IB Math: Vector Method for the Angle Between Two Lines | IB数学：两直线夹角的向量求法

📚 IB Math: Vector Method for the Angle Between Two Lines | IB数学：两直线夹角的向量求法

In IB Mathematics, especially in Analysis and Approaches and Applications and Interpretation, finding the angle between two straight lines is a classic vector problem. The vector method uses direction vectors and the dot product, giving a reliable way to find the angle without drawing a diagram.

在IB数学中，尤其是分析与方法（AA）和应用与解释（AI），求两条直线的夹角是非常经典的向量问题。向量法利用方向向量和点积，可以在不画图的情况下可靠地求出夹角。

---

1. Why Vector Method? | 为什么用向量法？

A straight line can be described completely by a fixed point and a direction vector. The angle between two lines depends only on their directions, not on their positions. This makes the vector method elegant and useful in both 2D and 3D.

一条直线可以由一个定点和方向向量完全描述。两条直线的夹角只取决于它们的方向，与位置无关。因此，向量法简洁高效，适用于二维和三维空间。

• Works in both 2D and 3D | 同时适用于二维和三维
• No need to draw accurate graphs | 无需精确作图
• Produces an exact angle using a calculator | 借助计算器即可得到精确角度

---

2. Direction Vectors of a Line | 直线的方向向量

For a line written in parametric form, L: r = a + λb, the vector b is a direction vector. If a line passes through two points A and B, a direction vector is AB = B − A.

对于参数形式的直线 L: r = a + λb，向量 b 就是方向向量。如果直线经过两点 A 和 B，那么方向向量可取 AB = B − A。

The common forms are shown below.

常见的直线形式如下表所示。

Form | 形式	Direction Vector | 方向向量
r = a + λb	b
(x − x₀)/p = (y − y₀)/q = (z − z₀)/r	(p, q, r)
Line through A and B	B − A

---

3. The Dot Product and the Cosine Formula | 点积与余弦公式

Let d₁ and d₂ be the direction vectors of two lines. The angle θ between the lines satisfies

设 d₁ 和 d₂ 分别是两条直线的方向向量。两直线的夹角 θ 满足

cos θ = |d₁ · d₂| / (|d₁||d₂|)

If d₁ = (x₁, y₁, z₁) and d₂ = (x₂, y₂, z₂), then the dot product is

若 d₁ = (x₁, y₁, z₁)，d₂ = (x₂, y₂, z₂)，则点积为

d₁ · d₂ = x₁x₂ + y₁y₂ + z₁z₂

The magnitude of a vector is |d| = √(x² + y² + z²).

向量的模为 |d| = √(x² + y² + z²)。

---

4. Choosing the Acute Angle | 取锐角

Lines have no direction, so the angle between two lines is conventionally the acute angle, with 0° ≤ θ ≤ 90°. The absolute value in the numerator ensures that we always obtain the acute angle.

直线没有方向，因此两条直线的夹角通常指锐角，即 0° ≤ θ ≤ 90°。分子上的绝对值保证了我们求出的一定是锐角。

If your calculator gives an obtuse value because you forgot the absolute value, subtract that value from 180°.

如果忘记使用绝对值，计算器给出的可能是钝角，此时用 180° 减去该值即可。

---

5. Worked Example 1 | 实例 1

Find the angle between L₁: r = (1, 2, 3) + t(1, −1, 2) and L₂: r = (0, 1, −2) + s(2, 1, −1).

求直线 L₁: r = (1, 2, 3) + t(1, −1, 2) 与 L₂: r = (0, 1, −2) + s(2, 1, −1) 的夹角。

The direction vectors are d₁ = (1, −1, 2) and d₂ = (2, 1, −1).

两条直线的方向向量为 d₁ = (1, −1, 2)，d₂ = (2, 1, −1)。

d₁ · d₂ = 1 × 2 + (−1) × 1 + 2 × (−1) = 2 − 1 − 2 = −1.

因此 d₁ · d₂ = 1 × 2 + (−1) × 1 + 2 × (−1) = 2 − 1 − 2 = −1。

|d₁| = √(1 + 1 + 4) = √6, and |d₂| = √(4 + 1 + 1) = √6.

|d₁| = √(1 + 1 + 4) = √6，|d₂| = √(4 + 1 + 1) = √6。

cos θ = |−1| / (√6 × √6) = 1/6

Therefore θ = cos⁻¹(1/6) ≈ 80.4°.

因此 θ = cos⁻¹(1/6) ≈ 80.4°。

Without the absolute value we would get 99.6°, which is the obtuse angle between the direction vectors, not the angle between the lines.

如果不加绝对值，我们会得到 99.6°，这是方向向量之间的钝角，而不是两直线之间的夹角。

---

6. Worked Example 2 | 实例 2

Find the angle between L₁: (x − 1)/2 = y/3 = (z + 2)/1 and the line L₂ passing through A(1, 1, 1) and B(2, −1, 3).

求直线 L₁: (x − 1)/2 = y/3 = (z + 2)/1 与经过 A(1, 1, 1)、B(2, −1, 3) 的直线 L₂ 的夹角。

For L₁, the direction vector is d₁ = (2, 3, 1). For L₂, a direction vector is d₂ = AB = (2 − 1, −1 − 1, 3 − 1) = (1, −2, 2).

L₁ 的方向向量为 d₁ = (2, 3, 1)。L₂ 的方向向量可取 d₂ = AB = (2 − 1, −1 − 1, 3 − 1) = (1, −2, 2)。

d₁ · d₂ = 2 × 1 + 3 × (−2) + 1 × 2 = 2 − 6 + 2 = −2.

所以 d₁ · d₂ = 2 × 1 + 3 × (−2) + 1 × 2 = 2 − 6 + 2 = −2。

|d₁| = √(4 + 9 + 1) = √14, and |d₂| = √(1 + 4 + 4) = 3.

|d₁| = √(4 + 9 + 1) = √14，|d₂| = √(1 + 4 + 4) = 3。

cos θ = |−2| / (√14 × 3) = 2 / (3√14)

θ = cos⁻¹(2 / (3√14)) ≈ 79.7°.

因此 θ = cos⁻¹(2 / (3√14)) ≈ 79.7°。

---

7. Special Cases | 特殊情况

Some cases can be checked quickly before applying the formula.

有些特殊情况可以在套用公式前快速判断。

Case | 情况	Condition | 条件	Angle | 夹角
Parallel | 平行	d₁ = k d₂	0°
Perpendicular | 垂直	d₁ · d₂ = 0	90°
General | 一般情况	Use the cosine formula | 使用余弦公式	0° to 90° | 0° 到 90°

Two identical lines are also parallel, so their angle is 0°.

两条重合直线也属于平行，因此夹角为 0°。

---

8. Common Mistakes | 常见错误

Students often lose marks on straightforward vector questions by making small errors.

在简单的向量题中，学生常因小错误而失分。

• Forgetting the absolute value sign | 忘记加绝对值
• Using position vectors instead of direction vectors | 用位置向量代替方向向量
• Adding components incorrectly in the dot product | 计算点积时分量对应错误
• Using the angle between normals in 2D without checking the relationship | 在平面中直接用法向量夹角而不验证关系
• Forgetting to take the square root when finding magnitudes | 求模时忘记开根号

---

9. Practice Exercise | 练习

Find the angle between L₁: r = (0, 4, 1) + t(1, 2, −2) and L₂: r = (2, 2, 2) + s(−2, 1, 1).

求直线 L₁: r = (0, 4, 1) + t(1, 2, −2) 与 L₂: r = (2, 2, 2) + s(−2, 1, 1) 的夹角。

Solution: d₁ = (1, 2, −2) and d₂ = (−2, 1, 1).

解答：d₁ = (1, 2, −2)，d₂ = (−2, 1, 1)。

d₁ · d₂ = 1 × (−2) + 2 × 1 + (−2) × 1 = −2 + 2 − 2 = −2.

因此 d₁ · d₂ = 1 × (−2) + 2 × 1 + (−2) × 1 = −2 + 2 − 2 = −2。

|d₁| = √(1 + 4 + 4) = 3, and |d₂| = √(4 + 1 + 1) = √6.

|d₁| = √(1 + 4 + 4) = 3，|d₂| = √(4 + 1 + 1) = √6。

cos θ = |−2| / (3√6) = 2 / (3√6)

θ ≈ 74.2°.

因此 θ ≈ 74.2°。

---

10. Summary | 总结

To find the angle between two lines, first identify their direction vectors, then apply the dot product formula with the absolute value.

要求两条直线的夹角，首先确定它们的方向向量，然后代入带绝对值的点积公式。

cos θ = |d₁ · d₂| / (|d₁||d₂|)

This method works for 2D and 3D, covers parallel and perpendicular cases, and avoids the need for graphing. Always remember to report the acute angle.

该方法适用于二维和三维空间，能处理平行和垂直等特殊情况，也无需作图。切记最终报告的是锐角。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）