📚 Coordinate Representation of Displacement Vectors in 2D Planes | 二维平面位移向量的坐标表示
In physics and mathematics, a displacement vector describes the change in position of an object from its initial point to its final point. In a two-dimensional Cartesian coordinate system, this vector can be represented elegantly by its coordinates, allowing us to perform calculations with precision and clarity.
在物理和数学中，位移向量描述物体从起点到终点的位置变化。在二维笛卡尔坐标系中，这个向量可以通过其坐标简洁地表示出来，使我们能够精确而清晰地进行计算。
This article explores the coordinate representation of displacement vectors in 2D planes, covering fundamental concepts, notation, vector operations, and their applications — all essential knowledge for IB Mathematics.
本文深入探讨二维平面中位移向量的坐标表示，涵盖基本概念、记号、向量运算及其应用——这些对于IB数学考试都是至关重要的知识。
1. Position Vectors and Displacement Vectors | 位置向量与位移向量
A position vector locates a point relative to the origin O(0, 0). For a point A(x₁, y₁), the position vector is written as OA = (x₁, y₁). It tells us how far and in which direction the point lies from the origin.
位置向量确定一个点相对于原点O(0, 0)的位置。对于点A(x₁, y₁)，位置向量记为OA = (x₁, y₁)。它告诉我们该点距原点的距离和方向。
A displacement vector describes the change in position between two points. If an object moves from point A(x₁, y₁) to point B(x₂, y₂), the displacement vector AB is the difference between the position vectors of B and A.
位移向量描述两点之间的位置变化。如果物体从点A(x₁, y₁)移动到点B(x₂, y₂)，则位移向量AB等于B和A的位置向量之差。
AB = (x₂ − x₁, y₂ − y₁)
Note that displacement is independent of the path taken — only the initial and final positions matter.
注意，位移与所经过的路径无关——仅与起点和终点的位置有关。
2. Unit Vectors in the 2D Coordinate Plane | 二维坐标平面中的单位向量
The Cartesian coordinate system defines two fundamental unit vectors: i = (1, 0) along the positive x-axis, and j = (0, 1) along the positive y-axis. Each has a magnitude of exactly 1 unit.
笛卡尔坐标系定义了两个基本单位向量：i = (1, 0) 沿x轴正方向，j = (0, 1) 沿y轴正方向。它们的模长都恰好为1个单位。
Any displacement vector in 2D can be expressed as a linear combination of these unit vectors:
二维空间中的任何位移向量都可以表示为这两个单位向量的线性组合：
v = x·i + y·j, written as v = (x, y)
Here, x is the horizontal component (along i) and y is the vertical component (along j). This dual notation — component form (x, y) and unit vector form x·i + y·j — is used interchangeably in IB exams.
这里，x称为水平分量（沿i方向），y称为垂直分量（沿j方向）。这种双重表示法——坐标形式(x, y)和单位向量形式x·i + y·j——在IB考试中可互换使用。
3. Component Representation of Displacement | 位移的分量表示
Consider two points A(2, 3) and B(7, 5). The displacement vector from A to B is calculated by subtracting the coordinates of A from those of B:
考虑两个点A(2, 3)和B(7, 5)。从A到B的位移向量通过将B的坐标减去A的坐标来计算：
AB = (7 − 2, 5 − 3) = (5, 2)
This means the object moved 5 units in the positive x-direction and 2 units in the positive y-direction. The components (5, 2) fully describe the displacement.
这意味着物体沿x轴正方向移动了5个单位，沿y轴正方向移动了2个单位。分量(5, 2)完整地描述了这次位移。
Conversely, the displacement from B to A would be:
反过来，从B到A的位移则为：
BA = (2 − 7, 3 − 5) = (−5, −2)
Notice that BA = −AB. The negative sign reverses the direction of the displacement.
注意到BA = −AB。负号表示位移方向相反。
4. Magnitude of a Displacement Vector | 位移向量的模长
The magnitude (or length) of a displacement vector v = (x, y) is denoted by |v| and is calculated using the Pythagorean theorem:
位移向量v = (x, y)的模长（或长度）记作|v|，使用勾股定理计算：
|v| = √x² + y²
For example, the magnitude of AB = (5, 2) is:
例如，AB = (5, 2)的模长为：
|AB| = √5² + 2² = √25 + 4 = √29 ≈ 5.39 units
The magnitude represents the straight-line distance between the initial and final points, regardless of direction.
模长表示起点到终点的直线距离，与方向无关。
5. Direction of a Displacement Vector | 位移向量的方向
be described by the angle θ it makes with the positive x-axis. This angle is found using:
位移向量v = (x, y)的方向可以用它与x轴正方向所成的角θ来描述。该角度通过下式求得：
tan θ = y / x, so θ = tan⁻¹(y / x)
However, careful attention must be paid to the quadrant in which the vector lies. For vector (5, 2):
然而，必须注意向量所在的象限。对于向量(5, 2)：
θ = tan⁻¹(2/5) ≈ 21.8°
Since both components are positive, the vector lies in the first quadrant, and this angle is correct. For vectors in other quadrants, add 180° as appropriate to obtain the correct reference angle.
由于两个分量均为正，向量位于第一象限，该角度即为正确值。对于位于其他象限的向量，需要适当加上180°以获得正确的参考角。
6. Equal and Opposite Vectors | 相等向量与相反向量
Two displacement vectors are equal if and only if they have the same components, meaning both the magnitude and direction are identical. In coordinate form:
两个位移向量相等当且仅当它们具有相同的分量，即模长和方向完全相同。在坐标形式下：
If u = (x₁, y₁) and v = (x₂, y₂), then u = v ⇔ x₁ = x₂ and y₁ = y₂
The negative y) is −v = (−x, −y). It has the same magnitude but points in the exact opposite direction.
向量v = (x, y)的相反向量为−v = (−x, −y)。它们的模长相同但方向完全相反。
(−5, −2) = −AB. The zero displacement vector is 0 = (0, 0), representing no net movement.
例如，如果AB = (5, 2)，则BA = (−5, −2) = −AB。零位移向量为0 = (0, 0)，表示没有净移动。
7. Addition and Subtraction of Vectors in Coordinate Form | 向量的坐标形式的加法与减法
displacement vectors u = (x₁, y₁) and v = (x₂, y₂), their sum is computed component-wise:
给定两个位移向量u = (x₁, y₁)和v = (x₂, y₂)，它们的和按分量计算：
u + v = (x₁ + x₂, y₁ + y₂)
For subtraction:
对于减法：
u − v = (x₁ − x₂, y₁ − y₂)
Example: Let u = (3, 4) and v = (1, −2). Then:
例如：设u = (3, 4)，v = (1, −2)。则：
u + v = (3 + 1, 4 + (−2)) = (4, 2)
u − v = (3 − 1, 4 − (−2)) = (2, 6)
Geometrically, vector addition follows the triangle law or parallelogram law, while subtraction corresponds to adding the negative vector.
在几何上，向量加法遵循三角形法则或平行四边形法则，而减法对应于加上相反向量。
8. Scalar Multiplication of Vectors | 向量的标量乘法
v = (x, y) is multiplied by a scalar k (a real number), each component is multiplied by k:
当位移向量v = (x, y)乘以标量k（一个实数）时，每个分量都乘以k：
k·v = (kx, ky)
a factor of k. If k < 0, the direction reverses. If k = 0, the result is the zero vector.
若k > 0，方向不变，模长缩放k倍。若k < 0，方向反转。若k = 0，结果为零向量。
Example: If v = (3, −2), then:
例如：若v = (3, −2)，则：
2v = (6, −4), −3v = (−9, 6), ½v = (1.5, −1)
The magnitude of k·v equals |k| × |v|, which is a straightforward consequence of the component-wise multiplication.
k·v的模长等于|k| × |v|，这是分量乘法的直接结果。
9. Parallel (Collinear) Vectors | 平行（共线）向量
Two non-zero vectors u = (x₁, y₁) and v = (x₂, y₂) are parallel (or collinear) if one is a scalar multiple of the other:
两个非零向量u = (x₁, y₁)和v = (x₂, y₂)平行（或共线）当且仅当一个向量是另一个向量的标量倍：
u = k·v for some scalar k
Equivalently, their components are proportional:
等价地，它们的对应分量成比例：
x₁ / x₂ = y₁ / y₂, provided x₂ ≠ 0 and y₂ ≠ 0
llel because u = 2·v. Also, (4, 6) and (−2, −3) are parallel because (4, 6) = −2·(−2, −3).
例如：向量u = (4, 6)和v = (2, 3)平行，因为u = 2·v。同样，(4, 6)和(−2, −3)平行，因为(4, 6) = −2·(−2, −3)。
10. Unit Vector in the Direction of a Given Vector | 沿给定向量方向的单位向量
A unit vector in the direction of v = (x, y) is obtained by dividing v by its magnitude:
沿v = (x, y)方向的单位向量通过将v除以其模长来获得：
v̂ = v / |v| = (x/√x² + y², y/√x² + y²)
This unit vector has a magnitude of 1 and points in exactly the same direction as v.
该单位向量的模长为1，且指向与v完全相同的方向。
For example, for v = (3, 4), the magnitude is |v| = 5, so:
例如，对于v = (3, 4)，模长为|v| = 5，因此：
v̂ = (3/5, 4/5)
This concept is frequently tested in IB exams, especially when separating magnitude from direction in kinematics problems.
这一概念在IB考试中经常出现，特别是在运动学问题中需要将模长与方向分离时。
11. Displacement Vector between Two Points: Worked Example | 两点间位移向量：完整示例
A particle moves from point P(−1, 4) to point Q(3, −1). Find: (a) the displacement vector PQ; (b) its magnitude; (c) the direction angle; (d) the unit vector in that direction.
一个质点从点P(−1, 4)移动到点Q(3, −1)。求：(a) 位移向量PQ；(b) 其模长；(c) 方向角；(d) 该方向的单位向量。
(a) The displacement vector:
(a) 位移向量：
PQ = (3 − (−1), −1 − 4) = (4, −5)
(b) The magnitude:
(b) 模长：
|PQ| = √(4² + (−5)²) = √16 + 25 = √41 ≈ 6.40 units
(c) The direction angle. Since x > 0 and y < 0, the vector lies in the fourth quadrant:
(c) 方向角。由于x > 0且y < 0，向量位于第四象限：
θ = tan⁻¹(|−5|/4) = tan⁻¹(5/4) ≈ 51.3° below the +x axis, so θ ≈ −51.3° or 308.7°
(d) The unit vector:
(d) 单位向量：
PQ̂ = (4/√41, −5/√41)
12. Applications in Problem-Solving | 在解题中的应用
Coordinate representation of displacement vectors is essential in solving problems involving:
位移向量的坐标表示在解决以下类型的问题中至关重要：
- Kinematics: tracking position changes and velocities in a plane
- Forces: representing forces as vectors for resultant calculations
- Navigation: computing resultant displacements from multiple legs of a journey
- Geometry: proving properties of shapes using vector methods
- Relative motion: finding displacement between two moving objects
- 运动学：追踪平面内的位置变化和速度
- 力：将力表示为向量以进行合力计算
- 导航：计算多段旅程的合位移
- 几何：使用向量方法证明图形的性质
- 相对运动：求两个运动物体之间的位移
For instance, if a hiker walks 3 km east then 4 km north, the total displacement from the starting point is (3, 4), with a magnitude of 5 km at an angle of 53.1° north of east — a direct application of vector addition in coordinate form.
例如，如果一个徒步者先向东走3公里，再向北走4公里，则从出发点的总位移为(3, 4)，模长为5公里，方向为东偏北53.1°——这正是坐标形式向量加法的直接应用。
Mastering the coordinate representation of displacement vectors provides a powerful and systematic approach to solving two-dimensional problems across physics and mathematics. Practice converting between geometric descriptions, coordinate forms, and unit-vector notations to build fluency and confidence for the IB exams.
熟练掌握位移向量的坐标表示，为解决物理和数学中的二维问题提供了强大而系统的方法。多加练习在几何描述、坐标形式和单位向量记法之间的转换，以提升熟练度和自信心，从容应对IB考试。
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