If you think about the dot product in the following (somewhat odd way), the triple scalar product identity becomes a triviality: Dot product-ing replaces standard unit vectors with corresponding vector components:

$$ \epsilon_{ijk} a_j b_k \hat{e}_i \times \epsilon_{lmn} a_m c_n \hat{e}_l \\ = \epsilon_{rst} (\epsilon_{ijk} a_j b_k \hat{e}_i \cdot \hat{e}_s)(\epsilon_{lmn} a_m c_n \hat{e}_l \cdot \hat{e}_t) \hat{e}_r $$ The second step merely expresses the fact that the cross-product formula requires the s-th component of the first vector and the t-th ...

2022年5月17日 · Now by expanding the scalar triple product by determinants and arranging in both cases, I can easily prove that it remains same . But what I am asking is an more algebraic and elegant way . I mean , without just simply expanding and without geometrical interpretation. I meant , I want a prove involving only vectors like using properties of ...

2014年7月28日 · Scalar triple product in terms of scalar products only. Ask Question Asked 10 years, 5 months ago. ...

2020年6月19日 · The Wikipedia article for scalar triple product says the following: Geometrically, the scalar triple product $$\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})$$ is the (signed) volume of the parallelepiped defined by the three vectors given. Here, the parentheses may be omitted without causing ambiguity, since the dot product cannot be evaluated ...

2023年12月8日 · Scalar Triple/Box Product. Ask Question Asked 1 year, 2 months ago. Modified 1 year, 2 months ago. Viewed ...

2024年8月8日 · I am computing the scalar triple product of two vectors a, b, and their orthogonal cross product c. A special case of this is where a x b = c, which leads to the scalar triple product equalling 0. My textbook states: The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product:

2015年6月18日 · When you compute the triple product, you are doing as many calculus as finding out the equation and verifying that D belongs to the plane. $\endgroup$ – Martigan Commented Jan 19, 2016 at 11:37

Using the properties of the vector triple product and the scalar triple product,prove that. 0.

2021年4月15日 · In the book I'm reading, it claims that we also have an isomorphism with the space of $2$-form given by $$\omega_v^2(w_1,w_2)=(v,w_1,w_2),$$ where the right hand side is the triple scalar product. Notably they remark that the isomorphisms above do not depend on the choice of oriented basis, but they both do depend on the choice of scalar product.