The pseudoscalar doesn't make physical sense. What I'm interpreting wrong? In geometric algebra, if we take 3 basis vectors, $e_1, e_2, e_3$, physically interpreted as units of length (like in meters), then the pseudoscalar $I$ has units of volume:
$e_{1 [m]} e_{2 [m]} e_{3 [m]} =I_{[m^3]}$
But then, the duality bivector-vector does not makes physical sense. Length units do not match:
$A_{[m^2]}  =I_{[m^3]}v_{[m]}$
that's because $e_{i [m]} e_{i [m]} =1_{[m^0]}$
How is this problem solved in mathematical physics?
 A: Consider the following projective-rejective split of a vector $ \mathbf{v} $ along the direction of a different vector $ \mathbf{u} $
$$   \mathbf{v} = \left( { \frac{\mathbf{u}}{\left\lVert {\mathbf{u}} \right\rVert} \cdot \mathbf{v} } \right) \frac{\mathbf{u}}{\left\lVert {\mathbf{u}} \right\rVert} + \left( { \frac{\mathbf{u}}{\left\lVert {\mathbf{u}} \right\rVert} \times \mathbf{v} } \right) \times \frac{\mathbf{u}}{\left\lVert {\mathbf{u}} \right\rVert}.$$
It should be clear that we want the unit vector $ \hat{\mathbf{u}} = \mathbf{u}/\left\lVert {\mathbf{u}} \right\rVert $ to be dimensionless, because if you didn't then
$$\mathbf{v} = \left( { \hat{\mathbf{u}} \cdot \mathbf{v} } \right) \hat{\mathbf{u}} + \left( { \hat{\mathbf{u}} \times \mathbf{v} } \right) \times \hat{\mathbf{u}},$$
would have dimensions of $ \textrm{some-units} $ on the left, but on the right you'd have $ {\textrm{some-units}}^3 $.
Instead, if let the units ride along coefficients instead of the unit vectors, and this sort of trouble is resolved.  You could have, for example
$$   \mathbf{x} = \left( { 10\, \textrm{m} } \right) \mathbf{e}_1,$$
$$   \left\lVert {\mathbf{x}} \right\rVert = \sqrt{\mathbf{x} \cdot \mathbf{x}} = \sqrt{ \left( {10\, \textrm{m}} \right)^2 \mathbf{e}_1 \cdot \mathbf{e}_1 } = 10\, \textrm{m},$$
and the unit vector will still be dimensionless
$$   \mathbf{e}_1 = \frac{\mathbf{x}}{\left\lVert {\mathbf{x}} \right\rVert} = \frac{\left( {10\, \textrm{m}} \right) \mathbf{e}_1}{10\, \textrm{m}} = \mathbf{e}_1.$$
As well as resolving all sorts of troubles for plain old dot-product spaces, dimensionless unit vectors resolve the dimension problems that would exist in the geometric algebra pseudoscalar dual products you mentioned.
