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2

We have the following identities for vectors $\vec{x}$,$\vec{y}$,$\vec{z}$: $$\vec{x}\cdot(\vec{y}\times\vec{z})=\vec{y}\cdot(\vec{z}\times\vec{x})=\vec{z}\cdot(\vec{x}\times\vec{y})$$ and $$\vec{x}\times (\vec{y}\times\vec{z})=\vec{y}\;(\vec{x}\cdot \vec{z})-\vec{z}\;(\vec{x}\cdot \vec{y})$$ Let $\vec{a}\times\vec{b} = \vec{x}$: ...

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Note that $[\overrightarrow a \times \overrightarrow b] \perp \overrightarrow a, \overrightarrow b$ so that $\overrightarrow a \times [\overrightarrow a \times \overrightarrow b]$ is a vector of magnitude $|\overrightarrow a| |\overrightarrow a \times \overrightarrow b| = a |\overrightarrow a \times \overrightarrow b|$ that is both $\perp \overrightarrow a$ ...

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Note: one need to be careful when working in spherical coordinates, as there are two conventions used for the notation of the polar and azimuthal angles. See https://en.wikipedia.org/wiki/Spherical_coordinate_system and the 2 pictures near the top. In this answer I am using the physicist's convention, since it complies better with your own notation. To ...

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Generally speaking, I don't think it's helpful to think of vectors as "starting somewhere". It can be helpful e.g. when you're adding vectors; then you can imagine one "starting" where the other one "ends"; but more often than not I'd say it's best not to think in these terms. In your case, if you insist on wanting to visualize the vector as "starting ...

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The bigger picture is this: Since the quaternion norm $|\cdot|$ is multiplicative, that is, because $$|\dot{q}_1 \dot{q}_2| = |\dot{q}_1| |\dot{q}_2|,$$ the set of unit quaternions forms a group, which we usually denote $SU(2)$ (a special unitary group) or sometimes $Spin(3)$ (a spin group). As a topological space, the set of unit quaternions is just the ...

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No, $\def\bfa{{\bf a}}\def\bfb{{\bf b}}\def\bfc{{\bf c}}\bfa \times \bfb \times \bfc$ simply doesn't make sense for vectors in $\Bbb R^3$, precisely because $\times$ is nonassociative. The cross product does, however, satisfy the so-called Jacobi identity, $$\bfa \times (\bfb \times \bfc) + \bfb \times (\bfc \times \bfa) + \bfc \times (\bfa \times \bfb) = ... 1 No, there is no such convention and, since the cross product is not associative, "\vec{a}\times\vec{b}\times\vec{c}" simply has no meaning. You must write either (\vec{a}\times\vec{b})\times\vec{c} or \vec{a}\times(\vec{b}\times\vec{c}) to distinguish the two. 4 The notation \vec{a}\times \vec{b}\times \vec{c} would be meaningful if and only if you could say that$$(\vec{a}\times \vec{b})\times \vec{c}=\vec{a}\times(\vec{b}\times \vec{c})$$for any three vectors \vec{a},\vec{b} and \vec{c} but it is not true in general as you have noted in your answer. In fact, due to the non-associativity of the products it ... 0 You do it the same way as in 3D, but you do a 4*4 matrix, whose rows are the three vectors, and the vector (w, x, y, z). Then, find the determinate of them. The usual rules apply: the exchange of any pair of vectors will produce the negative, etc, but a cyclic permutation of A, B, C, will not. The dot product of V(A,B,C) and D gives the volume of a ... 1 Eckmann's definition of a cross product on a finite-dimensional, positive definite, real inner product space \def\bfv{{\bf v}}(\Bbb V, \langle\,\cdot\, , \,\cdot\,\rangle) (say, \dim \Bbb V = n) is a map$$\times: \Bbb V^r \to \Bbb Vfor some r \in \{1, \ldots, n\} satisfying \begin{align} \langle \times(\bfv_1, \ldots, \bfv_r), \bfv_a \rangle &= ... 0 It is a matter of convention deriving from the convention on orientation of x,y,z axis (  \vec i, \vec j, \vec k) , so that \vec k= \vec i \times \vec j= \left |\begin{matrix} \vec i&\vec j&\vec k\\ 1&0&0\\0&1&0 \end{matrix} \right| I don't know why the right hand was chosen as positive, but i love to think that this has ... 0 This is just a convention. We could use the opposite convention, and everything would work fine. However, the usual convention has the following nice property. For any two linearly independent vectors u,v, if we let A denote the matrix whose columns are u,v,u\times v, then \det A>0. 0 There is no mathematical way to distinguish between the two directions, they are perfectly equivalent. But in the physical applications, there is a difference. Therefore by convention we use the right hand rule. Why right and not left could be for any of the following reasons: Historically the right hand was considered "better" than the left In ... 0 Yes, it is. Another motivation is the following: in three dimensions (and only in three dimensions) a plane has only one normal direction. The angle between the normal direction and any vector v has a geometric interpretation as "exposure" of the plane to v. For example, a photo film is maximally exposed to the sun if its normal vector is parallel to ... 1 Counterexample:\mathbf{x}:=(1,0,...,0)^T,\mathbf{y}:=(0,...,0,1)^T\implies \mathbf z:=\mathbf{x}\times\mathbf{y}=(0,0,0,0,0,1,0)^T\\ R_{ij}:={1\over\sqrt{2}}(\delta_{i1}\delta_{j1}+\delta_{i2}\delta_{j2}-\delta_{i1}\delta_{j2}+\delta_{i2}\delta_{j1})+\delta_{ij}(1-\delta_{i1})(1-\delta_{i2})\\ R\mathbf{x}={1\over\sqrt{2}}(1,1,0,0,0,0,0)^T\\ ...

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Look. I'm not a mathematician, but I have a perspective which can explain why the cross product of two vectors is another vector perpendicular to them. It is not a proof but it will help make that idea fimilar. One can understand cross product in this way:imagine a line segment that makes colorful marks wherever it moves on a paper. Now make it move in the ...

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