Geometric algebra cross product

[krcmd1]

Homework Statement [/b]

my text (Geometric Algebra for Physicists, by Doran and Lasenby), p. 69, deals with rotating frame {fsubk} (I assume in 3D)

d/dt (fsubk) = omega X fsubk omega being angular velocity

then

omega X fsubk = (-I omega) dot fsubk = fsubk dot (I omega), where I is pseudoscalar and (I omega) is I think the geometric product of I and omega.

I don't understand these algebraic steps. Can someone explain?

I did follow earlier explanations of why with vectors a,b that a X b = -I(a wedge b)

If I have posted this to the wrong forum, would the moderators kindly forward it to a more appropriate one?


Thank you all very much!

Ken Cohen

 

[member 553083]

Homework Equations d/dt (fsubk) = omega X fsubk omega being angular velocityomega X fsubk = (-I omega) dot fsubk = fsubk dot (I omega), where I is pseudoscalar and (I omega) is I think the geometric product of I and omega.The Attempt at a Solution The first equation is just the derivative of the vector fsubk with respect to time. The second equation is saying that the vector omega X fsubk is equal to the dot product of (-I omega) and fsubk and also equal to the dot product of fsubk and (I omega). This is just the result of a few algebraic manipulations, using the fact that for any vectors a and b, a X b = -I(a wedge b) and a dot b = I(a dot b).