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Because $\vec{a}$, $\vec{b}$ and $\vec{a}\wedge\vec{b}$ are a right hand triple of vectors. And, if $(\vec{a},\vec{b},\vec{c})$ is a right hand (ordered) triple of vectors, then also $(\vec{b},\vec{a},-\vec{c})$ is, while $(\vec{b},\vec{a},\vec{c})$ is not. Indeed not only cross product is not commutative, further it is anticommutative. See, also, the ...

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The magnitude of the resulting vector is a function of the angle between the vectors you are multiplying. The key issue is that the angle between two vectors is always measured in the same direction (by convention, counterclockwise). Try holding your left thumb and index finger in an L shape. Measuring counterclockwise, the angle between your thumb and ...

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Since two vectors are perpendicular to any two non parallel vectors, and these vectors are in opposite directions, it makes sense to decide which one is to be the result of the cross product. So the convention was adopted to follow a right hand triad, as opposed to a left hand triad.

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As others have said, if that's what the textbook says then it's wrong because the vector product is not associative . This may have arisen due to some confusion between $\times$ and $\cdot$, though. Generally, there is a formula for the vector triple product in terms of the scalar product, given by $$\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec ... 0 There must be some mistake in the book since the cross product is not associative see https://proofwiki.org/wiki/Cross_Product_is_not_Associative And a\times a=0 0 There must be some mistake in the book since the cross product is not associative see https://proofwiki.org/wiki/Cross_Product_is_not_Associative 0 You're right. The first step is incorrect. 0 \newcommand{vec}[1]{\hat{\bf #1}} In the linked paper, Equation (18),$$\left(\frac{\vec W_2-(\vec W_1\cdot\vec W_2)\vec W_1} {|\vec W_2-(\vec W_1\cdot\vec W_2)\vec W_1|}\right) = A \left(\frac{\vec V_2-(\vec V_1\cdot\vec V_2)\vec V_1} {|\vec V_2-(\vec V_1\cdot\vec V_2)\vec V_1|}\right), $$is apparently not derived from ... 0 You have$$ \begin{align} \mathbf W_2-(\mathbf W_1\cdot\mathbf W_2)\mathbf W_1 &= A \mathbf V_2-(A \mathbf V_1\cdot A\mathbf V_2)A \mathbf V_1 \\ &= A \mathbf V_2-(\mathbf V_1\cdot \mathbf V_2)A \mathbf V_1 \\ &= A (\mathbf V_2-(\mathbf V_1\cdot \mathbf V_2) \mathbf V_1) \end{align} $$where the first equals sign is by definition, the second is ... 0 To get a gemoetric interpretation, convert the above relation into geometric algebra. Whence; $$a \times (b \times c) = -a \rfloor (b \wedge c)$$ The b \wedge c (\equiv X) part is a bivector; it is to be understood as an area element in the bc plane. Then a \rfloor X is a left contraction of a onto X, which is dot ... 0 Up to a rotation around the \hat x axis, you can always represent the punctual situation as you did in your drawing. Now: \hat r lies in the plane of the sheet of paper (the one spanned by the radius \vec R and by \hat x), while d\vec l is perpendicular to the sheet of paper. Hence, to all the vectors lying in it. 0 take a look at the point were the vector 'r' meets the circular loop....at that point this 'r' is perpendicular to the tangent of that circular loop at that point.... in the law the term 'dl ' is the current carrying element ..... here it's that tangent now it's clear that the angle is 90 ...and it's true for all the points on that loop 1 The dot product is:$$ \mathbf V \cdot \mathbf U = V_rU_r + V_\phi U_\phi + V_\theta U_\theta $$For the cross product take the determinant of the matrix with unit vectors and components, the same way you would for cartesian coordinates:$$ \mathbf V \times \mathbf U = \begin{vmatrix} \mathbf{\hat r}&\mathbf{\hat \phi}& \mathbf{\hat \theta}\\ V_r ...

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Although the map $(x,y)\to *(x\wedge y)$ doesn't generalize to a map $\mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}^n$ in general, not every cross-product has to be generated as such. In fact, the cross-product on $\mathbb{R}^3$ is unique modulo sign, so you could think of the construction you outline as just an effect of that uniqueness: anything that ...

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