📚 Constructing Line Equations in 3D Space | 三维空间直线方程的构建方法
In three-dimensional space, a straight line is uniquely determined by a fixed point and a direction vector. Unlike lines in 2D, a 3D line cannot be described by a single linear equation; instead, it requires a vector, parametric, or symmetric equation system. This article explores the core methods for constructing these equations, tailored to IB Mathematics Analysis and Approaches HL.
在三维空间中，一条直线由一定点和一方向向量唯一确定。与二维直线不同，三维直线不能由单个线性方程描述，而需要使用向量方程、参数方程或对称方程系统。本文将围绕IB数学分析与方法HL的考点，系统讲解这些方程的构建方法。
1. Core Idea: Point + Direction Vector | 核心思想：点 + 方向向量
To build any line equation in 3D, you need two ingredients: a known point \(A\) on the line (with position vector \(\mathbf{a}\)) and a direction vector \(\mathbf{d}\) parallel to the line. Every point \(P\) on the line can then be reached by starting at \(A\) and moving a scalar multiple of \(\mathbf{d}\).
要构建三维空间中的任意直线方程，需要两个要素：直线上已知一点 \(A\)（位置向量为 \(\mathbf{a}\)）以及平行于直线的方向向量 \(\mathbf{d}\)。直线上任意一点 \(P\) 都可以从 \(A\) 出发，沿 \(\mathbf{d}\) 方向移动某个标量倍数而得到。
P = A + t·d or r = a + t·d (t ∈ ℝ)
Here \(t\) is a parameter. Different values of \(t\) correspond to different points on the line. This underlying relation is the basis of all three standard equation forms.
其中 \(t\) 是参数。不同的 \(t\) 值对应直线上不同的点。这一基本关系是三种标准方程形式的基础。
2. Vector Equation of a Line | 直线的向量方程
The vector equation directly states the position vector \(\mathbf{r}\) of a general point on the line:
向量方程直接给出直线上一般点的位置向量 \(\mathbf{r}\)：
r = a + t·d
For example, if a line passes through \(A(1, -2, 3)\) and has direction vector \(\mathbf{d} = (4, 1, -2)\), its vector equation is
例如，若一条直线经过点 \(A(1, -2, 3)\)，方向向量为 \(\mathbf{d} = (4, 1, -2)\)，则其向量方程为
r = (1, -2, 3) + t(4, 1, -2)
Here \(\mathbf{r} = (x, y, z)\). In IB, you may write this as \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix}\).
这里 \(\mathbf{r} = (x, y, z)\)。在IB中，你可以写成 \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix}\)。
3. Parametric Equations | 参数方程
By equating components in the vector equation, we obtain three separate equations for \(x\), \(y\) and \(z\) in terms of the parameter \(t\):
将向量方程中的各分量分别相等，即可得到关于参数 \(t\) 的三个独立方程：
x = x₀ + t·d₁, y = y₀ + t·d₂, z = z₀ + t·d₃
Using the previous example, the parametric equations are:
使用前面的例子，参数方程为：
x = 1 + 4t, y = -2 + t, z = 3 – 2t
Parametric form is extremely useful because it gives explicit coordinates for every point on the line. It also simplifies calculations of intersections and distances.
参数形式非常实用，因为它给出了直线上每个点的显式坐标，同时简化了交点与距离的计算。
4. Symmetric (Cartesian) Equations | 对称（笛卡尔）方程
If each component of the direction vector is non-zero, we can eliminate the parameter \(t\) by solving each parametric equation for \(t\) and equating the results:
若方向向量的每个分量均不为零，我们可以解出每个参数方程中的 \(t\)，再令它们相等，从而消去参数 \(t\)：
\(\frac{x – x₀}{d₁} = \frac{y – y₀}{d₂} = \frac{z – z₀}{d₃}\)
For the example above, the symmetric equations are
对于上述例子，对称方程为
\(\frac{x – 1}{4} = \frac{y + 2}{1} = \frac{z – 3}{-2}\)
Note that if a directional component is zero, the corresponding numerator must equal zero as well. For instance, if \(d₂ = 0\), then \(y = y₀\) and the symmetric form becomes a pair of equalities plus a separate condition.
注意若方向向量的某分量为零，则对应分子也必须等于零。例如，若 \(d₂ = 0\)，则 \(y = y₀\)，对称形式变为一组等量关系外加一个独立条件。
5. Building a Line from Two Points | 由两点构建直线
Given two distinct points \(A\) and \(B\), the direction vector can be found by subtracting their position vectors: \(\mathbf{d} = \mathbf{b} – \mathbf{a}\). Then choose either point to write the equation.
已知两个不同点 \(A\) 和 \(B\)，方向向量可通过它们的位置向量相减得到：\(\mathbf{d} = \mathbf{b} – \mathbf{a}\)。然后选择其中任意一点写出方程。
Let \(A(2, -1, 5)\) and \(B(0, 3, -1)\). Then
设 \(A(2, -1, 5)\)，\(B(0, 3, -1)\)。则
\(\mathbf{d} = \mathbf{b} – \mathbf{a} = (0-2, 3-(-1), -1-5) = (-2, 4, -6)\)
Using point \(A\), the vector equation is \(\mathbf{r} = (2, -1, 5) + t(-2, 4, -6)\). Symmetric form:
以点 \(A\) 为基础，向量方程为 \(\mathbf{r} = (2, -1, 5) + t(-2, 4, -6)\)。对称形式为：
\(\frac{x – 2}{-2} = \frac{y + 1}{4} = \frac{z – 5}{-6}\)
Always check that both points satisfy the final equation. This verifies that the direction vector is correct.
务必验证两个点是否都满足最终方程，以确认方向向量无误。
6. Direction Ratios and Direction Cosines | 方向比与方向余弦
Direction ratios are the components of any vector parallel to the line. They are not unique; multiplying by a non-zero scalar gives another valid set. For example, \(\mathbf{d} = (2, -4, 6)\) and \(\mathbf{d} = (1, -2, 3)\) represent the same direction.
方向比是平行于直线的任意向量的分量。它们不唯一；乘以非零标量仍得到有效的一组方向比。例如，\(\mathbf{d} = (2, -4, 6)\) 与 \(\mathbf{d} = (1, -2, 3)\) 表示同一直线方向。
Direction cosines are the cosines of the angles between the line and the positive \(x\), \(y\), and \(z\) axes. They are obtained by normalising the direction vector:
方向余弦是直线与 \(x\)、\(y\)、\(z\) 轴正方向夹角之余弦，由方向向量单位化得到：
\(\cos α = \frac{d₁}{|\mathbf{d}|}, \cos β = \frac{d₂}{|\mathbf{d}|}, \cos γ = \frac{d₃}{|\mathbf{d}|}\)
These satisfy \(\cos²α + \cos²β + \cos²γ = 1\). In IB, direction cosines sometimes appear in vector geometry problems, so be ready to compute them from a given direction vector.
这些方向余弦满足 \(\cos²α + \cos²β + \cos²γ = 1\)。在IB考试中，方向余弦偶见于向量几何题中，因此应熟练掌握从给定方向向量计算它们。
7. Converting Between Different Equation Forms | 不同方程形式之间的转换
The three forms (vector, parametric, symmetric) are interchangeable. To convert from symmetric to vector form, set the common ratio equal to \(t\), then express \(x\), \(y\), \(z\) as linear expressions in \(t\).
三种形式（向量、参数、对称）可以相互转换。从对称形式到向量形式，令公共比值等于 \(t\)，然后将 \(x\)、\(y\)、\(z\) 表示为 \(t\) 的线性表达式。
For example, given \(\frac{x – 1}{2} = \frac{y}{-3} = \frac{z + 4}{1}\), set this equal to \(t\):
例如，已知 \(\frac{x – 1}{2} = \frac{y}{-3} = \frac{z + 4}{1}\)，令它等于 \(t\)：
\(x = 1 + 2t, y = -3t, z = -4 + t\)
This gives the vector equation \(\mathbf{r} = (1, 0, -4) + t(2, -3, 1)\). Notice that the fixed point corresponds to \(t = 0\).
由此得到向量方程 \(\mathbf{r} = (1, 0, -4) + t(2, -3, 1)\)。注意 \(t = 0\) 时对应定点。
8. Parallel and Coincident Lines | 平行直线与重合直线
Two lines are parallel if their direction vectors are scalar multiples of each other. If they are parallel and share a common point, they are coincident (the same line).
若两直线的方向向量成比例，则它们平行。若平行且共有一个公共点，则它们重合（即同一条直线）。
Given \(L₁: \mathbf{r} = (1, 2, 3) + s(2, -1, 4)\) and \(L₂: \mathbf{r} = (3, 1, 7) + t(-4, 2, -8)\), we see \(\mathbf{d₂} = -2\mathbf{d₁}\), so the lines are parallel. Since \((3,1,7)\) lies on \(L₁\)? Check: set \(s=1\) gives \((3,1,7)\), yes, hence they are actually coincident.
已知 \(L₁: \mathbf{r} = (1, 2, 3) + s(2, -1, 4)\) 和 \(L₂: \mathbf{r} = (3, 1, 7) + t(-4, 2, -8)\)，观察得 \(\mathbf{d₂} = -2\mathbf{d₁}\)，所以两直线平行。又 \((3,1,7)\) 是否在 \(L₁\) 上？检验：令 \(s=1\) 得 \((3,1,7)\)，是的，因此它们实际重合。
To test whether parallel lines are identical, substitute the coordinate of a known point from one line into the equation of the other line. This is a common IB exam question.
要判断两条平行直线是否重合，可将其中一条直线的已知点代入另一条直线的方程中。这是IB考试的常见题型。
9. Intersection and Skew Lines | 相交直线与异面直线
To find whether two non-parallel 3D lines intersect, solve the parametric equations simultaneously. Equate the \(x\), \(y\), and \(z\) components using different parameter names, then solve for the parameters and check consistency.
要判断两条不平行三维直线是否相交，可联立参数方程。使用不同参数名令 \(x\)、\(y\)、\(z\) 分量相等，解出参数并验证一致性。
For example, \(L₁: (x,y,z) = (1,2,3) + t(1,0,-1)\) and \(L₂: (x,y,z) = (3,0,2) + s(0,1,1)\). Equating gives:
例如，\(L₁: (x,y,z) = (1,2,3) + t(1,0,-1)\) 与 \(L₂: (x,y,z) = (3,0,2) + s(0,1,1)\)。令各分量相等：
\(1 + t = 3\), \(2 = s\), \(3 – t = 2 + s\). From the first, \(t = 2\); from the second, \(s = 2\); the third gives \(3 – 2 = 1\), but \(2 + 2 = 4\). Contradiction, so the lines do not intersect.
\(1 + t = 3\)，\(2 = s\)，\(3 – t = 2 + s\)。由第一个得 \(t = 2\)；第二个得 \(s = 2\)；第三个给出 \(3 – 2 = 1\)，但 \(2 + 2 = 4\)。矛盾，因此两直线不相交。
In 3D, non-parallel lines that do not meet are called skew lines. This concept is fundamental in spatial geometry and often appears in IB Paper 3.
在三维空间中，不平行但不相交的直线称为异面直线。这一概念是空间几何的基础，常出现在IB Paper 3中。
10. Worked Example: Constructing a Line from a Point and Direction | 综合例题：由点和方向构建直线
Let us apply the methods to a typical IB-style problem. A line passes through point \(P(3, -1, 2)\) and is parallel to the line \(\frac{x – 2}{4} = \frac{y + 5}{-2} = \frac{z}{7}\). Find the parametric equations of the line.
让我们用典型IB题来应用上述方法。一条直线经过点 \(P(3, -1, 2)\)，且平行于直线 \(\frac{x – 2}{4} = \frac{y + 5}{-2} = \frac{z}{7}\)。求该直线的参数方程。
Solution: The given line has direction vector \(\mathbf{d} = (4, -2, 7)\). Since our line is parallel, it can use the same direction vector. With point \(P\), the parameter equations are:
解：已知直线的方向向量为 \(\mathbf{d} = (4, -2, 7)\)。因为所求直线平行于已知直线，所以可采用相同的方向向量。以点 \(P\) 为准，参数方程为：
\(x = 3 + 4t, y = -1 – 2t, z = 2 + 7t\)
We can also write the symmetric equations as \(\frac{x – 3}{4} = \frac{y + 1}{-2} = \frac{z – 2}{7}\).
对称方程为 \(\frac{x – 3}{4} = \frac{y + 1}{-2} = \frac{z – 2}{7}\)。
Always remember: the direction vector of a line is not unique; any non-zero scalar multiple works. However, the equation structure remains standard, and you must carefully read the problem to decide whether the line passes through a given point or has a given direction.
永远记住：直线方向向量不唯一；任何非零标量倍均有效。但方程的结构保持标准，须仔细审题以确定直线经过给定点还是具有给定方向。
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