📚 Magnitude and Direction of Vectors | 向量的大小与方向
In A-Level Mathematics and Mechanics, a vector is a quantity that has both magnitude and direction, whereas a scalar has magnitude only. This article explains how to calculate the magnitude and direction of a vector, how to convert between component form and magnitude-direction form, and how these ideas are used with forces, velocities and displacements.
在 A-Level 数学和力学中，向量是既有大小又有方向的量，而标量只有大小。本文将解释如何计算向量的大小和方向，如何在分量形式与大小-方向形式之间转换，以及这些概念如何应用于力、速度和位移。
1. Scalars, Vectors and Notation | 标量、向量与记号
A scalar is fully described by a single number and a unit. Common scalar quantities include distance, speed, mass, time, temperature and energy. For example, saying that a car travels at 60 km h⁻¹ gives only its speed, not its direction.
标量只需一个数值和单位就能完整描述。常见的标量包括距离、速率、质量、时间、温度和能量。例如，说一辆车以 60 km h⁻¹ 行驶，只给出了速率，没有给出方向。
A vector needs both magnitude and direction. Displacement, velocity, acceleration, force and momentum are vector quantities. Saying that a car travels at 60 km h⁻¹ due north is a vector statement because the direction is specified.
向量需要同时给出大小和方向。位移、速度、加速度、力和动量都是向量。说一辆车以 60 km h⁻¹ 向正北行驶，就是向量表述，因为方向已被说明。
Vectors are usually printed in bold, such as a, or underlined when handwritten, such as a. The magnitude of a vector is written |a| or simply a, and it is never negative. Two vectors are equal only if they have the same magnitude and the same direction, no matter where they are drawn on a diagram.
向量通常用粗体印刷，如 a，手写时加下划线，如 a。向量的大小写作 |a|，或简写为 a，它永远不是负数。两个向量相等，当且仅当它们大小相同、方向相同，与它们在图上画在什么位置无关。
2. Component Form and Column Vectors | 分量形式与列向量
In two dimensions, a vector can be written as a = x i + y j, where i and j are the standard unit vectors along the positive x-axis and the positive y-axis. The numbers x and y are called the components of the vector.
在二维中，向量可以写作 a = x i + y j，其中 i 和 j 分别是沿正 x 轴和正 y 轴的标准单位向量。数 x 和 y 称为该向量的分量。
For example, a = 3i + 4j has a horizontal component 3 and a vertical component 4. The same vector can be written as a column vector with entries 3 and 4:
例如，a = 3i + 4j 的水平分量为 3，竖直分量为 4。同一个向量也可写成列向量，元素为 3 和 4：
a = ( 3 , 4 )ᵀ
If A = (x₁, y₁) and B = (x₂, y₂), then the vector from A to B is AB = (x₂ − x₁)i + (y₂ − y₁)j. This is useful in coordinate geometry and in mechanics problems involving position vectors.
若 A = (x₁, y₁)，B = (x₂, y₂)，则从 A 到 B 的向量为 AB = (x₂ − x₁)i + (y₂ − y₁)j。这在坐标几何和涉及位置向量的力学问题中很有用。
3. Magnitude of a 2D Vector | 二维向量的大小
The magnitude of a vector a = x i + y j is written |a| and is found using Pythagoras' theorem. It represents the length of the directed line segment from the tail to the head of the vector.
向量 a = x i + y j 的大小写作 |a|，用勾股定理求得。它表示从向量起点到终点的有向线段长度。
|a| = √x² + y²
For example, the magnitude of a = 3i + 4j is |a| = √3² + 4² = √25 s the classic 3-4-5 right-angled triangle, and it appears frequently in A-Level questions.
例如，a = 3i + 4j 的大小为 |a| = √3² + 4² = √25 = 5。这就是经典的 3-4-5 直角三角形，在 A-Level 题目中经常出现。
The magnitude is always non-negative. It is zero only for the zero vector 0 = 0i + 0j. If a question asks for a distance, speed or force, the answer should be the magnitude, not the vector itself.
大小永远非负。只有零向量 0 = 0i + 0j 的大小才是零。如果题目问距离、速率或力的大小，答案应是模长，而不是向量本身。
4. Direction of a 2D Vector | 二维向量的方向
The direction of a vector is usually described by the angle θ that the vector makes with the positive x-axis, measured anticlockwise. The tangent of this angle is the ratio of the vertical component to the horizontal component:
向量的方向通常用它与正 x 轴所成的角 θ 来描述，按逆时针方向测量。该角的正切等于竖直分量与水平分量之比：
tan θ = y / x
For a = 3i + 4j so θ = tan⁻¹(4 / 3) ≈ 53.1°. This means the vector points 53.1° above the positive x-axis.
对于 a = 3i + 4j，tan θ = 4 / 3，所以 θ = tan⁻¹(4 / 3) ≈ 53.1°。这意味着该向量指向正 x 轴上方 53.1° 的方向。
In navigation and mechanics, direction is sometimes given as a bearing, measured clockwise from north. If θ is the anticlockwise angle from the positive x-axis, then the bearing is 90° − θ when the vector lies in the first quadrant. Always read the wording of the question carefully to decide which convention is expected.
在导航和力学中，方向有时用方位角表示，即从正北开始顺时针测量。若 θ 是从正 x 轴逆时针测量的角，当向量在第一象限时，方位角为 90° − θ。务必仔细阅读题目措辞，判断应采用哪种约定。
5. From Magnitude and Direction to Components | 由大小和方向求分量
If a vector has magnitude r and direction θ measured anticlockwise from the positive x-axis, its components are found by resolving:
若向量的大小为 r，方向为从正 x 轴逆时针测量的 θ，则其分量可通过分解得到：
x = r cos θ y = r sin θ
Therefore the vector can be written as a = r cos θ i + r sin θ j. This conversion is essential when a force or velocity is given by its size and direction.
因此该向量可写作 a = r cos θ i + r sin θ j。当一个力或速度以大小和方向给出时，这种转换非常关键。
a force of 10 N acting at 30° above the positive x-axis has components Fₓ = 10 cos 30° = 5√3 ≈ 8.66 N and Fy = 10 sin 30° = 5 N. So F ≈ 8.66i + 5j N.
例如，一个 10 N 的力沿正 x 轴上方 30° 方向作用，其分量为 Fₓ = 10 cos 30° = 5√3 ≈ 8.66 N，Fy = 10 sin 30° = 5 N。所以 F ≈ 8.66i + 5j N。
Remember that cos θ gives the component along the reference direction, while sin θ gives the perpendicular component. Drawing a right-angled triangle makes the correct choice obvious and reduces sign errors.
记住，cos θ 给出沿参考方向的分量，sin θ 给出垂直方向的分量。画一个直角三角形可以让正确选择变得显而易见，并减少符号错误。
6. Quadrants and tan⁻¹ Corrections | 象限与 tan⁻¹ 修正
ause tan θ = tan(θ + 180°). Therefore, after calculating tan⁻¹(y / x), you must check the quadrant of the vector and adjust the angle if necessary.
计算器上的反正切函数 tan⁻¹ 返回 −90° 到 90° 之间的角。它不能区分相反方向，因为 tan θ = tan(θ + 180°)。因此，在计算 tan⁻¹(y / x) 之后，必须检查向量所在象限，必要时调整角度。
- First quadrant (x > 0, y > 0): θ = tan⁻¹(y / x). 中文：第一象限 (x > 0, y > 0)：θ = tan⁻¹(y / x)。
- Second quadrant (x < 0, y > 0): θ = tan⁻¹(y / x) + 180°. 中文：第二象限 (x < 0, y > 0)：θ = tan⁻¹(y / x) + 180°。
- Third quadrant (x < 0, y < 0): θ = tan⁻¹(y / x) + 180°. 中文：第三象限 (x < 0, y < 0)：θ = tan⁻¹(y / x) + 180°。
- Fourth quadrant (x > 0, y < 0): θ = tan⁻¹(y / x) + 360°, or use the negative angle. 中文：第四象限 (x > 0, y < 0)：θ = tan⁻¹(y / x) + 360°，或使用负角。
For example, for a = −3i + 4j 3) ≈ −53.1°. Since the vector lies in the second quadrant, θ ≈ −53.1° + 180° = 126.9°. For a = −3i − 4j, the third-quadrant correction gives θ ≈ 233.1°.
例如，对于 a = −3i + 4j，tan⁻¹(4 / −3) ≈ −53.1°。由于向量在第二象限，θ ≈ −53.1° + 180° = 126.9°。对于 a = −3i − 4j，第三象限修正后 θ ≈ 233.1°。
7. Unit Vectors and Normalisation | 单位向量与归一化
A unit vector is a vector of magnitude 1. The unit vector in the direction of a is obtained by dividing a by its magnitude:
单位向量是大小为 1 的向量。沿 a 方向的单位向量可通过将 a 除以其大小得到：
â = a / |a|
For a = 3i + 4j, |a| = 5, so the unit vector is â = (3 / 5)i + (4 / 5)j = 0.6i + 0.8j. It points in the same direction as a but has length 1.
对于 a = 3i + 4j，|a| = 5，所以单位向量为 â = (3 / 5)i + (4 / 5)j = 0.6i + 0.8j。它与 a 方向相同，但长度为 1。
Unit vectors are useful because they separate direction from magnitude. Any vector can be written as a = |a| â. This form appears in vector equations of lines and in mechanics when a force acts along a given direction.
单位向量很有用，因为它们把方向与大小分离开来。任意向量都可写作 a = |a| â。这种形式出现在直线的向量方程中，也出现在力学中力沿给定方向作用的问题里。
8. Magnitude and Direction in 3D | 三维向量的大小与方向
In three dimensions, a vector is written as a = x i + y j + z k, where k is the unit vector along the positive z-axis. Its magnitude extends Pythagoras' theorem to three dimensions:
在三维中，向量写作 a = x i + y j + z k，其中 k 是沿正 z 轴的单位向量。其大小将勾股定理推广到三维：
|a| = √x² + y² + z²
For example, for a = 2i − j + 2k, |a| = √(2² + (−1)² + 2²) = √9 = 3. The direction is described by the direction cosines:
例如，对于 a = 2i − j + 2k，|a| = √(2² + (−1)² + 2²) = √9 = 3。方向由方向余弦描述：
cos α = x / |a| cos β = y / |a| cos γ = z / |a|
α + cos²β + cos²γ = 1. For the example above, the direction cosines are 2/3, −1/3 and 2/3.
这里 α、β、γ 分别是向量与正 x、y、z 轴所成的角。它们满足恒等式 cos²α + cos²β + cos²γ = 1。对于上述例子，方向余弦为 2/3、−1/3 和 2/3。
9. Applications to Forces | 在力中的应用
Forces are vectors, so they can be added by adding their components. If several forces act on a particle, the resultant force R has components Rₓ = ΣFₓ and Ry = ΣFy. Its magnitude and direction are then found in the usual way:
力是向量，因此可以通过相加分量来相加。若有多个力作用在一个质点上，合力 R 的分量为 Rₓ = ΣFₓ，Ry = ΣFy。然后按通常方法求其大小和方向：
|R| = √Rₓ² + Ry² tan θ = Ry / Rₓ
For example, an 8 N force acting due east and a 6 N force acting due north give a resultant of magnitude √8² + 6² = 10 N. Its direction is tan⁻¹(6 / 8) ≈ 36.9° north of east.
例如，一个 8 N 的力向正东，一个 6 N 的力向正北，合力大小为 √8² + 6² = 10 N。其方向为 tan⁻¹(6 / 8) ≈ 36.9°，即东偏北 36.9°。
If the resultant force is zero, the forces are in equilibrium. This means the sum of the components in every direction is zero. Resolving forces into perpendicular components is often the fastest way to solve equilibrium problems involving inclined planes, strings and pulleys.
若合力为零，则这些力处于平衡状态。这意味着在每个方向上的分量之和都为零。将力分解为互相垂直的分量，通常是解决涉及斜面、绳子和滑轮的平衡问题的最快方法。
10. Applications to Velocity and Displacement | 在速度与位移中的应用
Velocity is a vector; speed is its magnitude. If an object moves with velocity v = vₓi + vyj, its speed is |v| = √vₓ² + vy². The displacement after time t is s = vt, provided the velocity is constant.
速度是向量，速率是其大小。若物体以速度 v = vₓi + vyj 运动，其速率为 |v| = √vₓ² + vy²。若速度恒定，经过时间 t 后的位移为 s = vt。
Relative velocity is also a vector calculation. The velocity of A relative to B is vAB = vA − vB. For
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