Vector Product in ijk notation Consider two vectors,

AandB,We can

definea new vectorC,C=AxBC = A B sin |C| = |A| |B| sin

Remember, tho', that

Cis avector. ThedirectionofCisperpendicularto the plane define by vectorsAandB.With the fingers of your right hand, rotate the first vector A into the second vector B. The direction your thumb points is the direction of the new vector C.(In this example, vectorsAandBare in the plane of this page or screen and the new vectorCpointsintothe page or screen)

With this basic definition of the vector product, what can we do with vectors

AandBwritten in component ori,j,knotation?A= A_{x}i+ A_{y}j+ A_{z}k

B= B_{x}i+ B_{y}j+ B_{z}k

C=AxB

C= [A_{x}i+ A_{y}j+ A_{z}k] x [B_{x}i+ B_{y}j+ B_{z}k]

C= A_{x}B_{x}(ixi) + A_{x}B_{y}(ixj) + A_{x}B_{z}(ixk) ++ A

_{y}B_{x}(jxi) + A_{y}B_{y}(jxj) + A_{y}B_{z}(jxk) ++ A

_{z}B_{x}(kxi) + A_{z}B_{x}(kxj) + A_{x}B_{x}(kxk)What are these cross products of the unit vectors?

ixi= 0

ixj=kixk= -j

jxi= -k

jxj= 0

jxk=i

kxi=j

kxj= -i

kxk= 0So our cross product becomes

C= A_{x}B_{x}( 0 ) + A_{x}B_{y}(k) + A_{x}B_{z}(- j) ++ A

_{y}B_{x}(- k) + A_{y}B_{y}( 0 ) + A_{y}B_{z}(i) ++ A

_{z}B_{x}(j) + A_{z}B_{x}(- i) + A_{x}B_{x}( 0 )

C=i(A_{y}B_{z}- A_{z}B_{y}) +j(A_{z}B_{x}- A_{x}B_{z}) +k(A_{x}B_{y}- A_{y}B_{x})That is certainly a good enough form to remember and we could stop here. But this expression can

alsobe remembered as adeterminant,

Vector Product (1)TorqueReturn to ToC, Ch11, Rolling Motion(c) Doug Davis, 2001; all rights reserved