📚 3D Vector Geometry: Reflecting Light Sources | 三维向量几何：光源的反射
In A-level Further Mathematics, three-dimensional vector geometry extends the ideas of lines and planes into space. An elegant application of these concepts is the modelling of light rays reflecting from flat surfaces, such as mirrors or still water. By combining parametric vector equations with the dot product, you can trace light sources, find reflected rays, and determine the apparent position of an image behind a reflecting plane.
在A-level进阶数学中，三维向量几何将直线与平面的概念拓展到空间。这些概念的一个优美应用是模拟光线在平面镜或平静水面等平坦表面上的反射。通过将参数向量方程与数量积相结合，你可以追踪光源、求反射线，并确定反射平面后方像的视在位置。
1. Recap of 3D Coordinate Systems | 三维坐标系回顾
Three-dimensional space is described using three mutually perpendicular axes: the x-axis, y-axis and z-axis. A point P is written as (x, y, z) and its position vector is typically p = xi + yj + zk, where i, j and k are the unit vectors along the axes.
三维空间用三条互相垂直的坐标轴来描述：x轴、y轴和z轴。点P写作(x, y, z)，其位置向量通常为p = xi + yj + zk，其中i、j和k是沿坐标轴方向的单位向量。
Distances and midpoints obey the familiar extensions of 2D formulas. The distance between points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is √[(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²].
距离与中点的计算公式是二维情形的自然延伸。点A(x₁, y₁, z₁)与B(x₂, y₂, z₂)之间的距离为√[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]。
2. Vector Equations of Lines in 3D | 三维空间中的直线向量方程
A straight line in 3D can be expressed in parametric vector form: r = a + λb, where a is the position vector of a known point on the line, b is a direction vector parallel to the line, and λ is a real parameter.
三维空间中的直线可用参数向量形式表示：r = a + λb，其中a是直线上已知点的位置向量，b是与直线平行的方向向量，λ为实数参数。
If a line passes through points with position vectors p and q, a direction vector is q − p. The line can also be written as r = p + λ(q − p).
如果直线经过位置向量为p和q的两点，则一个方向向量为q−p。该直线也可写作r = p + λ(q−p)。
3. Equations of Planes | 平面方程
A plane can be defined in vector form by r · n = a · n = d, where n is a normal vector perpendicular to the plane, and a is the position vector of any fixed point in the plane.
平面的向量形式定义为r · n = a · n = d，其中n是垂直于平面的法向量，a是平面内任一固定点的位置向量。
In Cartesian form, this becomes ax + by + cz = d, where n = ai + bj + ck. The normal vector is read directly from the coefficients of x, y and z.
在笛卡尔形式中，方程化为ax + by + cz = d，其中n = ai + bj + ck。法向量可直接从x、y、z的系数读出。
For reflection problems, identifying the unit normal vector is crucial. If n is any normal vector, the unit normal is n̂ = n / |n|.
在反射问题中，确定单位法向量的方向至关重要。若n为任意法向量，则单位法向量为n̂ = n / |n|。
4. Dot Product and Angle Between Vectors | 数量积与向量夹角
For two vectors u and v, the dot product is u · v = |u||v| cos θ, where θ is the angle between them. This product is central to calculating angles and projections.
对于两个向量u和v，数量积为u · v = |u||v| cos θ，其中θ是二者的夹角。该乘积是计算角度和投影的核心工具。
If the vectors are expressed in components, u · v = u₁v₁ + u₂v₂ + u₃v₃. The angle can be found from cos θ = (u · v) / (|u||v|).
若向量以分量表示，u · v = u₁v₁ + u₂v₂ + u₃v₃。可由cos θ = (u · v) / (|u||v|)求出夹角。
5. Angle Between a Line and a Plane | 直线与平面的夹角
The angle between a line and a plane is defined as the complement of the angle between the line’s direction vector b and the plane’s normal n. If α is the angle between the line and the plane, then sin α = |b · n| / (|b||n|).
直线与平面的夹角定义为该直线的方向向量b与平面法向量n之间夹角的余角。若α为线面角，则sin α = |b · n| / (|b||n|)。
This relationship is useful when checking whether an incident ray is parallel to the plane or verifying the law of reflection.
当检验入射线是否与平面平行或验证反射定律时，这一关系十分有用。
6. Laws of Reflection in Vector Form | 向量形式的反射定律
When a light ray strikes a smooth plane, the angle of incidence equals the angle of reflection, and the incident ray, reflected ray and the normal all lie in the same plane. In vector terms, if i is a unit vector along the incident ray (pointing towards the surface) and n̂ is the unit normal pointing into the mirror, then the direction of the reflected ray r is given by:
当光线射到光滑平面上时，入射角等于反射角，且入射线、反射线与法线共面。用向量表述，若i是沿入射方向的单位向量（指向表面），n̂是指向镜面内部的单位法向量，则反射线的方向向量r由下式给出：
r = i − 2 (i · n̂) n̂
This formula reflects the velocity vector across the surface; the component parallel to the mirror is unchanged, while the perpendicular component reverses sign.
该公式将速度向量关于表面进行镜面反射；平行于镜面的分量保持不变，而垂直于镜面的分量改变符号。
Note: it is essential that the incident direction vector i points from the light source towards the plane, and n̂ points from the plane toward the side from which the light comes. Many mistakes arise from inconsistent sign conventions.
注意：入射方向向量i必须从光源指向平面，n̂必须从平面指向光线来源的一侧。符号惯例不一致是常见的错误来源。
7. Reflecting a Point in a Plane | 点关于平面的反射
The image of a point L (the light source) in a plane can be found by dropping a perpendicular to the plane and extending it an equal distance on the other side. If l is the position vector of L, and the plane has equation r · n̂ = d (with n̂ unit normal), then the foot of the perpendicular F is f = l + tn̂, where t = d − l · n̂.
点L（光源）在平面中的像可通过作垂线并等距延长至另一侧来求得。若l为L的位置向量，平面方程为r · n̂ = d（n̂为单位法向量），则垂足F为f = l + tn̂，其中t = d − l · n̂。
The image point L’ has position vector l’ = l + 2tn̂ = l + 2(d − l · n̂)n̂. This is the virtual image of the light source behind the mirror.
像点L’的位置向量为l’ = l + 2tn̂ = l + 2(d − l · n̂)n̂。这正是光源在镜面后方的虚像。
8. Finding the Reflected Ray from a Light Source | 求光源的反射线
Given a point light source L and a plane mirror, we often need the equation of the reflected ray that leaves the mirror after hitting a specific point of incidence P. First find the incident direction vector i = (vector from L to P) / |LP|. Then compute the reflected direction using r = i − 2(i · n̂)n̂. The reflected ray is R(λ) = p + λr, where p is the position vector of P.
已知点光源L和一个平面镜，通常需要求出光线在入射点P击中镜面后反射线的方程。首先找到入射方向向量i = (从L到P的向量) / |LP|。然后用r = i − 2(i · n̂)n̂计算反射方向。反射线为R(λ) = p + λr，其中p是P的位置向量。
If the point of incidence is unknown, you may need to solve for the intersection of the incoming line from L towards the plane, set λ to find the hit point, then proceed as above.
若入射点未知，你可能需要先将从L指向平面方向的入射线与平面方程联立，解出λ以求得碰撞点，再按上述步骤进行。
9. Determining the Virtual Image and Its Uses | 虚像的确定及其用途
For many ray-tracing problems, it is easier to work with the virtual image L’ of the light source. The reflected ray from point P appears to come from L’. In fact, the line from L’ to P is exactly the continuation of the reflected ray, just on the opposite side of the mirror.
对许多光线追踪问题，借助光源的虚像L’会使求解更简单。从P点出发的反射线看起来来自L’。实际上，从L’到P的连线正是反射线在镜面后方的延长线。
Thus, to find the reflected ray without computing the reflection formula, you can simply join L’ and P and use that direction vector (pointing from the mirror into the real world). This approach is particularly elegant in exam problems asking for the path of a ray after reflection.
因此，无需套用反射公式便可求得反射线：只需连接L’与P，并将该方向向量（从镜面指向真实世界）作为反射方向。在要求给出反射后光线路径的考题中，这一方法尤为简洁。
10. Worked Example 1: Light Reflecting from a Mirror Plane | 例题一：光从镜面反射
Problem: A light source is at L(2, 1, 3). A mirror lies in the plane x + y + z = 6. A ray from L strikes the plane at P(1, 2, 3). Find the vector equation of the reflected ray.
题目：光源位于L(2, 1, 3)。镜面在平面x + y + z = 6内。从L出发的光线在P(1, 2, 3)处击中平面。求反射线的向量方程。
Solution: The plane’s normal is n = i + j + k. Check that P lies on the plane: 1+2+3=6, correct. The incident direction from L to P is LP = (−1, 1, 0). Its magnitude is √2. The unit incident vector i = (−1/√2, 1/√2, 0). Unit normal n̂ = (1/√3, 1/√3, 1/√3). Compute i · n̂ = (−1/√2)(1/√3) + (1/√2)(1/√3) + 0 = 0. Thus the reflected direction r = i − 2(0)n̂ = i. The reflected ray is R(λ) = p + λi = (1,2,3) + λ(−1/√2, 1/√2, 0).
解：平面的法向量为n = i + j + k。验证P在平面上：1+2+3=6，正确。从L到P的入射方向为LP = (−1, 1, 0)，模长为√2。单位入射向量i = (−1/√2, 1/√2, 0)。单位法向量n̂ = (1/√3, 1/√3, 1/√3)。计算i · n̂ = (−1/√2)(1/√3) + (1/√2)(1/√3) + 0 = 0。因此反射方向r = i − 2(0)n̂ = i。反射线为R(λ) = p + λi = (1,2,3) + λ(−1/√2, 1/√2, 0)。
Interpretation: Since the incident vector is perpendicular to the normal, it lies in the plane, so the ray just glances along the mirror. The reflected ray follows the same direction.
解读：由于入射向量与法向量垂直，光线位于平面内，因此光线只是沿着镜面擦过。反射线沿同一方向前进。
11. Worked Example 2: Image and Reflected Ray Without P Given | 例题二：未给P点时的像与反射线
Problem: A point light at (0, 0, 5) shines towards the plane z = 1. Find the virtual image and the equation of the reflected ray that passes through the point (2, 2, 0).
题目：点光源在(0, 0, 5)处，朝向平面z = 1发光。求虚像方程以及经过点(2, 2, 0)的反射线方程。
Solution: Plane z = 1 has unit normal n̂ = k (pointing upward, but we take n̂ = −k if light comes from above? Light from z=5 down to z=1, so the light approaches from the +z side. The plane’s normal pointing towards the source should be −k. However for image formula we can use any normal orientation as long as sign is consistent. Let’s use plane equation r · k = 1. The distance from L to plane along k is 5 − 1 = 4. So image L’ = L − 2·4·k = (0,0,5) − 8k = (0,0,−3).
解：平面z = 1的单位法向量可取n̂ = k（方向朝上，但光线从z=5射向z=1，所以光源从+z侧来。为使法向量指向光源，应取−k。但对于像的公式，只要符号自洽即可。我们用平面方程r · k = 1。L到平面沿k的距离为5 − 1 = 4。因此像L’ = L − 2·4·k = (0,0,5) − 8k = (0,0,−3)。
Now any reflected ray appears to come from L’. The required ray passes through (2,2,0) and seems to emanate from (0,0,−3). The direction vector from L’ to the point is (2,2,3). So the reflected ray is r(λ) = (2,2,0) + λ(2,2,3). (We start the ray at the hit point on the mirror; we could find the exact intersection with z=1, but the equation is acceptable in this form for many exam boards.)
此时任何反射线看起来都来自L’。所需的反射线经过(2,2,0)且似乎从(0,0,−3)发出。从L’到该点的方向向量为(2,2,3)。因此反射线为r(λ) = (2,2,0) + λ(2,2,3)。（光线从镜面击中点出发；我们可以求出与z=1的交点，但对许多考纲而言，上述形式也可接受。）
To be precise, find intersection of line from L’ to (2,2,0) with z=1. Parametric: x=2+2λ, y=2+2λ, z=0+3λ. Set z=1 gives 3λ=1 so λ=1/3. Point on mirror is (2+2/3, 2+2/3, 1) = (8/3, 8/3, 1). Reflected ray is then r = (8/3, 8/3, 1) + μ(2,2,3).
精确做法是，求L’到(2,2,0)的直线与z=1的交点。参数方程：x=2+2λ, y=2+2λ, z=0+3λ。令z=1得3λ=1，故λ=1/3。镜面上击中点为(8/3, 8/3, 1)。反射线为r = (8/3, 8/3, 1) + μ(2,2,3)。
12. Exam Tips and Common Pitfalls | 考试技巧与常见错误
Always draw a clear sketch to establish the direction of the normal relative to the incoming light. Consistently define your unit normal n̂ as pointing from the surface towards the incident side.
务必画出清晰的示意图，以确定法线相对于入射光的方向。始终将单位法向量n̂定义为从表面指向入射侧。
Check that your incident vector points towards the plane, not away. If you use the light’s position vector and the point of incidence, the vector from source to plane is correct.
检查入射向量是否指向平面，而非背离平面。若你使用光源位置向量和入射点，从光源到平面的向量方向是正确的。
When finding the image, the formula l’ = l + 2tn̂ assumes n̂ is unit and t is the signed distance from the point to the plane. For plane r·n̂ = d, the signed distance is d − l·n̂.
求像点时，公式l’ = l + 2tn̂要求n̂为单位向量，且t为点到平面的有号距离。对于平面r·n̂ = d，有号距离为d − l·n̂。
Be careful with radicals: you may leave direction vectors un-normalised if the question only asks for a vector equation. However, when computing dot products for reflection, using a unit normal is essential; otherwise scale the formula accordingly or normalise first.
处理根式时要小心：若题目仅要求向量方程，方向向量可以不单位化。但在为反射计算数量积时，必须使用单位法向量；否则需相应缩放公式，或先单位化。
Remember that the reflected ray is a half-line starting at the point of incidence; the parameter λ ≥ 0.
记住反射线是从入射点出发的射线，参数λ ≥ 0。
Practise identifying whether the final answer requires the line in vector or Cartesian form. Being fluent in changing between forms saves time.
多练习判断答案要求的是向量形式还是笛卡尔形式。熟练进行形式转换能节省时间。
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