📚 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
ion 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
≤ 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
nd 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
₁: (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。
= (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
ind 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。
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