I know the cross product follows the RHR. I know it has to be normal to the plane defined by both vectors, and can prove this. But I can't prove it obeys the RHR. I saw in another post that it is sufficient to prove that $\hat i \times \hat j= \hat k$ in order to say it always follows the RHR. I don't see why this is the case.
Can anyone prove to me why the cross product always obeys the RHR?
Thanks.
Here's an informal proof that if $u$ and $v$ are orthogonal unit vectors, then $w = u \times v$ follows the right hand rule:
Observe that it's true that for $u = i$ and $v = j$, it works. You could imagine placing a plaster cast of a right-hand with the index finger along $i$, the pinky extended along $j$, and the thumb pointing along $k$.
Observe that for any rotation $R$ and any vectors $p$ and $q$, $Rp \times Rq$ is $R(p \times q)$. [That's just a computation with a lot of algebra.]
Observe that you can always find a rotation $R$ that sends $i$ to $u$. It'll send $j$ to some vector $v'$ perpendicular to $u$. Now rotate about the vector $u$ until $v'$ aligns with $v$. Call that $S$. The operation "rotate by $R$ and then $S$" can be applied not only to $i$ and $j$ but also to the entire plaster hand. By item 2, the vector $z$ will be rotated, by this same pair of rotations, to the direction of the thumb.
Here's a slightly more formal proof, again for the case where $u$ and $v$ are unit vectors, perpendicular to each other (which I'll call a 2-frame in 3-space).
The entries of the cross product $w = u \times v$ are polynomials in the entries of $u$ and $v$, so thinking of $w$ as a function of $u$ and $v$, it's a continuous function.
Of the two unit vectors perpendicular to both $u$ and $v$, one satisfies the right hand rule and one does not. The distance between these two is exactly $2$ (because they're negatives of one another).
The right-hand-rule vector, $R(u, v)$, for any pair of vectors $u$ and $v$, varies continuously as a function of $u$ and $v$: if you rotate $u$ a little bit about $v$, the right-hand vector will also rotate a little bit. More formally, if $u'$ and $v'$ are distance less than 0.1 from $u$ and $v$ respectively, then $R(u', v')$ and $R(u, v)$ differ by less than $1$ (i.e., the distance between them is a vector of length less than one).
For $u = i, v = j$, the right-hand-rule vector $R(u, v) = k$.
So: for $u$ near $i$ and $v$ near $j$, the right-hand rule vector must be near $k$. Since $R(u,v)$ must be either $u \times v$ or $-u \times v$, and it's near $k$, it must be $u \times v$.
The set of $(u,v)$ pairs for which $R(u,v) = u \times v$ is an open set in the space of all 2-frames in 3-space, by an extension of the argument in part 4. It's also a closed set, so it must be a connected component of the space of frames. Since any two frames can be connected by a path of frames, the whole space of 2-frames is path-connected, hence connected, so the connected component must be the whole space of 2-frames. In short: the set of frames $(u, v)$ where $R(u, v) = u \times v$ is all frames.
This is simply a convention following Hamilton's non-commutative quaternions where he set $ij=k$, $jk=i$ and $ki=j$. Taking $i$ as the thumb, $j$ as the index finger and $k$ as the middle finger, as with the convention for $x,y,z$ axes, gives the right hand rule.
If Hamilton had started with $ji=k$, $kj=i$ and $ik=j$ then there would be a left hand rule instead.
The vector part of quaternion multiplication led to the cross-product.
