How to compute signed angle between $n$-dimensional vectors? I am looking for clarification on this question. The accepted answer says that measuring a signed distance is impossible because the plane spanned by two vectors $u$ and $v$ does not have a canonical orientation. But what does orientation mean here? Is it just picking a particular vector in the orthocomplement of $\text{span}\{u,v\}$ or must ond pick several such vectors?
I imagine that an orientation for the plane can be picked arbitrarily for any two vectors. Once it is defined, how would one go about computing the signed angle between them?
 A: Put aside $n$-dimensionality for a moment.  Let's look at a plane spanned by the vectors $a$ and $b$.  There is an unsigned angle between $a$ and $b$.  To compute the signed angle between $a$ and $b$, we must decide (for instance) which way is the direction of increasing angle and which is decreasing angle.
When presented with the usual coordinate plane, oriented in the usual way, the direction of increasing angles is anticlockwise and the direction of decreasing angles is clockwise.  Notice that to make sense of that claim, we have to know which side of the plane we look at it from.  If you are looking at the plane from the "wrong" side, we must swap "anticlockwise" and "clockwise" to get the correct directions.
So, to assign signed angles to vectors on a plane, we must know from which side to view the plane so that anticlockwise angles are positive.  So there are only two cases to resolve.
Let's temporarily take things down one dimension to see how much information we must specify to resolve these two cases.  We have a line in 3-space.  We want to know which direction on the line is positive and which negative.  At a distinguished point on the line (its origin), there are infinitely many (unit) vectors perpendicular to the line.  No one of these is sufficient to resolve the orientation of the line.  (To see this, treat the space as a metric space then rotate the entire space, with the line embedded in it, around the perpendicular vector by $\pi$ radians.  The orientation of the line has reversed, but nothing else has changed.)  Our experience with vectors in 3-space tells us that a linearly independent pair of vectors in the orthocomplement of the line is sufficient to pick an orientation by the "right-hand rule" once we assign an order to the list of two vectors.
The same thing happens for your plane in $n$-space.  You need to select $n-2$ linearly independent vectors, $V$, from the orthocomplement of your plane.  Then specify an order on the list of $n$ vectors $V \cup \{u,v\}$.  This ordered list will produce a positively or negatively oriented coordinate system and the plane will inherit this orientation in the ordered basis of $u$ and $v$ in whatever relative order they appear in the $n$-vector list.  Of course, if you swap $u$ and $v$ in that list the signs of both orientations (of the $n$-space and of the plane) simultaneously reverse, meaning positive angles remain positive and negative angles remain negative.  The freedom of choice is present in the $n-2$ vectors and their order in the list since exchanging any pair of these reverses the orientation on the space but does not change the order of $\{u,v\}$ in the plane, so reverses the signs of all signed angles in the plane.
