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BASIC PROPERTIES OF CROSS PRODUCTS

PETER F. MCLOUGHLIN

Lemma 1. Suppose u, v and w are vectors in R n . If a cross product exists on R n

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 rst prove (1.1). By denition 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 denition 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 wehave, 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 withu:=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 Pythagoreanproperty 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

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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 u 1 = w + u and v 1 = 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 . Claimnow follows.

E-mail address : pmclough@csusb.edu