📚 Angles between Lines and Planes | 线与平面之间的夹角
In AQA A-Level Mathematics, vector geometry questions often ask you to find the angle between two lines, between a line and a plane, or between two planes. These angles are always measured as the acute angle, and they are usually found using direction vectors, normal vectors and the dot product.
在 AQA A-Level 数学中，向量几何题常要求你求两条直线之间、直线与平面之间或两个平面之间的夹角。这些夹角总是取锐角，通常借助方向向量、法向量和点积来计算。
1. Why Angles in 3D Matter | 为什么三维中的夹角很重要
In two dimensions, the angle between two lines is found from their gradients. In three dimensions, lines and planes are described by vector equations, so we need a reliable vector method. AQA exam questions regularly combine this topic with finding intersections, distances and Cartesian forms.
在二维中，两条直线之间的夹角可由斜率求得。在三维中，直线和平面用向量方程描述，因此我们需要可靠的向量方法。AQA 考试题常将这一内容与求交点、距离和笛卡尔形式结合考查。
A well-organised vector method gives you the acute angle quickly, without needing to draw a 3D diagram. This is especially useful when the line or plane is given in algebraic form.
一套清晰的向量方法能让你快速求出锐角，而不需要画出三维图形。当直线或平面以代数形式给出时，这一点尤其有用。
2. Direction Vectors and Normal Vectors | 方向向量与法向量
A line has the vector equation r = a + λd, where d is a direction vector. A plane can be written as r · n = k, where n is a normal vector perpendicular to the plane.
直线的向量方程为 r = a + λd，其中 d 是方向向量。平面可写成 r · n = k，其中 n 是垂直于该平面的法向量。
by Cartesian equations (x − x₁)/a = (y − y₁)/b = (z − z₁)/c, then the direction vector is d = (a, b, c). If a plane is given by ax + by + cz = d, then the normal vector is n = (a, b, c).
如果直线以笛卡尔形式 (x − x₁)/a = (y − y₁)/b = (z − z₁)/c 给出，则方向向量为 d = (a, b, c)。如果平面为 ax + by + cz = d，则法向量为 n = (a, b, c)。
Always write down the correct vector before applying any angle formula. This avoids the common mistake of using a plane point instead of its normal.
在应用任何夹角公式之前，一定要先写出正确的向量。这可以避免用平面上的点代替法向量这一常见错误。
3. Dot Product Reminder | 点积复习
For two vectors u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃), the dot product is u · v = u₁v₁ + u₂v₂ + u₃v₃. It is also equal to |u||v| cos θ, where θ is the angle between the two vectors.
对于两个向量 u = (u₁, u₂, u₃) 和 v = (v₁, v₂, v₃)，点积为 u · v = u₁v₁ + u₂v₂ + u₃v₃。它也等于 |u||v| cos θ，其中 θ 是两向量的夹角。
cos θ = (u · v) / (|u||v|)
If u · v f one vector is a scalar multiple of the other, they are parallel and the angle between them is 0° or 180°.
如果 u · v = 0，则两个向量垂直。如果一个向量是另一个向量的标量倍数，则它们平行，夹角为 0° 或 180°。
4. Angle between Two Lines | 两条直线之间的夹角
To find the angle between two lines, use their direction vectors d₁ and d₂. Since the angle between lines must be acute, take the absolute value of the dot product:
求两条直线之间的夹角时，使用它们的方向向量 d₁ 和 d₂。由于直线夹角必须为锐角，取点积的绝对值：
cos θ = |d₁ · d₂| / (|d₁||d₂|)
Here θ is the acute angle between the lines. If the dot product is negative, ignoring the sign gives the cosine of the acute angle rather than the obtuse supplement.
这里 θ 是两条直线之间的锐角。如果点积为负，忽略符号得到的是锐角的余弦值，而不是钝补角。
If d₁ · d₂ icular. If the direction vectors are parallel, the lines are parallel and the angle is 0°.
如果 d₁ · d₂ = 0，则两条直线垂直。如果方向向量平行，则两条直线平行，夹角为 0°。
5. Worked Example: Two Lines | 例题：两条直线
Find the acute angle between the lines r = (1, 0, 2) + λ(2, −1, 3) and r = (0, 3, 1) + μ(1, 2, −2).
求直线 r = (1, 0, 2) + λ(2, −1, 3) 与 r = (0, 3, 1) + μ(1, 2, −2) 之间的锐角。
Direction vectors are d₁ = (2, −1, 3) and d₂ = (1, 2, −2). The dot product is 2 × 1 + (−1) × 2 + 3 × (−2) = 2 − 2 − 6 = −6.
方向向量为 d₁ = (2, −1, 3) 和 d₂ = (1, 2, −2)。点积为 2 × 1 + (−1) × 2 + 3 × (−2) = 2 − 2 − 6 = −6。
The magnitudes are |d₁| = √(2² + (−1)² + 3²) = √14 and |d₂| = √(1² + 2² + (−2)²) = √9 = 3.
模长分别为 |d₁| = √(2² + (−1)² + 3²) = √14 和 |d₂| = √(1² + 2² + (−2)²) = √9 = 3。
cos θ = |−6| / (3√14) = 6 / (3√14) = 2 / √14
Therefore θ = cos⁻¹(2 / √14) ≈ 57.7° (to 1 decimal place).
因此 θ = cos⁻¹(2 / √14) ≈ 57.7°（保留 1 位小数）。
6. Angle between a Line and a Plane | 直线与平面之间的夹角
The angle between a line and a plane is defined as the angle between the line and its projection onto the plane. It is not the angle between the direction vector and the normal vector; it is the complement of that angle.
直线与平面之间的夹角定义为直线与其
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