Exercise 11A.3: Vector Operations and Magnitudes | 练习 11A.3：向量运算与模长

📚 Exercise 11A.3: Vector Operations and Magnitudes | 练习 11A.3：向量运算与模长

In IB Mathematics, vectors form a fundamental building block for geometry, mechanics, and later topics such as lines and planes. Exercise 11A.3 typically revisits the basic operations of vectors in component form: addition, scalar multiplication, and the calculation of magnitude. Mastering these skills now will save you time and prevent careless errors in exams. This article breaks down the key ideas and provides worked examples mirroring the style of the exercise.

在 IB 数学中，向量是几何、力学以及后续直线与平面等内容的基础。练习 11A.3 通常会回顾分量形式下向量的基本运算：加法、标量乘法以及模长的计算。现在掌握这些技能可以为你节省时间并避免考试中的粗心错误。本文将分解关键概念，并提供与练习风格一致的详细解题示例。

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1. What Are Vectors? | 什么是向量？

rs i and j: v = v₁ i + v₂ j. In three dimensions we add the k component: v = (v₁, v₂, v₃).

向量是既有大小又有方向的量，而标量只有大小。在二维空间中向量通常写成分量形式 v = (v₁, v₂) 或使用单位向量 i 和 j：v = v₁ i + v₂ j。在三维空间中我们增加 k 分量：v = (v₁, v₂, v₃)。

Two vectors are equal if and only if their corresponding components are identical. The zero vector (0, 0) has magnitude zero and no specific direction.

两个向量相等当且仅当它们的对应分量完全相同。零向量 (0, 0) 的模长为零且无特定方向。

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2. Vector Addition | 向量加法

₂ + b₂). Geometrically, vector addition corresponds to placing the tail of the second vector at the head of the first – the "tip‑to‑tail" method.

两个向量相加时，只需将它们的对应分量相加。若 a = (a₁, a₂) 且 b = (b₁, b₂)，则 a + b = (a₁ + b₁, a₂ + b₂)。从几何角度看，向量加法相当于将第二个向量的尾部放到第一个向量的箭头上——即“首尾相连”的方法。

a + b) + c = a + (b + c). You can also add vectors graphically by forming a parallelogram.

该运算满足交换律：a + b = b + a，以及结合律：(a + b) + c = a + (b + c)。你还可以通过画出平行四边形来图解加法。

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3. Scalar Multiplication | 标量乘法

tretches (or shrinks) its length and may reverse its direction if k is negative. Component‑wise: k·v = (k·v₁, k·v₂). For example, 3·(2, –1) = (6, –3).

用一个标量 k 去乘以一个向量，会拉伸（或缩短）其长度，若 k 为负还会反转方向。从分量角度看：k·v = (k·v₁, k·v₂)。例如 3·(2, –1) = (6, –3)。

Scalar multiplication is distributive: k(a + b) = k a + k b. Negative of a vector –v has the same magnitude as v but points in the opposite direction.

标量乘法满足分配律：k(a + b) = k a + k b。向量的负向量 –v 与 v 具有相同的模长但指向相反方向。

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4. Magnitude of a Vector | 向量的模长

The magnitude (or length) of a 2D vector v = (x, y) is found by the distance formula: |v| = √x² + y². In three dimensions for v = (x, y, z), the magnitude is √x² + y² + z². The magnitude is always a non‑negative scalar.

二维向量 v = (x, y) 的模长（或长度）由距离公式给出：|v| = √x² + y²。对于三维向量 v = (x, y, z)，模长为 √x² + y² + z²。模长始终是非负的标量。

e zero vector. The distance between two points A and B is the magnitude of the vector AB→.

模长为零的向量是零向量。两点 A 和 B 之间的距离就是向量 AB→ 的模长。

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5. Unit Vectors | 单位向量

A unit vector has magnitude 1. Any non‑zero vector v can be turned into a unit vector û in the same direction by dividing by its own magnitude: û = v / |v|. The standard basis vectors i, j, k are the unit vectors along the positive x‑, y‑ and z‑axes.

单位向量的模长为 1。任何非零向量 v 都可以通过除以其模长转化为同方向的单位向量 û：û = v / |v|。标准基向量 i、j、k 分别是沿正 x、y、z 轴方向的单位向量。

Finding a unit vector is a common exam request; double‑check your arithmetic to avoid losing marks on a simple calculation.

求单位向量是常见的考试要求；请仔细检查算术，以免在简单计算中失分。

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6. Position Vectors | 位置向量

ition vector is p = (x, y, z). The vector from point A to point B is given by AB→ = b − a.

位置向量以原点 O 为起点、以点 P 为终点，记为 OP→ 或简写为 p。其分量就是该点的坐标。对于 P(x, y, z)，位置向量为 p = (x, y, z)。从点 A 到点 B 的向量由 AB→ = b − a 给出。

This subtraction is crucial for solving geometry problems: always think “end minus start” when forming a vector between two points.

这种减法在解决几何问题时至关重要：建立两点间的向量时，始终记住“终点减起点”。

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7. Worked Example – Addition and Scalar Multiplication | 示例 – 加法与标量乘法

Given a = (3, –2, 1) and b = (–1, 4, 0), find 2a − 3b.

已知 a = (3, –2, 1) 和 b = (–1, 4, 0)，求 2a − 3b。

Solution:
2a = (6, –4, 2)
3b = (–3, 12, 0)
2a − 3b = (6 − (–3), –4 − 12, 2 − 0) = (9, –16, 2)

解：
2a = (6, –4, 2)
3b = (–3, 12, 0)
2a − 3b = (6 − (–3), –4 − 12, 2 − 0) = (9, –16, 2)

Always perform scalar multiplication before addition or subtraction, and work component by component to minimise mistakes.

务必先进行标量乘法再进行加法或减法，并逐分量计算以减少错误。

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8. Worked Example – Magnitude and Unit Vector | 示例 – 模长与单位向量

Find the magnitude of v = (4, –3, 0) and the unit vector in the same direction.

求 v = (4, –3, 0) 的模长以及同方向的单位向量。

Solution:
|v| = √(4² + (–3)² + 0²) = √16 + 9 + 0 = √25 = 5.
Unit vector û = v / |v| = (4/5, –3/5, 0) = (0.8, –0.6, 0).

解：
|v| = √(4² + (–3)² + 0²) = √16 + 9 + 0 = √25 = 5。
单位向量 û = v / |v| = (4/5, –3/5, 0) = (0.8, –0.6, 0)。

When the magnitude is a neat number, take a moment to verify your square root; many IB questions are designed to simplify nicely.

当模长是一个整齐的数时，花点时间验证开方结果；很多 IB 题目设计得能够美妙化简。

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9. Worked Example – Solving for Unknown Components | 示例 – 求解未知分量

The vector p = (2, k, –1) has magnitude 3. Find the possible values of k.

向量 p = (2, k, –1) 的模长为 3。求 k 的可能值。

Solution:
|p| = √(2² + k² + (–1)²) = √4 + k² + 1 = √k² + 5 = 3.
Square both sides: k² + 5 = 9 ⇒ k² = 4 ⇒ k = ±2.

解：
|p| = √(2² + k² + (–1)²) = √4 + k² + 1 = √k² + 5 = 3。
两边平方：k² + 5 = 9 ⇒ k² = 4 ⇒ k = ±2。

This pattern appears frequently – setting up an equation from the magnitude formula and then solving a simple quadratic.

这种模式经常出现——根据模长公式建立方程，然后解一个简单的二次方程。

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10. Vector Operations in Context: Displacement | 向量运算的应用：位移

Suppose a particle moves from A(1, 2) to B(5, –3). Its displacement vector is AB→ = (5 − 1, –3 − 2) = (4, –5). The distance travelled is the magnitude: √(4² + (–5)²) = √16 + 25 = √41.

假设一个质点从 A(1, 2) 移动到 B(5, –3)。其位移向量为 AB→ = (5 − 1, –3 − 2) = (4, –5)。经过的距离是模长：√(4² + (–5)²) = √16 + 25 = √41。

Remember: displacement is a vector, distance is a scalar. These two are easily confused; the vector tells you both how far and in which direction, while the magnitude only tells you how far.

记住：位移是向量，距离是标量。二者很容易混淆；向量既告诉你有多远又告诉你方向，而模长只告诉你多远。

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11. Common Mistakes & How to Avoid Them | 常见错误及如何避免

• Forgetting to square a negative component: (–4)² = 16, not –16. Apply brackets carefully.
• 忘记对负分量平方：(–4)² = 16，而不是 –16。小心使用括号。
• Mixing addition and scalar multiplication order: 2a + 3b means scalar multiples first, then add.
• 搞混加法与标量乘法的顺序：2a + 3b 表示先做标量乘法，再加。
• Calculating vector AB→ the wrong way round: AB→ = b − a, not a − b.
• 求向量 AB→ 时方向弄反：AB→ = b − a，而不是 a − b。
• Assuming the magnitude of a + b equals |a| + |b|: this is true only if the vectors are parallel and point in the same direction (triangle inequality).
• 误以为 |a + b| 等于 |a| + |b|：只有当两个向量平行且同向时才成立（三角不等式）。

Double‑checking these points will build strong foundational habits.

复查这些要点可以培养扎实的基础习惯。

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12. Practice Questions to Test Yourself | 自测练习题

Try these short exercises to consolidate the skills from Exercise 11A.3:

尝试以下简短的练习来巩固练习 11A.3 中的技能：

1. Given a = (1, −2, 3) and b = (0, 4, −1), find 3a + 2b.
2. 已知 a = (1, −2, 3) 和 b = (0, 4, −1)，求 3a + 2b。
3. Find the unit vector in the direction of v = (2, 2, −1).
4. 求 v = (2, 2, −1) 方向的单位向量。
5. If |(x, 3, −4)| = √26, find x.
6. 若 |(x, 3, −4)| = √26，求 x。
7. A point P has position vector (3, −5). Write down OP→ and find its length.
8. 点 P 的位置向量为 (3, −5)。写出 OP→ 并求其长度。

Answers: 1. (3, 2, 7) 2. (2/3, 2/3, −1/3) 3. x = ±1 4. |OP| = √34.

答案：1. (3, 2, 7) 2. (2/3, 2/3, −1/3) 3. x = ±1 4. |OP| = √34。

Smooth execution of these operations is expected in the IB exam, often embedded in larger vector questions involving lines and planes.

在 IB 考试中，通常要求流畅地执行这些运算，并且它们常嵌套在涉及直线和平面的大题中。

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