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It's clear geometrically that if you have two vectors in $\mathbb{R}^3$ a rotation will preserve their lengths and the angle between them. But how do you visualize that a rotation preserves orientation. What is the geometric explanation for why the change of basis matrix, from one orthonormal basis to another, has determinant $1$?

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  • $\begingroup$ What does orientation mean to you? $\endgroup$ – user137731 Apr 8 '15 at 16:09
  • $\begingroup$ a choice of equivalence class given by the sign of the change of basis matrix. I don't see intuitively why the change of basis matrix for a rotation is always 1. In particular, given two orthonormal vectors, there's 2 choices for a third. Whichever one you choose, the rotated basis will be in the same class. Why's that true only for an isometry which is a rotation? $\endgroup$ – user162520 Apr 8 '15 at 16:11
  • $\begingroup$ Rotations are NOT the only isometries which preserve orientations. See these helpful tables. $\endgroup$ – user137731 Apr 8 '15 at 16:21
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The way that I picture orientation (for a 2d object) is like this. Say you're given two noncollinear arrow vectors. Translate those two vectors so that their tails are touching. Like this:

enter image description here

Now consider the parallelogram that can be formed by those two vectors. Place a particle on the boundary of this parallelogram and constrain it to stay on the boundary and to traverse the boundary at some constant (and nonzero) angular rate. So if it starts at the origin (where the tails meet) it only has two choices, start moving along $A$ or start moving along $B$. These represent the two different orientations that this planar object can potentially have.

Now consider, if you were to rotate this parallelogram to some other position in $3$-space, will that affect the direction that this particle moves (if the particle were moving in what we'll call the "$A$" direction to begin with, will the particle suddenly start moving in the "$B$" direction if we rotate the entire parallelogram)?

That's how I visualize it.

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Make an orthogonal triple of vectors with your index and middle fingers and thumb, and freely rotate your hand. You can apply any rotation to your hand you like, but doing so never changes it into a left hand.

Edit We can be slightly more rigorous: One can continuously rotate any oriented orthogonal basis of vectors to any other such basis, and so one can for any rotations $A$ there is a continuous family of rotations $\gamma(t)$ connecting the identity $I$ and $A$ (more precisely, the space $SO(3)$ of rotations is path-connected). Now, the map $\det: SO(3) \to \{\pm 1\}$ is continuous, and hence so is $\det \circ \,\gamma$; but its codomain is discrete, so it must be constant. Now, $\det I = 1$, and so (for all rotations $A$), $\det A = 1$.

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  • $\begingroup$ What exactly does into a left hand mean? $\endgroup$ – user162520 Apr 8 '15 at 16:08
  • $\begingroup$ It simply means that a reversed orientation would look like the same triple of vectors but made with one's left hand, and as rotating one's right hand around shows, rotations simply don't do this. $\endgroup$ – Travis Apr 9 '15 at 11:55

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