📚 Vector Equations: Formulation and Solving | 向量方程的建立与求解
In vector geometry, a vector equation is an equation that describes a geometric locus such as a line or a plane using position vectors and direction vectors. It provides a compact and flexible algebraic method for solving problems involving points, lines, and planes in two or three dimensions.
在向量几何中，向量方程是用位置向量和方向向量描述直线或平面等几何轨迹的方程。它为求解二维或三维空间中涉及点、直线和平面的问题提供了一种紧凑而灵活的代数方法。
This article covers the formulation of vector equations for lines and planes, and demonstrates how to solve them to find intersections, test points, and apply them in typical examination problems. The techniques are essential for pure mathematics courses such as A-Level Mathematics and Further Mathematics.
本文将介绍直线与平面向量方程的建立方法，并通过例题演示如何求解这些方程以确定交点、检验点是否位于直线或平面上，以及处理考试中的典型问题。这些技巧是 A-Level 数学和进阶数学等纯数学课程的核心内容。
1. Position Vectors and Displacement Vectors | 位置向量与位移向量
Let O be the origin. The position vector of a point P is the vector OP, written as p. It gives the coordinates of P relative to O.
设 O 为原点。点 P 的位置向量是向量 OP，记作 p。它表示 P 相对于 O 的坐标。
If A and B have position vectors a and b, the displacement vector from A to B is AB = b − a. This vector points from A toward B and has magnitude equal to the distance AB.
若 A 和 B 的位置向量分别为 a 和 b，则由 A 到 B 的位移向量为 AB = b − a。该向量从 A 指向 B，其模等于 A、B 两点间的距离。
AB = b − a
This simple relation is the foundation for constructing line equations: the direction of a line is given by a displacement vector.
这个简单的关系是构造直线方程的基础：直线的方向由位移向量给出。
2. Vector Equation of a Line | 直线的向量方程
A straight line in space can be described by a fixed point A on the line and a direction vector d parallel to the line. For any point P on the line, the vector AP is a scalar multiple of d. Therefore the position vector r of P satisfies:
空间中一条直线可以用直线上一个定点 A 和一个平行于直线的方向向量 d 来描述。对于直线上任意一点 P，向量 AP 是 d 的标量倍。因此，P 的位置向量 r 满足：
r = a + λd
where a is the position vector of A, and λ is a real parameter. Every value of λ gives a point on the line; as λ varies over all real numbers, the entire line is traced out.
其中 a 是 A 的位置向量，λ 为实数参数。每一个 λ 值对应直线上一点；当 λ 取遍所有实数时，就描画出整条直线。
For example, if A = (1, 2, 3) and d = (2, −1, 4), then the line equation is r = (1, 2, 3) + λ(2, −1, 4).
例如，若 A = (1, 2, 3)，d = (2, −1, 4)，则直线方程为 r = (1, 2, 3) + λ(2, −1, 4)。
3. Establishing a Line Equation from Two Points | 由两点建立直线方程
When two points A(a) and B(b) lie on a line, the direction vector can be chosen as d = b − a. Then the equation of the line is:
当两点 A(a) 和 B(b) 位于同一直线上时，方向向量可取 d = b − a。则直线方程为：
r = a + λ(b − a)
This form uses A as the fixed point and the displacement from A to B as the direction. Equivalently, one may write r = (1 − λ)a + λb, which shows that any point on the segment AB corresponds to λ between 0 and 1.
这种形式以 A 为定点，以 A 到 B 的位移为方向。等价地，可写成 r = (1 − λ)a + λb，这表明线段 AB 上的任意点对应 λ 在 0 到 1 之间。
Worked example: Find the equation of the line through P(2, −1, 3) and Q(4, 3, −1).
例题：求经过 P(2, −1, 3) 和 Q(4, 3, −1) 的直线方程。
Direction vector PQ = (4−2, 3−(−1), −1−3) = (2, 4, −4). Using P as the fixed point:
方向向量 PQ = (4−2, 3−(−1), −1−3) = (2, 4, −4)。以 P 为定点：
r = (2, −1, 3) + λ(2, 4, −4)
4. Cartesian Form of a Line | 直线的笛卡尔坐标形式
Let r = (x, y, z), a = (x₁, y₁, z₁), and d = (l, m, n). Then the vector equation gives three scalar equations:
设 r = (x, y, z)，a = (x₁, y₁, z₁)，d = (l, m, n)。则向量方程给出三个标量方程：
x = x₁ + λl, y = y₁ + λm, z = z₁ + λn
Eliminating λ (when l, m, n are nonzero) yields the symmetric Cartesian form:
消去 λ（当 l、m、n 均不为零时）得到对称式笛卡尔形式：
(x − x₁) / l = (y − y₁) / m = (z − z₁) / n
If one of l, m, n is zero, the corresponding numerator is set equal to zero and the other two expressions are equated. For instance, if l = 0, then x = x₁ and (y − y₁) / m = (z − z₁) / n.
若 l、m、n 中有一个为零，则相应分子直接等于零，其余两个表达式相等。例如，若 l = 0，则 x = x₁，且 (y − y₁) / m = (z − z₁) / n。
This form is useful for checking whether a point lies on the line by substituting coordinates.
这种形式便于通过代入坐标检验点是否在直线上。
5. Parametric Equation of a Plane | 平面的参数方程
A plane in three-dimensional space can be described by a fixed point A and two non-parallel direction vectors b and c lying in the plane. The position vector r of any point in the plane can be written as:
三维空间中的平面可以用一个定点 A 和两个不平行且位于平面内的方向向量 b、c 来描述。平面内任意点的位置向量 r 可写成：
r = a + λb + μc
where λ and μ are real parameters. The plane is a two-dimensional surface, so two independent parameters are required.
其中 λ 和 μ 为实数参数。平面是二维曲面，因此需要两个独立的参数。
For example, if A = (1, 0, 2), b = (1, 1, 0), and c = (0, −1, 1), then the plane equation is r = (1, 0, 2) + λ(1, 1, 0) + μ(0, −1, 1).
例如，若 A = (1, 0, 2)，b = (1, 1, 0)，c = (0, −1, 1)，则平面方程为 r = (1, 0, 2) + λ(1, 1, 0) + μ(0, −1, 1)。
6. Normal Vector Form of a Plane | 平面的法向量形式
A more compact representation of a plane uses a vector n perpendicular to the plane, called a normal vector. For any point in the plane, the vector r − a is perpendicular to n, so their dot product is zero:
平面更紧致的表示方法使用垂直于平面的向量 n，称为法向量。对于平面内任意点，向量 r − a 与 n 垂直，因此内积为零：
(r − a) · n = 0
Expanding gives the standard form:
展开后得到标准形式：
r · n = a · n = d
Here d is a constant. This is called the scalar product form of the plane. If n = (a₁, b₁, c₁) and r = (x, y, z), the equation becomes a₁x + b₁y + c₁z = d.
其中 d 是常数。这称为平面的数量积形式。若 n = (a₁, b₁, c₁)，r = (x, y, z)，则方程化为 a₁x + b₁y + c₁z = d。
To find n from two direction vectors b and c, compute the vector product: n = b × c.
为从两个方向向量 b 和 c 求 n，计算向量积：n = b × c。
Worked example: Find the plane through A(1, 2, 3), B(2, 0, 4), C(3, 1, 2).
例题：求经过 A(1, 2, 3)、B(2, 0, 4)、C(3, 1, 2) 的平面方程。
Direction vectors: AB = (1, −2, 1), AC = (2, −1, −1). Their cross product is:
方向向量：AB = (1, −2, 1)，AC = (2, −1, −1)。两者叉积为：
n = AB × AC = (3, 3, 3)
Using A, d = a · n = 1×3 + 2×3 + 3×3 = 18, so the plane is r · (3, 3, 3) = 18, i.e. x + y + z = 6.
利用 A，d = a · n = 1×3 + 2×3 + 3×3 = 18，故平面为 r · (3, 3, 3) = 18，即 x + y + z = 6。
7. Solving Vector Equations: Point on a Line or Plane | 求解向量方程：检验点是否在直线或平面上
To test whether a point P with position vector p lies on a line, set r = p in the line equation and solve for λ. If a consistent value exists, the point is on the line.
检验点 P 是否在直线上时，令直线方程中 r = p，并解出 λ。若能找到一致的 λ 值，则该点在直线上。
For a plane, substitute r = p into the scalar product form r · n = d. If the equation holds, the point lies on the plane.
对于平面，将 r = p 代入数量积形式 r · n = d。若方程成立，则该点在平面上。
Example: Check whether P(3, 1, 2) lies on the line r = (1, 2, 3) + λ(2, −1, −1).
例：检验 P(3, 1, 2) 是否在直线 r = (1, 2, 3) + λ(2, −1, −1) 上。
We need (3, 1, 2) = (1 + 2λ, 2 − λ, 3 − λ). From the first component λ = 1. Then 2−1 = 1 and 3−1 = 2, so P is on the line.
需要 (3, 1, 2) = (1 + 2λ, 2 − λ, 3 − λ)。由第一分量得 λ = 1。此时 2−1 = 1，3−1 = 2，故 P 在直线上。
8. Intersection of Two Lines | 两条直线的交点
Given two lines r = a + λb and r = c + μd, their intersection is found by equating the two expressions:
已知两条直线 r = a + λb 与 r = c + μd，求交点时令两式相等：
a + λb = c + μd
This vector equation gives three scalar equations (x, y, z). Solve for λ and μ using two of them, then verify the third equation. If a unique solution exists, the lines intersect; if the third equation is not satisfied, they are skew.
这个向量方程给出三个标量方程（x、y、z 分量）。用其中两个方程求出 λ 和 μ，再检验第三个方程。若存在唯一解，则两直线相交；若第三个方程不满足，则两直线异面。
Example: Find the intersection of L₁: (2, 1, 0) + λ(1, −1, 2) and L₂: (1, 2, 3) + μ(2, 1, −1).
例：求直线 L₁: (2, 1, 0) + λ(1, −1, 2) 和 L₂: (1, 2, 3) + μ(2, 1, −1) 的交点。
Equating: 2 + λ = 1 + 2μ, 1 − λ = 2 + μ, 2λ = 3 − μ. From the first two, solve λ = 0.5, μ = 0.25. Substitute into the third: 2(0.5) = 1 and 3 − 0.25 = 2.75, not equal, so the lines are skew.
令两式相等：2 + λ = 1 + 2μ，1 − λ = 2 + μ，2λ = 3 − μ。由前两个方程解得 λ = 0.5，μ = 0.25。代入第三个方程：2(0.5) = 1，而 3 − 0.25 = 2.75，不相等，因此两直线异面。
9. Intersection of a Line and a Plane | 直线与平面的交点
To find the intersection of a line r = a + λb and a plane r · n = d, substitute the line expression into the plane equation:
求直线 r = a + λb 与平面 r · n = d 的交点时，将直线表达式代入平面方程：
(a + λb) · n = d
This gives a scalar equation in λ. If the coefficient of λ is nonzero, solve for λ and substitute back into the line equation to get the point. If the coefficient is zero and d equals a · n, the line lies in the plane; otherwise it is parallel to the plane.
这给出关于 λ 的标量方程。若 λ 的系数非零，则解出 λ 并代回直线方程得到交点。若系数为零且 d 等于 a · n，则直线位于平面内；否则直线与平面平行。
Example: Line r = (1, 0, 2) + λ(2, 1, −1), plane x + y + z = 5.
例：直线 r = (1, 0, 2) + λ(2, 1, −1)，平面 x + y + z = 5。
Substitute: (1+2λ) + (0+λ) + (2−λ) = 5 → 3 + 2λ = 5 → λ = 1. The intersection point is (3, 1, 1).
代入：(1+2λ) + (0+λ) + (2−λ) = 5 → 3 + 2λ = 5 → λ = 1。交点坐标为 (3, 1, 1)。
10. Applications: Distance from a Point to a Plane | 应用：点到平面的距离
The perpendicular distance from a point P(p) to a plane r · n = d is given by:
点 P(p) 到平面 r · n = d 的垂直距离为：
Distance = |p · n − d| / |n|
This formula is derived by projecting the vector from any point on the plane to P onto the normal direction.
该公式通过将平面上任意点到 P 的向量投影到法线方向得到。
For example, the distance from P(1, 2, 3) to plane 2x − y + 2z = 4 is |2(1) − 2 + 2(3) − 4| / √(4+1+4) = |2−2+6−4| / 3 = 2 / 3.
例如，P(1, 2, 3) 到平面 2x − y + 2z = 4 的距离为 |2(1) − 2 + 2(3) − 4| / √(4+1+4) = |2−2+6−4| / 3 = 2 / 3。
11. Common Pitfalls in Vector Equations | 向量方程中的常见误区
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Using two parallel direction vectors for a plane: The plane equation requires two non-parallel directions. If b and c are parallel, the equation degenerates to a line.
使用两个平行的方向向量表示平面：平面方程需要两个不共线的方向向量。若 b、c 平行，则方程退化描述一条直线。
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Forgetting to check the third equation when finding intersections of two lines: A pair of scalar equations may yield a solution that fails in the third component, meaning the lines are skew.
求两直线交点时忘记检验第三个方程：两个标量方程可能解出一组 λ、μ，但在第三个分量上不成立，说明两直线异面。
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Incorrectly computing the normal vector n = b × c: The cross product order matters; reversing the order changes the sign of n, but the plane equation remains equivalent if d is adjusted.
错误计算法向量 n = b ×
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