📚 3D3 How shadows form | 3D3 阴影的形成原理
In Further Mathematics, the study of how shadows form in three dimensions is a direct application of vector geometry. By modelling light sources as points or parallel rays, and surfaces as planes, we can use line–plane intersections to find the exact position of a shadow cast by a point, a line segment, or a more complex object. This topic bridges abstract vector equations with physically intuitive spatial reasoning, making it a favourite for exam questions that test both calculation and geometric understanding.
在进阶数学中，三维阴影形成的研究是向量几何的直接应用。通过将光源模型化为点或平行光线，将表面视为平面，我们可以利用直线与平面的交点来精确求出点、线段或更复杂物体的阴影位置。这一主题将抽象的向量方程与直观的空间推理连接起来，因此常出现在考查计算能力与几何理解的考试题中。
1. Shadows as Projections | 阴影即投影
A shadow is formed when an opaque object blocks light from a source, stopping rays from reaching a surface. Mathematically, finding the shadow of a single point reduces to finding the intersection of a straight line (the light ray) with a plane (the receiving surface). The type of light source – point or parallel – changes the line’s direction but not the fundamental intersection method.
当不透明物体遮挡光源发出的光线，阻止其到达某一表面时，便形成阴影。从数学上看，求一个点的阴影可归结为求一条直线（光线）与一个平面（接收面）的交点。光源的类型——点光源或平行光——会改变直线的方向，但不会改变基本的求交方法。
2. Point Light Source and Planes | 点光源与平面
Suppose we have a point light source at L, an opaque point P, and a receiving plane π with equation n·x = k, where n is a normal vector. The light ray that passes through P travels along the line through L and P. To cast a shadow, this line must continue beyond P in the direction from L to P. The shadow of P is the point Q where the extended line meets π.
设点光源位于 L，不透明点位于 P，接收平面 π 满足方程 n·x = k，其中 n 为法向量。穿过 P 的光线沿着经过 L 和 P 的直线传播。要形成影子，这条直线必须沿从 L 到 P 的方向延伸到 P 之外。P 的影子即该延长线与 π 的交点 Q。
3. Deriving the Shadow Point Formula | 推导阴影点公式
The line containing L and P can be expressed in parametric form as r = L + t(P – L), where t is a real parameter. Substituting this into the plane equation n·x = k gives:
n·(L + t(P – L)) = k
Solving for t yields:
t = (k – n·L) / (n·(P – L))
The shadow point is then Q = L + t(P – L). For a physically meaningful shadow, we require t > 1, which means Q lies on the opposite side of P from the light source. If 0 < t < 1, the intersection falls between L and P, so no real shadow is formed on that side of the plane. If the denominator n·(P – L) is zero, the ray is parallel to the plane and never meets it.
包含 L 和 P 的直线可用参数方程表示为 r = L + t(P – L)，其中 t 为实数参数。将其代入平面方程 n·x = k 得到：
n·(L + t(P – L)) = k
解出 t：
t = (k – n·L) / (n·(P – L))
阴影点为 Q = L + t(P – L)。要得到有物理意义的影子，需满足 t > 1，即 Q 位于 P 相对于光源的另一侧。若 0 < t < 1，则交点在 L 和 P 之间，平面这一侧不会形成真实的影子。若分母 n·(P – L) 为零，说明光线平行于平面，永不相交。
4. Parallel Light Projection | 平行光投影
When light comes from a very distant source, such as the sun, all rays can be considered parallel, travelling in a fixed direction vector d. For a point P, the ray is given by r = P + λd, with λ ≥ 0. Substituting into the plane n·x = k gives:
λ = (k – n·P) / (n·d)
The shadow point is Q = P + λd. A real shadow requires λ > 0, indicating the ray travels forward to hit the plane. If n·d = 0, the light rays are parallel to the plane and no shadow is cast.
当光线来自极远的光源（如太阳）时，所有光线可视为平行，沿恒定的方向向量 d 传播。对于点 P，光线方程为 r = P + λd，其中 λ ≥ 0。代入平面 n·x = k 得：
λ = (k – n·P) / (n·d)
阴影点为 Q = P + λd。真实的影子要求 λ > 0，表示光线向前照射到平面。若 n·d = 0，则光线平行于平面，无法投下影子。
The following table summarises the two types of projection.
下表总结了两种投影类型。
| Type | Line Equation | Parameter | Condition for shadow |
|---|---|---|---|
| Point light | r = L + t(P – L) | t = (k – n·L)/(n·(P – L)) | t > 1 |
| Parallel light | r = P + λd | λ = (k – n·P)/(n·d) | λ > 0 |
5. Shadow of a Line Segment | 线段的阴影
The shadow of a straight line segment AB is, in general, the line segment A’B’ connecting the shadows of its endpoints. This holds provided both endpoints cast valid shadows on the same plane and the entire segment lies between them. If one endpoint does not cast a shadow (e.g. t < 0 or denominator zero), the shadow may be a ray or completely absent. We can find A’ and B’ using the appropriate projection formula and then work with the straight line through them.
通常，线段 AB 的影子就是连接其端点影子 A’B’ 的线段。前提是两个端点都能在同一平面上投下有效的影子，且整条线段位于它们之间。若有一个端点无法形成影子（例如 t < 0 或分母为零），则影子可能是一条射线或完全不存在。我们可以用相应投影公式求出 A’ 和 B’，然后利用经过它们的直线来处理。
6. Shadows of Polygons and Solids | 多边形与立体的阴影
To find the shadow of a planar polygon, we project each vertex individually and then join the image points in the same order. Provided the polygon is flat and no vertex lies on the opposite side of the light source, the shadow will be a congruent or similar shape, possibly distorted if the plane is not parallel to the polygon. For a solid, one can consider the shadows of its faces or the silhouette as seen from the light source, but at A Level the focus is usually on projecting key points and edges.
要求平面多边形的影子，只需分别投影各项点，再按相同顺序连接像点即可。只要多边形是平面图形且没有顶点位于光源相反的一侧，影子将是与之全等或相似的形状；如果平面与多边形不平行，则可能产生变形。对于立体，可以考虑其各个面的影子或从光源方向看到的轮廓，但在 A Level 考试中，重点通常是投影关键点和棱边。
7. Special Cases and Degenerate Shadows | 特殊情况和退化阴影
Several special situations can occur: if the light source lies on the plane, every ray lies in the plane, so no unique shadow is formed; if the object point lies exactly on the plane, it is its own shadow (t = 1 or λ = 0); if the ray is parallel to the plane, the shadow is at infinity and is not drawn. Being able to identify these cases quickly can save time and prevent algebraic mistakes.
会出现几种特殊情况：若光源位于平面上，所有光线都在平面内，无法形成唯一的影子；若物体点恰好位于平面上，它就会被投影到自身（t = 1 或 λ = 0）；若光线平行于平面，影子在无穷远处，无法画出。能迅速识别这些情况可节省时间并避免代数错误。
8. Worked Example: Point Light | 示例：点光源
Problem: A point light source is located at L(1, 2, 3). An opaque point P is at (4, 5, 6). Find the shadow of P on the plane π: x + y + z = 20.
题目：点光源位于 L(1, 2, 3)，不透明点 P 在 (4, 5, 6)，求 P 在平面 π: x + y + z = 20 上的影子。
Solution: The normal vector is n = (1, 1, 1), and k = 20. Compute n·L = 1+2+3 = 6, and P – L = (3, 3, 3). Then n·(P – L) = 1×3 + 1×3 + 1×3 = 9.
t = (20 – 6) / 9 = 14/9 ≈ 1.556
Since t > 1, a real shadow is formed. The shadow point Q = L + t(P – L) = (1,2,3) + (14/9)(3,3,3) = (1 + 42/9, 2 + 42/9, 3 + 42/9) = (5 ⅔, 6 ⅔, 7 ⅔).
解答：法向量 n = (1, 1, 1)，k = 20。计算 n·L = 1+2+3 = 6，且 P – L = (3, 3, 3)，则 n·(P – L) = 1×3 + 1×3 + 1×3 = 9。
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