# Why u*v = cross(u,v)-dot(u,v) at quaternions?

Why is it that for quaternions, $u*v = \mathrm{cross}(u,v)-\mathrm{dot}(u,v)$?

I wonder for what reason they are equal to each others.

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@Jason: Here $u$ and $v$ are pure quaternions, which we identify with $\mathbb R^3$ and the scalar part of a quaternion we identify with scalars. – Eric O. Korman Feb 20 '11 at 4:14

Assume that $u = bi + cj + dk$ and $v = xi + yj + zk$ are imaginary quaternions (no real part). Then a straightforward computation using the identities $i^2 = j^2 = k^2 = ijk = -1$ gives $uv = -(bx + cy + dz) + (cz - dy)i + (dx - bz)j + (by - cx)k = - \langle u,v \rangle + u \times v$ with the usual identification $\mathbb{R}^{3} = \operatorname{Im}\mathbb{H}$.
In view of sweetser's answer below let me add that there is no difficulty in getting his "more general formula" from yours by recalling that the real quaternions are in the center of $\mathbb{H}$ (in fact they are equal to the center of $\mathbb{H}$). Namely, for real $a',b'$ we have $$(a' + u)(b'+v) = a'b' + a'v + b'u + uv = (a'b'-\langle u,v \rangle) + (a'v + b'u + u \times v),$$ as desired, where the two parentheses in the rightmost formula indicate real and imaginary parts, respectively. – t.b. Jun 16 '11 at 11:38
The quaternions written are called "pure quaternions", meaning the scalar value is zero. Let me write a quaternion as a scalar and a 3-vector, where the 3-vector has an arrow. Then: $(0, \vec{u})(0, \vec{v}) = (-u \cdot v, \vec{u} \times \vec{v})$ This is not very general because for a different inertial observer, the scalar will no longer be zero. In that case: $(a', \vec{u'})(b', \vec{v'}) = (a' b'- \vec{u'} \cdot \vec{v'}, a' \vec{v'} + b' \vec{u'} + \vec{u'} \times \vec{v'})$ If the 3-vectors represent a position in space, then the scalars are time. If the 3-vectors are 3-momentums, then the scalars are energy.