📚 The Triple Vector Product: Expansion and Simplification | 三重向量积的运算与化简
The vector triple product is an expression in which a vector is crossed with the result of another cross product, for example a × (b × c). It appears throughout geometry, mechanics and electromagnetism, and its simplification is a standard skill in IB Mathematics.
向量三重积是指一个向量与另一个叉积结果再做叉积的表达式，例如 a × (b × c)。它在几何、力学和电磁学中经常出现，如何化简它是 IB 数学中的标准技能。
1. What Is the Triple Vector Product? | 什么是三重向量积？
There are two common “triple products” involving three vectors. The scalar triple product a · (b × c) produces a number, while the vector triple product a × (b × c) produces another vector. The phrase “vector triple product” usually refers to the second type.
三个向量之间存在两种常见的“三重积”。标量三重积 a · (b × c) 得到一个数，而向量三重积 a × (b × c) 得到一个新的向量。“向量三重积”通常指后一种。
Because the cross product is not associative, the position of brackets completely changes the result. Thus a × (b × c) and (a × b) × c are generally different vectors.
由于叉积不满足结合律，括号的位置会完全改变结果。因此 a × (b × c) 与 (a × b) × c 通常不是同一个向量。
2. The BAC–CAB Rule | BAC–CAB 公式
The most important identity for simplifying a vector triple product is the BAC–CAB rule:
化简向量三重积最重要的恒等式是 BAC–CAB 规则：
a × (b × c) = b(a · c) − c(a · b)
On the right-hand side, a · c and a · b are scalars, so the result is a linear combination of b and c. The name comes from the order of letters: “BAC” minus “CAB”.
在等式右侧，a · c 与 a · b 都是标量，因此结果可以写成 b 和 c 的线性组合。公式名称源于字母顺序：“BAC”减去“CAB”。
For example, if a = (1, 2, 3), b = (0, 1, 2) and c = (2, 0, 1), then a · c = 5 and a · b = 8. Therefore a × (b × c) = 5b − 8c = (−16, 5, 2).
例如，若 a = (1, 2, 3)、b = (0, 1, 2)、c = (2, 0, 1)，则 a · c = 5 且 a · b = 8。所以 a × (b × c) = 5b − 8c = (−16, 5, 2)。
3. Why the Order Matters | 为什么顺序至关重要
Because the cross product is non-associative, the triple product also does not satisfy the usual associative property. In fact:
由于叉积不满足结合律，三重积也不满足通常的结合性质。事实上：
a × (b × c) = b(a · c) − c(a · b)
(a × b) × c = b(a · c) − a(b · c)
The two expressions are quite different. The second formula follows by applying the BAC–CAB rule to −c × (a × b), because (a × b) × c = −c × (a × b).
这两个表达式完全不同。第二个公式可以通过对 −c × (a × b) 使用 BAC–CAB 规则得到，因为 (a × b) × c = −c × (a × b)。
This is why brackets must never be omitted. Writing a × b × c is ambiguous and usually loses marks in an exam.
这正是括号绝不能省略的原因。直接写 a × b × c 会产生歧义，在考试中通常也会扣分。
4. A Sketch of the Proof | 公式证明思路
One way to prove the BAC–CAB rule is by direct expansion of components. Write a = (a₁, a₂, a₃), b = (b₁, b₂, b₃) and c = (c₁, c₂, c₃). Compute the components of b × c, then cross them with a, and finally compare with the corresponding components of b(a · c) − c(a · b).
证明 BAC–CAB 规则的一种方法是直接展开分量。设 a = (a₁, a₂, a₃)，b = (b₁, b₂, b₃)，c = (c₁, c₂, c₃)。先计算 b × c 的分量，再与 a 做叉积，最后与 b(a · c) − c(a · b) 的对应分量比较。
For example, the first component of a × (b × c) equals a₂(b₁c₂ − b₂c₁) − a₃(b₃c₁ − b₁c₃). After expanding, one obtains b₁(a₁c₁ + a₂c₂ + a₃c₃) − c₁(a₁b₁ + a₂b₂ + a₃b₃), which is exactly the first component of b(a · c) − c(a · b).
例如，a × (b × c) 的第一个分量为 a₂(b₁c₂ − b₂c₁) − a₃(b₃c₁ − b₁c₃)。展开后得到 b₁(a₁c₁ + a₂c₂ + a₃c₃) − c₁(a₁b₁ + a₂b₂ + a₃b₃)，这正是 b(a · c) − c(a · b) 的第一个分量。
This component-wise proof is easy to write but somewhat long. It is acceptable in an IB examination if you first state the result and then verify it with a simple numerical example.
这种分量证明容易书写，但篇幅较长。在 IB 考试中，如果你先写出结果，再用一个简单数值例子验证，通常也是可以接受的。
5. Geometric Interpretation | 几何意义
The BAC–CAB rule shows directly that a × (b × c) lies in the plane spanned by b and c, because it is a linear combination of b and c. Also, since the cross product is perpendicular to both of its factors, a × (b × c) is perpendicular to a.
BAC–CAB 规则直接表明 a × (b × c) 位于 b 与 c 张成的平面内，因为它是 b 和 c 的线性组合。同时，由于叉积垂直于它的两个因子，a × (b × c) 也与 a 垂直。
Geometrically, the vector triple product is a vector lying in the plane of b and c, perpendicular to a. Its size depends on the lengths of the vectors and the angles between them.
从几何上看，向量三重积是一个位于 b 与 c 所在平面上且垂直于 a 的向量。它的大小取决于各向量的长度以及它们之间的夹角。
This interpretation is useful in physics. For example, the Coriolis acceleration involves expressions of the form ω × (ω × r), which can be decomposed into radial and tangential components using exactly this identity.
这一几何解释在物理中很有用。例如，科里奥利加速度包含形如 ω × (ω × r) 的表达式，正是利用这一恒等式将其分解为径向和切向分量。
6. Special Cases and Degeneracies | 特殊情况与退化情形
Several special cases make the vector triple product easier to handle.
几种特殊情况会让向量三重积更容易处理。
- If b and c are parallel, then b × c = 0, so a × (b × c) = 0.
- 如果 b 与 c 平行，则 b × c = 0，因此 a × (b × c) = 0。
- If a is perpendicular to both b and c, then a · b = a · c = 0, so a × (b × c) = 0.
- 如果 a 同时垂直于 b 和 c，则 a · b = a · c = 0，因此 a × (b × c) = 0。
- If a is parallel to b, say a = kb, then a × (b × c) = kb(b · c) − c(kb · b) = k[(b · c)b − (b · b)c].
- 如果 a 与 b 平行，设 a = kb，则 a × (b × c) = kb(b · c) − c(kb · b) = k[(b · c)b − (b · b)c]。
These results are obtained directly by substituting special relationships into the BAC–CAB formula, so it is often faster to use the formula than to recompute the cross products by hand.
这些结果都是直接将特殊关系代入 BAC–CAB 公式得到的，因此使用公式通常比手工重算叉积更快。
7. Simplification Techniques | 化简技巧
When simplifying a vector expression, always keep the brackets and remember the non-associativity of the cross product. The following techniques are helpful.
化简向量表达式时，始终保留括号，并记住叉积不满足结合律。以下技巧很有用。
First, identify the “outer” vector and the two vectors inside the bracket. In a × (b × c), the outer vector is a, and the inner vectors are b and c. Apply BAC–CAB directly:
首先，确定“外部”向量和括号内的两个向量。在 a × (b × c) 中，外层向量是 a，内层向量是 b 与 c。直接应用 BAC–CAB：
a × (b × c) = b(a · c) − c(a · b)
Second, watch for repeated vectors. A very common special case is
第二，注意重复出现的向量。一个非常常见的特殊情形是
a × (a × b) = a(a · b) − b(a · a)
This follows by replacing c with b in the BAC–CAB rule, since a × (a × b) = a(a · b) − b(a · a).
这由在 BAC–CAB 规则中将 c 替换为 b 得到，因为 a × (a × b) = a(a · b) − b(a · a)。
Third, if the expression is long, expand it step by step. The cross product is distributive over addition and subtraction, so you may expand first and then collect like terms.
第三，如果表达式较长，就逐步展开。叉积对加减法满足分配律，因此可以先展开，再合并同类项。
Finally, learn the Jacobi identity, which is useful in advanced problems:
最后，记得雅可比恒等式，它在进阶问题中很有用：
a × (b × c) + b × (c × a) + c × (a × b) = 0
This identity can be verified by applying BAC–CAB to each term and adding the results.
这个恒等式可以通过对每一项使用 BAC–CAB 再相加来验证。
8. Worked Example 1: Direct Expansion | 示例1：直接展开计算
Let a = (1, 2, 3), b = (0, 1, 2) and c = (2, 0, 1). Compute a × (b × c) and verify the BAC–CAB rule.
设 a = (1, 2, 3)、b = (0, 1, 2)、c = (2, 0, 1)。计算 a × (b × c) 并验证 BAC–CAB 规则。
First compute b × c:
先计算 b × c：
b × c = (0, 1, 2) × (2, 0, 1) = (1×1 − 2×0, 2×2 − 0×1, 0×0 − 1×2) = (1, 4, −2)
Then cross with a:
再与 a 做叉积：
a × (1, 4, −2) = (2×(−2) − 3×4, 3×1 − 1×(−2), 1×4 − 2×1) = (−16, 5, 2)
Now use the formula: a · c = 1×2 + 2×0 + 3×1 = 5 and a · b = 1×0 + 2×1 + 3×2 = 8. Thus
现在用公式：a · c = 1×2 + 2×0 + 3×1 = 5，a · b = 1×0 + 2×1 + 3×2 = 8。于是
b(a · c) − c(a · b) = 5(0, 1, 2) − 8(2, 0, 1) = (0, 5, 10) − (16, 0, 8) = (−16, 5, 2)
Both methods give the same vector, confirming the identity.
两种方法得到同一个向量，从而验证了恒等式。
9. Worked Example 2: Using the Identity | 示例2：利用恒等式化简
Prove that (a × b) × (a × c) = (a · (b × c)) a.
证明 (a × b) × (a × c) = (a · (b × c)) a。
Set u = a × b. Then the left-hand side becomes u × (a × c). Applying BAC–CAB with outer vector u and inner vectors Published by TutorHao | IB Mathematics Revision Series | aleveler.com
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