A-Level Maths: Applications of Position Vectors | A-Level 数学：位置向量的应用

📚 A-Level Maths: Applications of Position Vectors | A-Level 数学：位置向量的应用

A position vector is a vector that starts at the origin and ends at a fixed point. It tells us exactly where a point is located in a coordinate system. In A-Level Mathematics, position vectors are used to solve problems involving geometry, motion, and forces, and they form a bridge between pure mathematics and applied mathematics.

位置向量是从原点出发并指向某个固定点的向量，用来准确描述点在坐标系中的位置。在 A-Level 数学中，位置向量被广泛用于解决几何、运动和力等问题，是连接纯数学与应用数学的重要桥梁。

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1. Position Vectors and Coordinates | 位置向量与坐标

In two dimensions, the position vector of a point P(x, y) is written as p = x i + y j. In three dimensions, it is written as p = x i + y j + z k. The coefficients x, y and z are the coordinates of P, and i, j, k are the unit vectors along the x-, y- and z-axes.

在二维平面中，点 P(x, y) 的位置向量写成 p = x i + y j。在三维空间中，它写成 p = x i + y j + z k。其中 x、y、z 是点 P 的坐标，而 i、j、k 分别是沿 x 轴、y 轴和 z 轴方向的单位向量。

For example, the point A(3, 4) has position vector a = 3 i + 4 j. This vector represents the displacement from the origin O to A.

例如，点 A(3, 4) 的位置向量是 a = 3 i + 4 j。这个向量表示从原点 O 到点 A 的位移。

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2. Displacement Vectors | 位移向量

The displacement vector from point A to point B is found by subtracting the position vector of A from the position vector of B:

从点 A 到点 B 的位移向量，由点 B 的位置向量减去点 A 的位置向量得到：

AB = b − a = (x₂ − x₁)i + (y₂ − y₁)j

This vector describes both the distance and the direction from A to B. It is independent of the origin, because it depends only on the relative positions of A and B.

这个向量同时描述了从 A 到 B 的距离和方向。它不依赖于原点的位置，因为只与 A、B 的相对位置有关。

a = 2i + j and B has position vector b = 5i + 6j, then AB = (5 − 2)i + (6 − 1)j = 3i + 5j.

如果 A 的位置向量为 a = 2i + j，B 的位置向量为 b = 5i + 6j，那么 AB = (5 − 2)i + (6 − 1)j = 3i + 5j。

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3. Magnitude and Unit Vectors | 模长与单位向量

ition vector p = x i + y j is the distance from the origin to the point P. It is given by:

位置向量 p = x i + y j 的模长就是从原点到点 P 的距离，计算公式为：

|p| = √x² + y²

In three dimensions, |p| = √x² + y² + z². The direction of a vector is often described using a unit vector.

在三维空间中，|p| = √x² + y² + z²。向量的方向通常用单位向量来描述。

A unit vector in the direction of p is:

与 p 方向相同的单位向量为：

p̂ = p / |p|

For example, if p = 3i + 4j, then |p| = 5 and p̂ = (3/5)i + (4/5)j.

例如，若 p = 3i + 4j，则 |p| = 5，因此 p̂ = (3/5)i + (4/5)j。

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4. Geometry: Midpoint of a Line Segment | 几何应用：线段的中点

The position vector of the midpoint M of a line segment joining points A and B is the average of their position vectors:

连接点 A 和点 B 的线段中点 M，其位置向量等于 A、B 位置向量的平均值：

m = (a + b) / 2

th two and three dimensions. For example, if a = i + 2j and b = 5i + 8j, then m = 3i + 5j.

这个结论在二维和三维空间中均成立。例如，若 a = i + 2j，b = 5i + 8j，则 m = 3i + 5j。

Notice that the midpoint formula is a special case of the section formula when the ratio is 1 : 1.

注意，中点公式是定比分点公式在 1 : 1 时的特殊情况。

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5. Dividing a Line Segment in a Given Ratio | 按定比分点

t P divides the line segment AB in the ratio AP : PB = m : n. The position vector of P is:

设点 P 将线段 AB 分成 AP : PB = m : n，则点 P 的位置向量为：

p = (n a + m b) / (m + n)

if A has position vector 2i + j, B has position vector 8i + 7j, and AP : PB = 1 : 2, then:

例如，A 的位置向量为 2i + j，B 的位置向量为 8i + 7j，且 AP : PB = 1 : 2，则：

p = (2(2i + j) + 1(8i + 7j)) / 3 = 4i + 3j

This formula is very useful in coordinate geometry and vector proofs.

这个公式在坐标几何和向量证明中非常有用。

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6. Relative Position Vectors | 相对位置向量

The position vector of B relative to A is the displacement vector from A to B:

点 B 相对于点 A 的位置向量，就是从 A 到 B 的位移向量：

b − a

This is often written as r_B/A or simply AB. It shows where B is located if we imagine A as the origin.

这通常记为 r_B/A 或直接写为 AB。它表示把 A 视为原点时，B 所在的位置。

In mechanics, if two particles have position vectors r₁ and r₂ at time t, then r₂ − r₁ gives the position of the second particle relative to the first. This is essential for solving collision and closest-approach problems.

在力学中，如果两个质点在时刻 t 的位置向量分别为 r₁ 和 r₂，那么 r₂ − r₁ 就表示第二个质点相对于第一个质点的位置。这是解决碰撞问题和最近距离问题的关键。

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

A straight line can be described using a fixed point A and a direction vector d. The position vector r of any point on the line is:

一条直线可以用一个固定点 A 和一个方向向量 d 来描述。直线上任意一点的位置向量 r 为：

r = a + t d

Here t is a scalar parameter. If the line passes through two points A and B, then d can be taken as b − a, so:

其中 t 是参数标量。如果直线经过两点 A 和 B，则 d 可取为 b − a，因此：

r = a + t(b − a)

ation is particularly useful for finding whether a point lies on a line or for finding intersection points.

当 t = 0 时，r = a；当 t = 1 时，r = b。这个方程在判断点是否在直线上以及求交点时特别有用。

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8. Motion: Position, Velocity and Acceleration | 运动学：位置、速度与加速度

In kinematics, the position vector of a particle is often given as a function of time, r(t). The velocity vector is the derivative of position with respect to time:

在运动学中，质点的位置向量通常表示为时间 t 的函数 r(t)。速度向量是位置对时间的导数：

v = dr/dt

The acceleration vector is the derivative of velocity with respect to time:

加速度向量是速度对时间的导数：

a = dv/dt = d²r/dt²

For example, if r(t) = t²i + 3tj, then v(t) = 2ti + 3j. At t = 2, the velocity is 4i + 3j.

例如，若 r(t) = t²i + 3tj，则 v(t) = 2ti + 3j。当 t = 2 时，速度为 4i + 3j。

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9. Distance Between Two Moving Particles | 两个运动质点之间的距离

If two particles have position vectors r₁(t) and r₂(t), their displacement is r₂ − r₁. The distance between them at time t is simply the magnitude:

如果两个质点的位置向量分别为 r₁(t) 和 r₂(t)，则它们的位移为 r₂ − r₁。任意时刻 t 两者之间的距离就是该向量的模长：

d(t) = |r₂(t) − r₁(t)|

To find the minimum distance, you can square the distance expression and then minimise with respect to t. This avoids working with square roots and is a common Edexcel exam technique.

要求最短距离，可以先对距离表达式取平方，然后对 t 求最小值。这样可以避免处理根号，是 Edexcel 考试中常见的技巧。

+ 2tj, then r₂ − r₁ = (4 − 2t)i + (t − 1)j. The squared distance is (4 − 2t)² + (t − 1)².

例如，若 r₁ = ti + (t + 1)j，r₂ = (4 − t)i + 2tj，则 r₂ − r₁ = (4 − 2t)i + (t − 1)j。距离平方为 (4 − 2t)² + (t − 1)²。

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10. Using Position Vectors in Geometric Proofs | 用位置向量进行几何证明

ition vectors of A, B and D be a, b and d. Then the position vector of C is c = b + d − a.

位置向量是证明几何结论的有力工具。例如，考虑平行四边形 ABCD。设 A、B、D 的位置向量分别为 a、b、d，则 C 的位置向量为 c = b + d − a。

The midpoint of AC has position vector:

AC 的中点位置向量为：

(a + c) / 2 = (a + b + d − a) / 2 = (b + d) / 2

The midpoint of BD has position vector:

BD 的中点位置向量为：

(b + d) / 2

Since the midpoints are equal, the diagonals of a parallelogram bisect each other.

因为两个中点位置相同，所以平行四边形的对角线互相平分。

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11. Scalar Product and Direction | 数量积与方向

The scalar product (dot product) of two position vectors a and b is defined as:

两个位置向量 a 和 b 的数量积定义为：

a · b = |a||b| cos θ

where θ is the angle between the vectors. If a = x₁i + y₁j and b = x₂i + y₂j, then:

其中 θ 是两个向量之间的夹角。如果 a = x₁i + y₁j，b = x₂i + y₂j，则：

a · b = x₁x₂ + y₁y₂

0, the vectors are perpendicular. This condition can be used to find unknown coordinates.

若 a · b = 0，则两个向量互相垂直。这个条件常用于求未知坐标。

ple, if a = 3i + 4j and b = 4i + qj, then a · b = 12 + 4q. Setting a · b = 0 gives q = −3.

例如，若 a = 3i + 4j，b = 4i + qj，则 a · b = 12 + 4q。令 a · b = 0，得 q = −3。

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12. Exam-Style Worked Example | 考试型例题

Points P and Q have position vectors p = 2i + 3j and q = 6i + yj. Given that |PQ| = 5, find the possible values of y.

点 P 和 Q 的位置向量分别为 p = 2i + 3j 和 q = 6i + yj。已知 |PQ| = 5，求 y 的可能值。

Solution / 解答:

PQ = q − p = (6 − 2)i + (y − 3)j = 4i + (y − 3)j.

PQ = q − p = (6 − 2)i + (y − 3)j = 4i + (y − 3)j。

|PQ| = √(4² + (y − 3)²) = √(16 + (y − 3)²).

|PQ| = √(4² + (y − 3)²) = √(16 + (y − 3)²)。

Since |PQ| = 5, we have 16 + (y − 3)² = 25, so (y − 3)² = 9.

因为 |PQ| = 5，所以 16 + (y − 3)² = 25，即 (y − 3)² = 9。

Hence y − 3 = ±3, which gives y = 6 or y = 0.

因此 y − 3 = ±3，解得 y = 6 或 y = 0。

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