# Tag Info

1

$T:\Bbb R^3 \to \Bbb R^3$ is given by $T(\mathbf v) = \mathbf a \times \mathbf v = \det\begin{bmatrix} \mathbf e_1 & \mathbf e_2 & \mathbf e_3 \\ a_1 & a_2 & a_3 \\ v_1 & v_2 & v_3 \\ \end{bmatrix}$ where $\{\mathbf e_1, \mathbf e_2, \mathbf e_3\}$ are the standard basis vectors for $\Bbb R^3$. For ...

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It is the same symbol used for different concepts; unfortunatelty there are too many "products" in mathematics. The "extension" of the product operation from numbers to vectors, gives rise to two different operations; see : Josiah Willard Gibbs (1839 – 1903), Vector Analysis (textbook by E.B.Wilson, first published in 1901 and based on the lectures that ...

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When you believe that the scalar product of two non-zero vectors is 0 exactly when they are perpendicular, then its easy. You just have to evaluate: $\mathbf{a}\cdot(\mathbf{b} \times \mathbf{c})$ for either $\mathbf{a} = \mathbf{b}$ or $\mathbf{a} = \mathbf{c}$. You can do this using the coordinate expansions you mentioned above or using algebraic ...

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If you learn index (summation) notation, then you'll be able to prove it without writing out every term: Consider an arbitrary $p$th coordinate of the vector $A\times (B\times C)$. Then \begin{align} [A\times (B\times C)]_p &= \varepsilon_{pqr}A_q(B\times C)_r \\ &= \varepsilon_{pqr}A_q\varepsilon_{rst}B_sC_t \\ &= ...

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You have expanded the left side. Now, expand the right side. With any luck, the two expansions will agree. A note: I would have written the terms as $(x, y, z)$ instead of using hats.

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