📚 3D1 Light and Dark | 3D1 光线与阴影
In Further Mathematics, the topic ‘3D1 Light and Dark’ explores how vector geometry can be used to model the behaviour of light rays, the formation of shadows, and the interaction between light sources and objects in three‑dimensional space. By treating light as a family of straight lines and using the parametric equations of lines and planes, we can precisely determine where shadows fall, whether a point is illuminated or in darkness, and how the shape of a shadow relates to the object casting it. This application of 3D vectors is not only a core part of the A‑Level Further Maths syllabus but also provides the mathematical foundation for computer graphics, optical design, and architectural visualisation.
在进阶数学中，“3D1 光线与阴影”这一专题利用向量几何来模拟光线的传播、阴影的形成，以及光源与三维空间中物体的相互作用。我们把光看作一族直线，运用直线和平面的参数方程，就能够精确地求出阴影落下的位置、判断某一点是被照亮还是处于黑暗中，还能分析阴影形状与产生阴影的物体之间的关系。这一三维向量的应用不仅是A‑Level进阶数学大纲的核心内容，也为计算机图形学、光学设计和建筑可视化提供了数学基础。
1. Understanding Light Rays as Vectors | 将光线理解为向量
A ray of light travels in a straight line from a light source until it hits an obstacle or a surface. In vector terms, this means we can represent a light ray using a parametric straight line equation: a fixed starting point (the source or a point on the object) plus a scalar multiple of a direction vector. This simple idea allows us to solve intersection problems algebraically.
一束光线从光源出发沿直线传播，直到遇到障碍物或表面。用向量的语言来说，我们可以用直线的参数方程来表示光线：一个固定起点（光源或物体上的点）加上方向向量的标量倍。这一简单的思想使我们能够用代数方法求解交点问题。
For instance, if a point light source is located at L and an object has a vertex at P, the ray that would pass through P and potentially cast a shadow has direction P − L. The ray equation is r = L + λ(P − L), with λ ≥ 0.
例如，如果点光源位于 L，物体有一个顶点在 P，那么穿过 P 并可能投下阴影的光线方向就是 P − L。光线方程为 r = L + λ(P − L)，其中 λ ≥ 0。
2. Representing Light Sources | 表示光源
In 3D vector geometry, we distinguish two common types of light source: the point light and the directional light. A point light, such as a small bulb, emits rays in all directions from a single coordinate L. A directional light, such as sunlight arriving from a very great distance, is modelled by a fixed direction vector d, and all rays are parallel to one another. Both can be handled using the ray equation, but the interpretation of the direction vector differs.
在三维向量几何中，我们区分两种常见的光源：点光源和方向光（平行光）。点光源，如小灯泡，从单个坐标 L 向所有方向发射光线。方向光，如来自极远处的阳光，用一个固定的方向向量 d 来模拟，所有光线都彼此平行。两者都可以用光线方程来处理，但方向向量的含义不同。
- Point light: ray equation r = L + λv, where v depends on the specific object point.
- Directional light: ray equation r = P + μd, with the same d for all points P on the object.
- 点光源：光线方程 r = L + λv，其中 v 取决于具体的物体点。
- 方向光：光线方程 r = P + μd，对于物体上所有点 P 使用相同的 d。
3. Equation of a Light Ray | 光线方程
The standard vector equation of a line in 3D, r = a + tb, is the workhorse of the Light and Dark topic. Here a is the position vector of a known point on the ray, b is the direction vector, and t is a real parameter. For light rays we restrict t ≥ 0, because light travels forward from the source.
三维空间中直线的标准向量方程 r = a + tb 是光线与阴影专题的基础工具。其中 a 是光线上已知点的位置向量，b 是方向向量，t 是实数参数。对于光线，我们限制 t ≥ 0，因为光从光源向前传播。
When we want to find where a ray hits a plane, we substitute the parametric form into the Cartesian or scalar product equation of the plane. Solving for t gives the point of intersection, provided t ≥ 0.
当我们想求光线与平面的交点时，只需将参数形式代入平面的笛卡尔方程或点积方程。解出 t 即得交点，前提是 t ≥ 0。
Ray: r = (x₀, y₀, z₀) + t(a, b, c), t ≥ 0
4. Objects and Shadows | 物体与阴影
An opaque object placed between a light source and a screen will block some of the light, creating an umbra – the fully shaded region. In our simplified mathematical model, the object is usually a set of points, line segments, or polygons. The shadow is the projection of these geometric elements onto a receiving surface (often a plane) along the direction of the light rays.
放置在光源和屏幕之间的不透明物体会遮挡部分光线，形成本影——完全阴影区域。在我们简化的数学模型中，物体通常是一组点、线段或多边形。阴影就是这些几何元素沿光线方向在接收表面（通常是一个平面）上的投影。
The crucial idea is that a point Q on the object generates a shadow point S on the plane precisely where the ray from the light source through Q meets the plane. Therefore, the whole shadow shape can be built by finding the shadows of all the object’s vertices.
关键思想是：物体上的点 Q 会在平面上产生一个阴影点 S，该点恰好是来自光源并经过 Q 的光线与平面的交点。因此，整个阴影形状可以通过找到物体所有顶点的阴影来构建。
5. Finding the Shadow of a Point | 求点的阴影
To find the shadow of a point P cast by a point light L on a plane Π, we first form the direction vector v = P − L. The ray is r = L + λv. Substituting this into the plane equation r·n = d (or ax + by + cz = d) yields an equation in λ. Solving gives the value of λ for which the ray meets the plane. The shadow point S is then r(λ).
对于点光源 L 在平面 Π 上投下点 P 的阴影，我们首先形成方向向量 v = P − L。光线为 r = L + λv。将其代入平面方程 r·n = d（或 ax + by + cz = d），得到关于 λ 的方程。解出 λ 即得光线与平面相遇时的参数值，阴影点 S 就是 r(λ)。
Example: L = (1, 2, 3), P = (4, 6, 3), plane z = 0. Then v = (3, 4, 0). The ray is (1+3λ, 2+4λ, 3+0λ). Setting z = 0 gives 3 = 0, which is impossible. This means the ray is parallel to the plane and never intersects it; the point P casts no shadow on that plane.
示例：L = (1, 2, 3)，P = (4, 6, 3)，平面 z = 0。那么 v = (3, 4, 0)。光线为 (1+3λ, 2+4λ, 3+0λ)。令 z = 0 得到 3 = 0，这是不可能的。这说明光线与平面平行，永不相交；点 P 在该平面上不会投下阴影。
When a valid λ > 0 is found, the shadow point is in the forward direction of the ray and physically meaningful.
当求出一个有效的 λ > 0 时，阴影点位于光线的正方向上，具有物理意义。
6. Shadow of a Line Segment | 线段的阴影
A line segment AB acts as an obstacle. Its shadow on a plane is typically another segment connecting the shadows of A and B, provided the entire segment lies between the source and the plane. The shadow segment is therefore SA−SB, where SA and SB are the shadow points of A and B respectively.
线段 AB 作为一个障碍物，它在平面上的阴影通常是一条连接 A 和 B 的阴影点的线段，前提是整个线段位于光源与平面之间。因此，阴影线段就是 SA−SB，其中 SA 和 SB 分别是 A 和 B 的阴影点。
If the segment is partially behind the light source, only the portion with λ ≥ 0 for rays from L counts. In exam problems, however, the objects are usually placed entirely between the source and the projection plane to avoid these complications.
如果线段部分地位于光源后方，那么只有从 L 出发的光线满足 λ ≥ 0 的部分才计入阴影。然而在考试题中，物体通常完全置于光源与投影平面之间，以避免这些复杂情况。
7. Shadows of Polygons | 多边形的阴影
When the opaque object is a polygon (typically a triangle or quadrilateral), its shadow on a plane is found by projecting each vertex. Because light rays are straight, a planar polygon casts a shadow that is also a polygon, with vertices in the same order – the projection preserves collinearity and the order of points along lines.
当不透明物体是多边形（通常是三角形或四边形）时，其在平面上的阴影可通过投影每个顶点得到。由于光线是直线，一个平面多边形投下的阴影也是一个多边形，且顶点顺序相同——投影保持共线性和直线上点的顺序。
Thus, to draw the shadow of triangle ABC, compute the shadow points A’, B’, C’ and join them to form triangle A’B’C’. This triangle represents the boundary of the umbra.
因此，要画出三角形 ABC 的阴影，只需计算出阴影点 A’、B’、C’，然后将它们连接成三角形 A’B’C’。这个三角形就代表了本影的边界。
If the polygon is not parallel to the projection plane, the shadow will be a distorted version of the original shape, but the area can still be calculated using vector cross products.
如果多边形不平行于投影平面，阴影将是原始形状的变形版本，但其面积仍然可以利用向量叉积来计算。
8. Intersection of a Ray with a Plane | 光线与平面的交点
The algebraic procedure for finding the intersection of a ray and a plane is central to the topic. Given a ray r = a + tb and a plane with equation r·n = p, substitute to obtain (a + tb)·n = p. Solve for t:
求解光线与平面交点的代数步骤是本专题的核心。给定光线 r = a + tb 和平面方程 r·n = p，代入得到 (a + tb)·n = p。解出 t：
t = (p − a·n) / (b·n), provided b·n ≠ 0
If b·n = 0, the ray is parallel to the plane and there is either no intersection or the ray lies entirely in the plane. For a valid shadow, we need t ≥ 0. This condition ensures the intersection occurs in front of the light source, not behind it.
如果 b·n = 0，光线与平面平行，则要么没有交点，要么光线完全落在平面上。要形成有效的阴影，我们需要 t ≥ 0。这一条件确保交点发生在光源的前方，而非后方。
In many exam questions the plane is a coordinate plane (e.g., x = 0, y = 0, z = 0), which simplifies the arithmetic considerably.
在许多考试题中，平面是坐标平面（如 x = 0，y = 0，z = 0），这会大大简化计算。
9. Special Case: Parallel Light (Directional Light) | 特殊情况：平行光（方向光）
Sunlight is often approximated by a directional light with fixed direction vector d. The ray from a point P on the object is then r = P + μd, μ ≥ 0. Notice that the starting point is on the object, not at a single source. The shadow of P on a plane r·n = p is found by solving (P + μd)·n = p for μ. The same formula for t applies with a = P and b = d.
太阳光常被近似为具有固定方向向量 d 的方向光。此时，来自物体上点 P 的光线为 r = P + μd，μ ≥ 0。注意，起点在物体上，而不是在一个光源点。P 在平面 r·n = p 上的阴影可通过解 (P + μd)·n = p 求出 μ。之前求 t 的公式同样适用，只需令 a = P，b = d。
With directional light, all object points use the same d, making the shadow a simple translation of the object’s vertices along d until they hit the plane. This often results in a parallelogram-shaped shadow for a quadrilateral.
对于方向光，所有物体点使用相同的 d，这使得阴影成为物体顶点沿 d 平移至平面上的结果。对于一个四边形，这通常会形成一个平行四边形状的阴影。
10. Worked Example: Point Light and a Triangle | 例题：点光源与三角形
Let a point light be at L(2, 3, 1) and a triangular opaque screen have vertices A(5, 6, 2), B(4, 8, 3), C(6, 7, 5). Find the shadow of this triangle on the floor, which is the plane z = 0.
设点光源位于 L(2, 3, 1)，一个三角形不透明屏幕的顶点为 A(5, 6, 2)、B(4, 8, 3)、C(6, 7, 5)。求该三角形在地面（平面 z = 0）上的阴影。
For each vertex we form the ray L to vertex, then set z = 0.
对每个顶点，我们构造从 L 到该顶点的光线，然后令 z = 0。
Vertex A: vA = A − L = (3, 3, 1). Ray: (2+3λ, 3+3λ, 1+λ). Set z: 1 + λ = 0 ⇒ λ = −1. But λ must be ≥ 0 for a forward ray. Since λ = −1 is negative, this means the ray from L through A goes downwards and hits the plane behind the light source – effectively, A is behind the source relative to the floor? Let us check: the light is at z=1, A at z=2. Both are above the floor. The direction vector has a positive z component (1), so as λ increases, z increases. The ray will never hit z = 0 in the forward direction. Thus A does not produce a shadow on the floor. The triangle is positioned above the light and the ray goes upward; the floor is below the light, so the triangle does not block the light going to the floor. This reveals a typical pitfall: we must check the sign of λ.
顶点 A：vA = A − L = (3, 3, 1)。光线：(2+3λ, 3+3λ, 1+λ)。令 z：1 + λ = 0 ⇒ λ = −1。但正向光线要求 λ ≥ 0。由于 λ = −1 为负，这意味着从 L 穿过 A 的光线向下走并与平面在光源后方相交——实际上，A 可能在地面投影无效。检查一下：光源 z=1，A 在 z=2，两者都在地板上方。方向向量的 z 分量为正（1），因此随着 λ 增大，z 增加，光线永远不会向前碰到 z=0。因此 A 不会在地面上产生阴影。该三角形位于光源上方，光线向上走；地面在光源下方，所以三角形并不遮挡照向地面的光。这揭示了一个典型陷阱：必须检查 λ 的符号。
Vertex B: vB = B − L = (2, 5, 2). Ray: (2+2λ, 3+5λ, 1+2λ). Set z = 0: 1 + 2λ = 0 ⇒ λ = −0.5, again negative, no forward shadow.
顶点 B：vB = B − L = (2, 5, 2)。光线：(2+2λ, 3+5λ, 1+2λ)。令 z = 0：1 + 2λ = 0 ⇒ λ = −0.5，同样为负，无正向阴影。
Vertex C: vC = C − L = (4, 4, 4). Ray: (2+4λ, 3+4λ, 1+4λ). Set z = 0: 1 + 4λ = 0 ⇒ λ = −0.25, negative.
顶点 C：vC = C − L = (4, 4, 4)。光线：(2+4λ, 3+4λ, 1+4λ)。令 z = 0：1 + 4λ = 0 ⇒ λ = −0.25，为负。
Conclusion: this triangle does not cast a shadow on the floor under this point light because all vertices are above the light and rays passing through them go upward. The light is blocked only if an object is between the light and the surface. For a shadow on the floor, the object must be below the light (z < 1) or the light must be positioned higher so that rays go downwards. Let us adjust the example to make it instructive: swap the light to L(2, 3, 5) and keep the same triangle A, B, C. Now light is above the triangle (z=5 > z=2,3). Compute shadows.
结论：这个三角形在此点光源下不会在地面投下阴影，因为所有顶点都在光源上方，穿过它们的光线向上走。只有当物体位于光源和表面之间时才会遮挡光线。要在地面产生阴影，物体必须在光源下方（z < 1）或光源位置更高使光线向下。为了让例题更有意义，我们将光源调整为 L(2, 3, 5)，三角形顶点保持不变。此时光源在三角形上方（z=5 > z=2,3）。重新计算阴影。
New L(2,3,5). For A(5,6,2): vA = (3,3,−3). Ray: (2+3λ, 3+3λ, 5−3λ). Set z=0: 5 − 3λ = 0 ⇒ λ = 5/3 (positive). Shadow A’ = (2+5, 3+5, 0) = (7, 8, 0). For B(4,8,3): vB = (2,5,−2). Ray: (2+2λ, 3+5λ, 5−2λ). Set z=0: λ=2.5 → B’ = (2+5, 3+12.5, 0) = (7, 15.5, 0). For C(6,7,5): vC = (4,4,0). Ray: (2+4λ, 3+4λ, 5+0λ). z is always 5, never 0 → no intersection. So only A and B produce finite shadow points; C’s ray is parallel to the floor. The shadow would be just the line segment A’B’. This shows that even a triangle can cast a degenerate shadow if some rays miss the plane.
新光源 L(2,3,5)。对于 A(5,6,2)：vA = (3,3,−3)。光线：(2+3λ, 3+3λ, 5−3λ)。令 z=0：5−3λ=0 ⇒ λ=5/3（正）。阴影 A’ = (2+5, 3+5, 0) = (7, 8, 0)。对于 B(4,8,3)：vB = (2,5,−2)。光线：(2+2λ, 3+5λ, 5−2λ)。令 z=0：λ=2.5 → B’ = (7, 15.5, 0)。对于 C(6,7,5)：vC = (4,4,0)。光线：(2+4λ, 3+4λ, 5+0λ)。z 永远为 5，不会变 0 → 无交点。因此只有 A 和 B 产生有限阴影点；C 的光线与地面平行。阴影将只是线段 A’B’。这表明即使是一个三角形，如果某些光线与平面不相交，也可能产生退化的阴影。
11. Hard and Soft Shadows – A Mathematical View | 硬阴影与软阴影——数学视角
The Light and Dark topic in Further Maths typically deals only with point light sources and directional lights, which produce hard, sharply defined shadows (umbra only). In reality, an extended light source creates a penumbra – a region of partial shadow – leading to soft edges. While you will not be asked to compute penumbras in A‑Level exams, understanding the geometric difference helps: a hard shadow boundary is formed by a single ray from a point source; a soft shadow involves a family of rays from different points of an extended source.
进阶数学中的“光线与阴影”专题通常只涉及点光源和方向光，它们产生边缘清晰分明的硬阴影（仅有本影）。现实中，面光源会产生半影——部分阴影区域——从而形成柔和的边缘。虽然A‑Level考试不会要求计算半影，但理解其几何差异仍有帮助：硬阴影边界由点光源的单根光线形成；软阴影则涉及从面光源上不同点发出的一族光线。
In vector terms, computing a soft shadow would require finding the envelope of shadows from multiple point sources. This is beyond the syllabus but forms a natural extension for those interested in computer graphics.
用向量的语言来说，计算软阴影需要求出多个点光源产生的阴影的包络。这超出了考试大纲，但对于对计算机图形学感兴趣的同学来说，是一个自然的延伸。
12. Summary and Key Exam Tips | 总结与考试技巧
The Light and Dark problems in 3D vector geometry reduce to a systematic process: set up the ray equation, substitute into the plane equation, solve for the parameter, and check λ ≥ 0. Always draw a quick sketch to visualise the relative positions of the light, object and shadow plane. Care with signs is essential – a negative λ means the intersection is behind the light source and should be discarded. When dealing with polygons, project vertices one by one and connect them; if a vertex’s ray is parallel to the plane, the shadow may be open or degenerate, and you should state this clearly in your solution.
三维向量几何中的“光线与阴影”问题归结为一个系统化的过程：建立光线方程，代入平面方程，解出参数，并检查 λ ≥ 0。始终画一个快速草图来直观把握光源、物体和阴影平面的相对位置。符号问题至关重要——λ 为负意味着交点在光源后方，应被舍弃。处理多边形时，逐点投影顶点并连接起来；如果某个顶点的光线与平面平行，阴影可能是不封闭的或退化的，你应在解答中清楚地说明这一点。
Remember, the mathematics is exact: a point is in shadow if the straight line from the source to that point is blocked. This principle underpins all questions on this topic. Mastering it will not only earn you marks in the exam but also deepen your appreciation of how pure mathematics elegantly describes the physical world.
请记住，数学是精确的：如果从光源到某一点的直线被遮挡，该点就处于阴影中。这一原理是本专题所有问题的基础。掌握它不仅能帮助你在考试中得分，还能加深你对纯粹数学如何优雅地描述物理世界的理解。
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