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  "body": "\n## Proof\n\nWe expand both sides and compare them.\n\n1. Let $a = (a_1, a_2, a_3)$, $b = (b_1, b_2, b_3)$ and $c = (c_1, c_2, c_3)$. Let $M$ be the $3 \\times 3$ matrix with the rows $a$, $b$ and $c$.\n2. By the definition of the [[cross product|0N5N15HWZ7]], $b \\times c = (b_2 c_3 - b_3 c_2, \\; b_3 c_1 - b_1 c_3, \\; b_1 c_2 - b_2 c_1)$.\n3. By the definition of the [[dot product|0S52TNNTH9]],\n$$a \\cdot (b \\times c) = a_1 (b_2 c_3 - b_3 c_2) + a_2 (b_3 c_1 - b_1 c_3) + a_3 (b_1 c_2 - b_2 c_1).$$\n4. By the [[Laplace expansion along the first row|0C08BQ3M05]], $\\det M = a_1 M_{11} - a_2 M_{12} + a_3 M_{13}$. Here $M_{1j}$ is the [[minor|0XPKR18APA]]: the [[determinant|00KDZRW4C0]] of $M$ without row $1$ and column $j$.\n5. By the [[formula for a $2 \\times 2$ determinant|0ZN2J42SK7]], the minors are\n$$M_{11} = b_2 c_3 - b_3 c_2, \\quad M_{12} = b_1 c_3 - b_3 c_1, \\quad M_{13} = b_1 c_2 - b_2 c_1.$$\n6. Thus $\\det M = a_1 (b_2 c_3 - b_3 c_2) - a_2 (b_1 c_3 - b_3 c_1) + a_3 (b_1 c_2 - b_2 c_1)$.\n7. The middle term is $-a_2 (b_1 c_3 - b_3 c_1) = a_2 (b_3 c_1 - b_1 c_3)$. So the expression in step 6 is the expression in step 3.\n8. Hence $a \\cdot (b \\times c) = \\det M$, which is the statement.\n\n## Idea\n\nThe coordinates of $b \\times c$ are exactly the signed $2 \\times 2$ minors that the Laplace expansion along the first row multiplies by $a_1$, $a_2$ and $a_3$. So the dot product with $a$ is this expansion term by term.\n",
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