as the magnitude of k (⃗u ×⃗v) and ⃗u × (k⃗v). Using this yields equality in 2. In the case where k < 0,
everything works the same way except the vectors are all pointing in the opposite direction and you must
multiply by |k| when comparing their magnitudes.
The distributive laws, 3. and 4., are much harder to establish. For now, it suffices to notice that if we
know that 3. is true, 4. follows. Thus, assuming 3., and using 1.,
(⃗v + ⃗w) ×⃗u = −⃗u × (⃗v + ⃗w)
= −(⃗u ×⃗v +⃗u × ⃗w)
=⃗v ×⃗u + ⃗w ×⃗u
♠