7 Shortest Distances | 七种最短距离

📚 7 Shortest Distances | 七种最短距离

In AQA A-Level Mathematics, distance problems appear both in plane geometry and in vectors. The shortest distance between two objects is always measured along the common perpendicular direction. This article presents seven classic shortest-distance calculations that you must master, from the distance between two points to the distance between two skew lines.

在 AQA 数学 A-Level 考试中，距离问题既出现在平面几何中，也出现在向量部分。两个物体之间的最短距离总是沿着公共垂直方向测量。本文整理了七种你必须掌握的最短距离计算方法，从两点间距离一直到两条异面直线之间的距离。

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1. Distance Between Two Points (2D) | 两点之间的距离（平面）

The most basic shortest distance is the straight-line distance between two points P(x₁, y₁) and Q(x₂, y₂). Construct a right-angled triangle with horizontal difference x₂ − x₁ and vertical difference y₂ − y₁, then apply Pythagoras’ theorem.

最基本的最短距离是两点 P(x₁, y₁) 与 Q(x₂, y₂) 之间的直线距离。以水平差 x₂ − x₁ 和垂直差 y₂ − y₁ 构造直角三角形，然后使用勾股定理。

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Worked example: For P(1, 2) and Q(4, 6), d = √[(4 − 1)² + (6 − 2)²] = √(9 + 16) = 5.

例如：对于 P(1, 2) 和 Q(4, 6)，d = √[(4 − 1)² + (6 − 2)²] = √(9 + 16) = 5。

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2. Distance from a Point to a Line (2D) | 点到直线的距离（平面）

Given a line written in general form ax + by + c = 0 and a point (x₁, y₁), the shortest distance is measured along the perpendicular from the point to the line. This formula is derived from the area of triangles and the length of the perpendicular.

已知直线的一般式为 ax + by + c = 0，点为 (x₁, y₁)，最短距离沿点到直线的垂线方向测量。该公式由三角形面积和垂线长度推导而来。

d = |ax₁ + by₁ + c| / √(a² + b²)

Worked example: line 3x + 4y − 5 = 0 and point (1, 2). Then d = |3×1 + 4×2 − 5| / √(3² + 4²) = |3 + 8 − 5| / 5 = 6/5 = 1.2.

例如：直线 3x + 4y − 5 = 0 和点 (1, 2)。则 d = |3×1 + 4×2 − 5| / √(3² + 4²) = |3 + 8 − 5| / 5 = 6/5 = 1.2。

If the line is given as y = mx + c, rewrite it as mx − y + c = 0, so a = m, b = −1.

如果直线以 y = mx + c 给出，可改写为 mx − y + c = 0，即 a = m，b = −1。

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3. Distance Between Two Parallel Lines | 两条平行线之间的距离

Two parallel lines in 2D have the same normal vector, so they can be written as ax + by + c₁ = 0 and ax + by + c₂ = 0. To find the shortest distance, choose any point on one line and apply the point-to-line formula to the other line.

平面中两条平行线具有相同的法向量，因此可写成 ax + by + c₁ = 0 和 ax + by + c₂ = 0。要求它们之间的最短距离，可在其中一条直线上任取一点，再用点到直线的距离公式求它到另一条直线的距离。

d = |c₂ − c₁| / √(a² + b²)

Worked example: for the lines 2x − y + 1 = 0 and 2x − y − 3 = 0, d = |−3 − 1| / √(2² + (−1)²) = 4/√5.

例如：对于直线 2x − y + 1 = 0 和 2x − y − 3 = 0，d = |−3 − 1| / √(2² + (−1)²) = 4/√5。

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4. Distance from a Point to a Plane (3D) | 点到平面的距离（三维）

For a plane written as ax + by + cz + d = 0 and a point (x₀, y₀, z₀), the shortest distance is along the normal vector (a, b, c). The formula is an extension of the 2D point-to-line distance.

对于平面 ax + by + cz + d = 0 和点 (x₀, y₀, z₀)，最短距离沿平面的法向量 (a, b, c) 方向。这个公式是二维点到直线距离公式的三维推广。

d = |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²)

Worked example: plane x + 2y − 2z − 6 = 0 and point (1, 1, 1). Then d = |1 + 2 − 2 − 6| / √(1² + 2² + (−2)²) = |−5| / 3 = 5/3.

例如：平面 x + 2y − 2z − 6 = 0 和点 (1, 1, 1)。则 d = |1 + 2 − 2 − 6| / √(1² + 2² + (−2)²) = |−5| / 3 = 5/3。

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5. Distance Between Two Parallel Planes | 两个平行平面之间的距离

Parallel planes share the same normal vector, so they can be written as ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0. Pick a point on one plane and apply the point-to-plane formula to the other plane.

两个平行平面具有相同的法向量，因此可写成 ax + by + cz + d₁ = 0 和 ax + by + cz + d₂ = 0。在其中一个平面上取一点，再用点到平面的距离公式求它到另一个平面的距离。

d = |d₂ − d₁| / √(a² + b² + c²)

Worked example: for the planes x + y + z − 1 = 0 and x + y + z + 3 = 0, d = |3 − (−1)| / √(1² + 1² + 1²) = 4/√3.

例如：对于平面 x + y + z − 1 = 0 和 x + y + z + 3 = 0，d = |3 − (−1)| / √(1² + 1² + 1²) = 4/√3。

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6. Distance from a Point

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