show that if the vectors u and v are identified with pure qu

Solution

Quaternions are elements of the 4-dimensional space Q formed by the real axis and three imaginary axes i ,j, k which are orthogonal to each other and obey Hamilton rule.

Hamilton rule is i² = j²= k= ijk = -1.

Quaternions whose real part is zero are caleed pure quaternions.

To multiply quaternions we should know how to multiply i,j,k by each other.

We know that i²= -1, j²= -1, k²= -1. also we know that ijk= -1.

consider ijk= -1.multiply both sides with i,we get i²kj = -jk = -i.similarly multiply both sides with j, we get ik= -kj= -j, also multiply both sides with k, we get ij = -ji =-k.

Let u = a₁i+b₁j+c₁k and v= a₂i+b^₂j+c₂k.

u.v = a₁a₂+b₁b₂=c₁c₂

This is equal to scalar product of uv*, vu*,u*v, v*u.

So u.v= (u*v+v*u) /2

The cross product of this is

uxv = (b₁c₂-c₁b₂ )i + (c₁a₂-a₁c₂)j +(a₁b₂-b₁a₂)k.This is equal to the vector part of the uv as quternions,as well as vector part of the -uv.

So uxv = (uv-vu)/2.

The dot product of two pure quaternions u,v can be defined naturally as the scalar product u.v = (u.v ,0) = (-uv+vu)/2.

while their cross prduct can be defined as purequaternion is

uxv = (0,uxv) = (uv-vu ) /2.