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  "body": "\n## Statement\n\nLet $a$ and $b$ be vectors in $\\mathbb{R}^3$, and let $\\times$ be the [[cross product|0N5N15HWZ7]]. Then\n$$b \\times a = -(a \\times b).$$\n\n## Remarks\n\nWith $b = a$ the statement gives $a \\times a = -(a \\times a)$, thus $a \\times a = 0$ for each vector $a$.\nThe proof is a direct computation: each component of $b \\times a$ is the component of $a \\times b$ with the two products in the other order.\nThe cross product is thus not commutative. For example, $e_1 \\times e_2 = e_3$ but $e_2 \\times e_1 = -e_3$.\n\n## Explanations\n\n### L1 human\n\nIf you swap the two factors of a cross product, the result points the other way. Its length stays the same. The right-hand rule shows this: swapping the factors turns your hand over.\n",
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