3D Vectors: Intersections and Closest Approach | 三维向量：交线与最近距离

📚 3D Vectors: Intersections and Closest Approach | 三维向量：交线与最近距离

Vector geometry is one of the most heavily examined topics in Edexcel A-Level Mathematics, particularly in Paper 3 (Pure Mathematics 3) and the Further Mathematics papers. This article focuses on two essential skills: finding the point of intersection between two lines, and determining the shortest distance between skew lines or from a point to a line.

向量几何是 Edexcel A-Level 数学中考查频率极高的专题，尤其在 Paper 3（纯数学 3）和进阶数学试卷中。本文聚焦两大核心技能：求两条直线的交点，以及求异面直线间或点到直线的最短距离。

1. Vector Equation of a Line | 直线的向量方程

In three dimensions, a straight line can be described by a position vector to a known point on the line, plus a scalar multiple of a direction vector. The standard form is r = a + λb, where a is the position vector of a fixed point, b is a non-zero direction vector parallel to the line, and λ is a real parameter.

在三维空间中，一条直线可用线上已知点的位置向量加上方向向量的标量倍来表示。标准形式为 r = a + λb，其中 a 是固定点的位置向量，b 是与直线平行的非零方向向量，λ 为实参数。

r = a + λb = (a₁ + λb₁, a₂ + λb₂, a₃ + λb₃)

For example, the line passing through A(1, 2, 3) with direction vector b = (2, −1, 4) has equation r = (1, 2, 3) + λ(2, −1, 4). Every value of λ gives a distinct point on the line.

例如，过点 A(1, 2, 3) 且方向向量为 b = (2, −1, 4) 的直线方程为 r = (1, 2, 3) + λ(2, −1, 4)。每一个 λ 值对应直线上的一个不同点。

2. Intersection of Two Lines | 两条直线的交点

When two lines intersect, there exists a point that satisfies both vector equations simultaneously. Given L₁: r = a + λb and L₂: r = c + μd, we equate the components to form a system of three equations in two unknowns (λ and μ).

当两条直线相交时，存在一个点同时满足两个向量方程。设 L₁: r = a + λb，L₂: r = c + μd，我们令各分量相等，构成含两个未知数（λ 和 μ）的三元方程组。

If a unique solution exists for all three component equations, the lines intersect at that point. If only two equations are consistent but the third fails, the lines are skew (in 3D). If no consistent solution exists for any pair, the lines are parallel or coincident.

若三个分量方程都有唯一解，则两直线在该点相交。若仅两个方程一致而第三个不成立，则两直线为异面直线（三维中）。若任何一对方程都无一致解，则两直线平行或重合。

a₁ + λb₁ = c₁ + μd₁ ①
a₂ + λb₂ = c₂ + μd₂ ②
a₃ + λb₃ = c₃ + μd₃ ③

3. Worked Example: Finding the Intersection | 求交点的完整示例

Consider L₁: r = (1, 2, 3) + λ(2, −1, 1) and L₂: r = (4, 1, 2) + μ(1, 2, −1). Equating components yields:

考察 L₁: r = (1, 2, 3) + λ(2, −1, 1) 与 L₂: r = (4, 1, 2) + μ(1, 2, −1)。令各分量相等可得：

1 + 2λ = 4 + μ
2 − λ = 1 + 2μ
3 + λ = 2 − μ

From the first equation, μ = 2λ − 3. Substituting into the second equation: 2 − λ = 1 + 2(2λ − 3) = 1 + 4λ − 6 = 4λ − 5. Thus 7 = 5λ, giving λ = 7/5. Then μ = 2(7/5) − 3 = 14/5 − 15/5 = −1/5.

由第一个方程得 μ = 2λ − 3。代入第二个方程：2 − λ = 1 + 2(2λ − 3) = 1 + 4λ − 6 = 4λ − 5。于是 7 = 5λ，得 λ = 7/5。进而 μ = 2(7/5) − 3 = 14/5 − 15/5 = −1/5。

Now verify with the third equation: LHS = 3 + 7/5 = 22/5, RHS = 2 − (−1/5) = 11/5. Since 22/5 ≠ 11/5, the lines do not intersect despite the first two equations working.

用第三个方程验证：左边 = 3 + 7/5 = 22/5，右边 = 2 − (−1/5) = 11/5。由于 22/5 ≠ 11/5，尽管前两个方程有解，两直线并不相交。

4. Skew Lines and the Closest Approach | 异面直线与最近距离

Two non-parallel lines in 3D that do not intersect are called skew lines. The shortest distance between them is measured along the line segment that is perpendicular to both lines. This segment is uniquely determined and can be found using vector projection.

三维空间中两条不平行且不相交的直线称为异面直线。它们之间的最短距离沿同时垂直于两条直线的线段来度量。该线段是唯一确定的，可通过向量投影法求得。

Given L₁: r = a + λb and L₂: r = c + μd, the vector connecting a point on L₁ to a point on L₂ is w = (c + μd) − (a + λb). The shortest distance occurs when w is perpendicular to both direction vectors b and d.

设 L₁: r = a + λb，L₂: r = c + μd，连接 L₁ 上一点到 L₂ 上一点的向量为 w = (c + μd) − (a + λb)。当 w 同时垂直于两个方向向量 b 和 d 时，距离最短。

Equivalently, the shortest distance is the magnitude of the projection of (c − a) onto the direction perpendicular to both lines, i.e., onto the cross product b × d.

等价地，最短距离等于向量 (c − a) 在垂直于两条直线的方向（即叉积 b × d 方向）上的投影的模长。

d = |(c − a) · (b × d)| / |b × d|

5. Deriving the Distance Formula | 推导距离公式

The vector n = b × d is perpendicular to both lines. The scalar projection of the difference vector (c − a) onto the unit vector n/|n| gives the perpendicular distance between the lines.

向量 n = b × d 同时垂直于两条直线。差向量 (c − a) 在单位向量 n/|n| 上的标量投影即为两直线间的垂直距离。

The absolute value is used because the projection may be negative depending on the orientation of the lines. This formula works for any pair of skew lines.

取绝对值是因为投影的正负取决于直线的方向。该公式适用于任意一对异面直线。

For example, with L₁: r = (1, 2, 3) + λ(2, −1, 1) and L₂: r = (4, 1, 2) + μ(1, 2, −1), we have c − a = (3, −1, −1).

例如，对于 L₁: r = (1, 2, 3) + λ(2, −1, 1) 和 L₂: r = (4, 1, 2) + μ(1, 2, −1)，有 c − a = (3, −1, −1)。

Compute b × d = (2, −1, 1) × (1, 2, −1) = ((−1)(−1) − (1)(2), (1)(1) − (2)(−1), (2)(2) − (−1)(1)) = (1 − 2, 1 + 2, 4 + 1) = (−1, 3, 5).

计算 b × d = (2, −1, 1) × (1, 2, −1) = ((−1)(−1) − (1)(2), (1)(1) − (2)(−1), (2)(2) − (−1)(1)) = (1 − 2, 1 + 2, 4 + 1) = (−1, 3, 5)。

Then (c − a) · (b × d) = (3)(−1) + (−1)(3) + (−1)(5) = −3 − 3 − 5 = −11. The magnitude is 11, and |b × d| = √(1 + 9 + 25) = √35. Hence d = 11/√35.

于是 (c − a) · (b × d) = (3)(−1) + (−1)(3) + (−1)(5) = −3 − 3 − 5 = −11。模为 11，而 |b × d| = √(1 + 9 + 25) = √35。因此 d = 11/√35。

6. Distance from a Point to a Line | 点到直线的距离

A related problem asks for the shortest distance from a point P with position vector p to a line L: r = a + λb. The closest point Q on the line is found by minimising |p − r(λ)|².

一个相关问题求点 P（位置向量为 p）到直线 L: r = a + λb 的最短距离。直线上的最近点 Q 通过最小化 |p − r(λ)|² 来求得。

Alternatively, the vector from P to any point on the line is p − a − λb. At the closest point, this vector is perpendicular to b. Setting (p − a − λb) · b = 0 gives the optimal λ.

另一种方法是，从 P 到直线上任意点的向量为 p − a − λb。在最近点处，该向量垂直于 b。令 (p − a − λb) · b = 0 即可求出最优 λ。

λ = (p − a) · b / |b|²

Once λ is known, the distance is |p − q|, where q = a + λb is the foot of the perpendicular from P to the line.

求出 λ 后，距离即为 |p − q|，其中 q = a + λb 是 P 到直线的垂足。

7. Worked Example: Point-to-Line Distance | 点到直线距离示例

Find the distance from P(4, 1, −2) to the line L: r = (1, 2, 3) + λ(2, −1, 1).

求点 P(4, 1, −2) 到直线 L: r = (1, 2, 3) + λ(2, −1, 1) 的距离。

First, p − a = (3, −1, −5). Compute (p − a) · b = (3)(2) + (−1)(−1) + (−5)(1) = 6 + 1 − 5 = 2. Also |b|² = 4 + 1 + 1 = 6. Hence λ = 2/6 = 1/3.

首先，p − a = (3, −1, −5)。计算 (p − a) · b = (3)(2) + (−1)(−1) + (−5)(1) = 6 + 1 − 5 = 2。又 |b|² = 4 + 1 + 1 = 6。因此 λ = 2/6 = 1/3。

The foot of the perpendicular is q = (1, 2, 3) + (1/3)(2, −1, 1) = (5/3, 5/3, 10/3). The distance is |p − q| = √[(4 − 5/3)² + (1 − 5/3)² + (−2 − 10/3)²] = √[(7/3)² + (−2/3)² + (−16/3)²] = √(49 + 4 + 256)/3 = √309/3.

垂足为 q = (1, 2, 3) + (1/3)(2, −1, 1) = (5/3, 5/3, 10/3)。距离为 |p − q| = √[(4 − 5/3)² + (1 − 5/3)² + (−2 − 10/3)²] = √[(7/3)² + (−2/3)² + (−16/3)²] = √309/3。

8. Minimisation via Calculus | 利用微积分求最小值

An alternative to the projection method is to treat the squared distance as a function of the parameters and use differentiation. For two lines, let D(λ, μ) = |w(λ, μ)|², where w = c + μd − a − λb.

投影法的替代方案是将距离平方视为参数的函数并用微分求解。对于两条直线，设 D(λ, μ) = |w(λ, μ)|²，其中 w = c + μd − a − λb。

At the minimum, ∂D/∂λ = 0 and ∂D/∂μ = 0. This yields a 2×2 linear system:

在最小值处，∂D/∂λ = 0 且 ∂D/∂μ = 0。这产生一个 2×2 线性方程组：

|b|²λ − (b·d)μ = (c−a)·b
−(b·d)λ + |d|²μ = −(c−a)·d

Solving this system gives the parameters of the closest points on both lines. This method is particularly useful in mechanics problems where time-dependent positions are involved.

求解该方程组即得两条直线上最近点的参数。此方法在涉及时间相关位置的力学问题中尤为有用。

9. Application: Relative Motion of Ships | 应用：船只相对运动

Consider two ships moving with constant velocities. Ship A has position vector (2, 1) km at t = 0 and velocity (4, 3) km/h. Ship B has position (−3, 5) km at t = 0 and velocity (2, −1) km/h. Their positions at time t are A(t) = (2+4t, 1+3t) and B(t) = (−3+2t, 5−t).

考虑两艘匀速运动的船只。船 A 在 t = 0 时的位置向量为 (2, 1) 千米，速度为 (4, 3) 千米/小时。船 B 在 t = 0 时的位置为 (−3, 5) 千米，速度为 (2, −1) 千米/小时。它们在时间 t 的位置为 A(t) = (2+4t, 1+3t) 和 B(t) = (−3+2t, 5−t)。

The relative position of B with respect to A is B − A = (−5−2t, 4−4t). The squared distance is |B − A|² = (−5−2t)² + (4−4t)² = (25+20t+4t²) + (16−32t+16t²) = 41 − 12t + 20t².

B 相对 A 的位置为 B − A = (−5−2t, 4−4t)。距离平方为 |B − A|² = (−5−2t)² + (4−4t)² = (25+20t+4t²) + (16−32t+16t²) = 41 − 12t + 20t²。

Differentiating gives d/dt = 40t − 12 = 0, so t = 0.3 h. The minimum distance is √(41 − 3.6 + 1.8) = √39.2 ≈ 6.26 km.

求导得 d/dt = 40t − 12 = 0，故 t = 0.3 小时。最小距离为 √(41 − 3.6 + 1.8) = √39.2 ≈ 6.26 千米。

10. Common Pitfalls and Exam Tips | 常见错误与应试技巧

A frequent error is assuming that two lines in 3D always intersect if they are not parallel. In fact, skew lines are the norm; intersection requires the existence of a single pair (λ, μ) satisfying all three component equations simultaneously.

一个常见错误是认为三维空间中不平行的两条直线一定相交。事实上，异面直线才是常态；相交要求存在唯一一对 (λ, μ) 同时满足所有三个分量方程。

Always verify your solution in all three equations. When using the distance formula, remember to take the absolute value of the dot product before dividing by |b × d|. Check that your direction vectors are correct and not scalar multiples of each other.

务必在所有三个方程中验证解。使用距离公式时，记得先对点积取绝对值再除以 |b × d|。检查方向向量是否正确，且不是彼此的标量倍数。

Write the vector equation of a line consistently: r = a + λb.
For intersection problems, solve two equations first, then verify the third.
For skew lines, apply the cross-product projection formula directly.
For point-to-line distance, find the foot of the perpendicular first.

一致地写出直线向量方程：r = a + λb。
求交点时先解两个方程，再验证第三个。
求异面直线距离时，直接应用叉积投影公式。
求点到直线距离时，先求垂足。

11. Advanced: Intersection with a Plane | 进阶：直线与平面的交点

Although the focus of this article is on lines, a closely related skill is finding where a line intersects a plane. If the plane is given in Cartesian form ax + by + cz = d, substitute the components of the line r = (x₀ + λp, y₀ + λq, z₀ + λr) into the plane equation and solve for λ.

尽管本文聚焦于直线，但一个密切相关的技能是求直线与平面的交点。若平面以笛卡尔形式 ax + by + cz = d 给出，将直线 r = (x₀ + λp, y₀ + λq, z₀ + λr) 的分量代入平面方程，解出 λ 即可。

If the denominator a·b = ap + bq + cr equals zero, the line is parallel to the plane. If additionally the constant terms satisfy the plane equation, the line lies in the plane.

若分母 a·b = ap + bq + cr 等于零，则直线与平面平行。若常数项还满足平面方程，则直线位于平面内。

λ = (d − a·a₀) / (a·b)

Here a₀ = (x₀, y₀, z₀) is the position vector of the known point on the line, and b = (p, q, r) is the direction vector.

这里 a₀ = (x₀, y₀, z₀) 是直线上已知点的位置向量，b = (p, q, r) 是方向向量。

12. Practice Problems | 巩固练习

The following problems test the full range of skills covered in this article. Attempt them before checking the solutions.

以下问题全面测试本文涉及的技能。请先尝试解答，再核对答案。

Problem \| 题目	Answer \| 答案
1. Do L₁: r = (3,1,0) + λ(1,2,1) and L₂: r = (0,0,2) + μ(2,1,3) intersect?	Solve 3+λ=2μ, 1+2λ=μ, λ=2+3μ. From first, μ = (3+λ)/2. Substitute: 1+2λ = (3+λ)/2 → 2+4λ = 3+λ → 3λ = 1 → λ = 1/3, μ = 5/3. Third: λ = 2+3μ = 2+5 = 7, false. No intersection.
2. Distance between L₁: r = (1,0,2) + λ(1,1,0) and L₂: r = (0,1,3) + μ(0,1,1)?	b×d = (1,1,0)×(0,1,1) = (1,−1,1). c−a = (−1,1,1). Dot = −1−1+1 = −1. Distance = 1/√3.
3. Distance from (2,−1,3) to r = (0,0,0) + λ(1,2,2)?	λ = (p·b)/\|b\|² = (2−2+6)/9 = 6/9 = 2/3. q = (2/3, 4/3, 4/3). Distance = √[(4/3)² + (−7/3)² + (5/3)²] = √90/3 = √10.
4. Find the intersection of r = (1,0,1) + λ(1,1,2) with plane x + 2y − z = 5.	(1+λ) + 2λ − (1+2λ) = 5 → 1+λ+2λ−1−2λ = λ = 5. Point: (6, 5, 11).

Mastering vector geometry requires practice. Focus on understanding the geometric meaning behind each formula; this will help you adapt to unfamiliar problem forms in the exam.

掌握向量几何需要多加练习。重点理解每个公式背后的几何意义，这将帮助你在考试中应对新颖的题型。

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3D Vectors: Intersections and Closest Approach | 三维向量：交线与最近距离