📚 Coplanarity of Vectors | 向量的共面性
Vectors are among the most visual topics in A-Level Mathematics. When we move from two dimensions into three, a new geometric question appears: do three given vectors lie in the same flat plane? This property is called coplanarity, and AQA examiners test it through the scalar triple product, linear combinations and the coordinate geometry of planes. In this article we develop every required method from first principles, with fully worked AQA-style examples.
向量是A-Level数学中最具直观性的主题之一。当从二维进入三维空间时，一个新的几何问题出现了：给定的三个向量是否位于同一个平面内？这个性质称为共面性。AQA考官会通过标量三重积、线性组合以及平面的坐标几何来考查这一概念。本文将从头推导每一个必备方法，并提供完整的AQA风格例题。
1. What Does "Coplanar" Mean? | “共面”的含义
Two vectors are always coplanar, because any two non-parallel vectors define a plane by themselves, and parallel vectors lie on infinitely many planes. The interesting case involves three vectors. Three vectors a, b and c are called coplanar if they are all parallel to one and the same plane. Geometrically, if we draw them from a common origin, their terminal points lie in a single plane through that origin.
两个向量始终是共面的，因为任意两个不平行向量本身就能确定一个平面，而平行向量则位于无穷多个平面中。真正值得关注的情况涉及三个向量。三个向量 a、b、c 称为共面，当且仅当它们都平行于同一个平面。从几何上看，若将它们从同一原点画出，则它们的终点位于经过该原点的同一个平面内。
s c lies in the plane spanned by a and b. This simple idea is the key to the algebraic tests below.
等价地，三个向量共面当且仅当其中一个向量可以写成另外两个向量的线性组合。例如，c = 2a + 3b 意味着 c 位于由 a 和 b 张成的平面内。这个简单的想法正是下面各种代数检验方法的关键。
2. The Scalar Triple Product Test | 标量三重积检验法
versely, if a·(b×c) = 0 and b and c are not parallel, then a lies in the plane of b and c.
检验共面性最有力的工具是标量三重积 a·(b×c)。回顾 b×c 得到的是垂直于 b、c 所在平面的法向量。若 a 也平行于该平面，则 a 必垂直于法向量，因此它们的点积为零。反之，若 a·(b×c) = 0 且 b、c 不平行，则 a 位于 b、c 所确定的平面内。
ponent form, the scalar triple product equals the determinant of the matrix with these rows:
将向量写成分量形式 a = (a₁, a₂, a₃)，b = (b₁, b₂, b₃)，c = (c₁, c₂, c₃)，则标量三重积等于以这些向量为行向量的矩阵的行列式：
a·(b×c) = det = | a₁ a₂ a₃ ; b₁ b₂ b₃ ; c₁ c₂ c₃ |
Three vectors are coplanar if and only if this determinant equals zero. This test is quick, reliable and forms the backbone of most AQA questions on the topic.
三个向量共面当且仅当该行列式为零。这一检验快速可靠，是AQA相关题目中最核心的方法。
3. The Linear Combination Method | 线性组合方法
μb, assuming a and b are not parallel. Writing this in components gives three simultaneous equations:
另一种方法是求解方程组。三个向量 a、b、c 共面当且仅当存在标量 λ 和 μ 使得 c = λa + μb（假设 a、b 不平行）。将其写成分量形式，得到三个联立方程：
c₁ = λa₁ + μb₁, c₂ = λa₂ + μb₂, c₃ = λa₃ + μb₃
Use the first two equations to solve for λ and μ. If the resulting values also satisfy the third equation, the vectors are coplanar; if not, they are not coplanar.
用前两个方程解出 λ 和 μ。若求出的值也满足第三个方程，则向量共面；若不满足，则不共面。
This method is especially useful when one vector contains an unknown parameter, because it turns the question into a simple substitution problem instead of a determinant expansion.
当某个向量含有未知参数时，这种方法尤其好用，因为它把问题转化为简单的代入运算，而不是展开行列式。
4. Coplanarity of Four Points | 四点的共面性
Often the exam asks whether four points A, B, C and D lie in the same plane. The key is to translate points into vectors. Define:
考试中常会问四个点 A、B、C、D 是否位于同一平面内。关键在于把点转化为向量。定义：
AB = b − a, AC = c − a, AD = d − a
The four points are coplanar if and only if the three vectors AB, AC and AD are coplanar, because all three share the common starting point A. Therefore the test is:
四个点共面当且仅当三个向量 AB、AC、AD 共面，因为它们都以 A 为共同起点。因此检验条件为：
(AB × AC)·AD = 0
This single equation is the entire method. Notice that it is essential to use vectors starting from the same point; using position vectors of A, B, C, D directly would only test whether the origin lies in their plane, which is a different question.
这一个方程就是完整的方法。注意必须使用从同一点出发的向量；直接使用 A、B、C、D 的位置向量只能检验原点是否位于它们的平面内，这是完全不同的问题。
5. The Equation of a Plane | 平面的向量方程
Understanding planes helps you see why coplanarity tests work. A plane through point A with position vector a, spanned by two non-parallel direction vectors u and v, has equation:
理解平面的方程有助于看清共面性检验的原理。过点 A（位置向量为 a）、由两个不平行方向向量 u、v 张成的平面，其方程为：
r = a + λu + μv
. If three vectors a, b, c are coplanar, they can all be written as such expressions with the same two directions — which is exactly the linear combination condition.
平面上任意点的位置向量 r 都具有这种形式。等价地，法向量 n = u × v 给出类似直角坐标形式 n·(r − a) = 0。若三个向量 a、b、c 共面，则它们都能写成具有相同两个方向的上述表达式——这正是线性组合条件。
tors u = b − a and v = c − a, then compute n = u × v. The plane equation is n·(r − a) = 0.
要求过三点 A、B、C 的平面方程，先构造方向向量 u = b − a 和 v = c − a，再计算 n = u × v。平面方程为 n·(r − a) = 0。
6. Worked Example 1: Applying the Scalar Triple Product | 例题1：应用标量三重积
Question. Show that the vectors a = i + 2j − k, b = 2i − j + 3k and c = 3i + j + 2k are coplanar.
题目。证明向量 a = i + 2j − k，b = 2i − j + 3k，c = 3i + j + 2k 共面。
Solution. and expand. With a₁ = 1, a₂ = 2, a₃ = −1, b₁ = 2, b₂ = −1, b₃ = 3, c₁ = 3, c₂ = 1, c₃ = 2:
解答。将向量写成行列式的行并展开。令 a₁ = 1，a₂ = 2，a₃ = −1，b₁ = 2，b₂ = −1，b₃ = 3，c₁ = 3，c₂ = 1，c₃ = 2：
det = 1[(−1)(2) − (3)(1)] − 2[(2)(2) − (3)(3)] + (−1)[(2)(1) − (−1)(3)]
= 1(−2 − 3) − 2(4 − 9) + (−1)(2 + 3) = −5 + 10 − 5 = 0
, 1, 2) = c. Hence c = a + b, confirming coplanarity through a linear combination as well.
由于标量三重积 a·(b×c) = 0，三个向量共面。从几何上看，可验证线性组合：c = a + b，即 (1+2, 2−1, −1+3) = (3, 1, 2) = c。因而 c = a + b，同样验证了共面性。
7. Worked Example 2: Finding an Unknown Parameter | 例题2：求未知参数
Question. j + k and c = 3i + pj + 4k, find the value of p for which the three vectors are coplanar.
题目。已知 a = i + j + 2k，b = 2i − j + k，c = 3i + pj + 4k，求使三个向量共面的 p 值。
Solution. Set up the determinant and require it to equal zero:
解答。构造行列式并令其等于零：
| 1 1 2 ; 2 −1 1 ; 3 p 4 | = 0
3)] = −(8 − 3) = −5. The third minor is (2)(p) − (−1)(3) = 2p + 3, multiplied by 2. Hence:
沿第一行展开。第一个余子式为 (−1)(4) − (1)(p) = −4 − p。第二个余子式带负号，为 −[(2)(4) − (1)(3)] = −(8 − 3) = −5。第三个余子式为 (2)(p) − (−1)(3) = 2p + 3，乘以 2。因此：
det = (−4 − p) − 5 + 2(2p + 3) = −4 − p − 5 + 4p + 6 = 3p − 3
etting det = 0 gives 3p − 3 = 0, so p = 1. With p = 1, all three vectors lie in one plane.
令 det = 0，得 3p − 3 = 0，故 p = 1。当 p = 1 时，三个向量位于同一平面内。
8. Worked Example 3: Plane Through Three Points | 例题3：过三点的平面方程
Question. Find the equation of the plane passing through A(1, 0, 2), B(2, 1, 1) and C(0, 2, 3).
题目。求过点 A(1, 0, 2)、B(2, 1, 1)、C(0, 2, 3) 的平面方程。
Solution. Form two direction vectors in the plane:
解答。构造平面内的两个方向向量：
AB = (2−1, 1−0, 1−2) = (1, 1, −1), AC = (0−1, 2−0, 3−2) = (−1, 2, 1)
Compute the normal vector n = AB × AC:
计算法向量 n = AB × AC：
n = ( (1)(1) − (−1)(2), (−1)(−1) − (1)(1), (1)(2) − (1)(−1) ) = (3, 0, 3)
) = 0 with a = (1, 0, 2): 3(x − 1) + 0(y − 0) + 3(z − 2) = 0. Expanding and dividing by 3:
利用 n·(r − a) = 0，其中 a = (1, 0, 2)：3(x − 1) + 0(y − 0) + 3(z − 2) = 0。展开并除以 3：
x + z = 3
C gives 0 + 3 = 3. All three points satisfy the equation, confirming the plane is correct.
检验：A 代入得 1 + 2 = 3，B 代入得 2 + 1 = 3，C 代入得 0 + 3 = 3。三个点均满足方程，验证平面方程正确。
9. Common Exam Pitfalls | 常见考点陷阱
Students lose marks in AQA exams for several consistent reasons. First, they confuse parallel vectors with coplanar vectors. Two parallel vectors are always coplanar with any third vector; only three non-parallel vectors can fail to be coplanar. Second, sign errors in determinant expansion are extremely common, especially with the middle term, which must carry a minus sign.
学生在AQA考试中常因几个固定原因失分。第一，混淆平行向量与共面向量。两个平行向量与任意第三个向量总是共面的；只有三个不平行向量才可能不共面。第二，展开行列式时符号错误极为常见，尤其是中间项必须带负号。
Third, when testing the coplanarity of four points, students sometimes substitute the position vectors of the points into the scalar triple product without first forming AB, AC and AD. This tests whether the origin is coplanar with the points, not whether the points themselves are coplanar. Always subtract the same reference point.
第三，检验四点共面时，学生有时不先构造 AB、AC、AD，而直接把各点的位置向量代入标量三重积。这检验的是原点是否与这些点共面，而不是这些点本身是否共面。务必先减去同一个参考点。
Fourth, when using the linear combination method, students forget to check the third equation. Two of the component equations may give a consistent pair of values even when the vectors are not coplanar; all three equations must be satisfied simultaneously.
第四，使用线性组合方法时，学生忘记检查第三个方程。即使向量不共面，前两个分量方程也可能给出一组一致的 λ、μ 值；必须三个方程同时被满足才行。
10. Summary of Key Results | 关键结论总结
× AC)·AD = 0. The plane through A, B, C has normal n = AB × AC and equation n·(r − a) = 0.
三个向量 a、b、c 共面当且仅当 a·(b×c) = 0，等价于它们分量的行列式为零，也等价于其中一个向量是另外两个的线性组合。四个点 A、B、C、D 共面当且仅当 (AB × AC)·AD = 0。过 A、B、C 的平面，其法向量 n = AB × AC，方程为 n·(r − a) = 0。
The table below summarises the methods and when to use each one.
下表总结了各方法及其适用时机。
| Method | 方法 | Test | 检验条件 | Best used when | 最佳使用场合 |
| Scalar triple product 标量三重积 | a·(b×c) = 0 | All vectors fully known 所有向量完全已知 |
| Linear combination 线性组合 | c = λa + μb has a solution | Unknown parameter present 含未知参数 |
| Four points 四点 | (AB × AC)·AD = 0 | Points given as coordinates 给出点坐标 |
Whatever method you choose, always state the zero result explicitly and conclude with a full sentence, because AQA marking schemes award the communication mark only when the conclusion is clearly linked to the calculation.
无论选择哪种方法，都要明确写出结果为零的结论，并用完整句子作出判定，因为AQA评分标准只在结论与计算过程清晰关联时才会给表达分。
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