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  "body": "\n## Proof\n\nWe compute both dot products from the coordinate formula.\n\n1. Let $a = (a_1, a_2, a_3)$ and $b = (b_1, b_2, b_3)$. By the definition of the [[cross product|0N5N15HWZ7]], $a \\times b = (a_2 b_3 - a_3 b_2, \\; a_3 b_1 - a_1 b_3, \\; a_1 b_2 - a_2 b_1)$.\n2. By the definition of the [[dot product|0S52TNNTH9]],\n$$a \\cdot (a \\times b) = a_1 (a_2 b_3 - a_3 b_2) + a_2 (a_3 b_1 - a_1 b_3) + a_3 (a_1 b_2 - a_2 b_1).$$\n3. We expand the right side. It is $a_1 a_2 b_3 - a_1 a_3 b_2 + a_2 a_3 b_1 - a_1 a_2 b_3 + a_1 a_3 b_2 - a_2 a_3 b_1$.\n4. The six terms cancel in pairs: $a_1 a_2 b_3$ with $-a_1 a_2 b_3$, $a_1 a_3 b_2$ with $-a_1 a_3 b_2$, and $a_2 a_3 b_1$ with $-a_2 a_3 b_1$. So $a \\cdot (a \\times b) = 0$.\n5. In the same way,\n$$b \\cdot (a \\times b) = b_1 (a_2 b_3 - a_3 b_2) + b_2 (a_3 b_1 - a_1 b_3) + b_3 (a_1 b_2 - a_2 b_1).$$\n6. Expanded, this is $a_2 b_1 b_3 - a_3 b_1 b_2 + a_3 b_1 b_2 - a_1 b_2 b_3 + a_1 b_2 b_3 - a_2 b_1 b_3$.\n7. Again the six terms cancel in pairs. So $b \\cdot (a \\times b) = 0$.\n\n## Idea\n\nEach coordinate of $a \\times b$ is a $2 \\times 2$ determinant built from $a$ and $b$. The dot product with $a$ is the expansion of a $3 \\times 3$ determinant with the rows $a$, $a$, $b$. A determinant with two equal rows is zero, which is why the terms cancel in pairs.\n",
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