BASIC PROPERTIES OF CROSS PRODUCTS
PETER F. MCLOUGHLIN
Lemma 1. Suppose u, v and w are vectors in Rn . If a cross product exists on Rn
then it must have the following properties:
(1.1) w · (u × v) = −u · (w × v)
(1.2) u × v = −v × u which implies u × u = 0
(1.3) v × (v × u) = (v · u)v − (v · v)u
(1.4) w × (v × u) + (w × v) × u = 2(w · u)v − (u · v)w − (w · v)u
Proof. We first prove (1.1). By definition of cross product we have, 0 = (u + w) ·
((u+w)×v) = (u+w)·(u×v +w×v) = u·(u×v)+u·(w×v)+w·(u×v)+w·(w×v).
This implies u · (w × v) = −w · (u × v).
We now prove (1.2). By definition of cross product, (u × u) · (u × u) = (u · u)2 − (u ·
u)2 = 0. It follows u × u = 0. Next using the bilinearity of the cross product we
have, 0 = (u + v) × (u + v) = (u × u) + (u × v) + (v × u) + (v × v) = (u × v) + (v × u).
Hence, we must have u × v = −v × u.
Let’s next prove (1.3). Using the Pythagorean property of the cross product with
u:=u+w we have,
((u + w) × v) · ((u + w) × v) = ((u + w) · (u + w))(v · v) − ((u + w) · v)2
Next using the bilinearity of the dot product we have,
((u + w) · (u + w))(v · v) − ((u + w) · v)2
=(u · u)(v · v) + 2(u · w)(v · v) + (w · w)(v · v) − (u · v)2 − 2(u · v)(w · v) − (w · v)2
(1)
Also, by the bilinearity of the cross product and the dot product, and the Pythagorean
property of the cross product we have,
((u + w) × v) · ((u + w) × v) = (u × v + w × v) · (u × v + w × v)
=(u × v) · (u × v) + 2(w × v) · (u × v) + (w × v) · (w × v)
=(u · u)(v · v) − (u · v)2 + 2(w × v) · (u × v) + (w · u)(v · v) − (w · v)2
(2)
Now equating (1) and (2) and using (1.1) and (1.2) we have,
w · (v × (v × u)) = −(u × v) · (w × v)
=(u · v)(w · v) − (u · w)(v · v) = w · ((u · v)v − (v · v)u) for all vectors w.
It follows we must have, v × (v × u) = (v · u)v − (v · v)u.
Lastly let’s prove (1.4). Using the bilinear property of the cross product, (1.2), and
(1.3) implies
1
2
PETER F. MCLOUGHLIN
(w + u) × (((w + u) × (v + u)) = (w + u) × (w × v + w × u + u × v + u × u) =
w × (w × v) + w × (u × v) + w × (w × u) + w × (u × u) + u × (w × v) + u × (u × v) + u ×
(w × u) + u × (u × u) = w × (u × v) + u × (w × v) − (w · w)v + (w · v)w − (w · w)u +
(w · u)w − (u · u)v + (u · v)u + (u · u)w − (u · w)u
(3)
Next applying (1.3) with u1 = w + u and v1 = v + u implies
(w + u) × (((w + u) × (v + u)) = ((w + u) · (v + u))(w + u) − ((w + u) · (w + u))(v + u) =
((w · v) + (w · u) + (u · v) + (u · u))w + ((w · v) − (w · u) + (u · v) − (w · w))u − ((w ·
w) + 2(w · u) + (u · u))v
(4)
Now equating (3) and (4) we get, w×(u×v)+u×(w×v) = (u·v)w+(w·v)u−2(w·u)v.
By (1.2) we have, w × (u × v) + u × (w × v) = −w × (v × u) − (w × v) × u. Claim
now follows.
E-mail address: pmclough@csusb.edu