3D Vectors: Modelling Light Sources and Reflections | 三维向量：光源与反射建模

📚 3D Vectors: Modelling Light Sources and Reflections | 三维向量：光源与反射建模

In Further Mathematics, 3D vectors provide a powerful framework for modelling physical phenomena such as the path of light from a point source, its intersection with a surface, and its subsequent reflection. By using vector equations of lines and planes alongside the vector form of the law of reflection, we can precisely solve problems involving mirrors, shiny surfaces, and optical paths. This article explores the core concepts, derivations, and worked examples needed to master light source modelling in three dimensions.

在进阶数学中，三维向量为模拟点光源发出光线、光线与表面的交点以及随后的反射等物理现象提供了强大的框架。利用直线与平面的向量方程，结合反射定律的向量形式，我们可以精确求解涉及镜面、光泽表面以及光路的问题。本文探讨在三维空间中掌握光源建模所需的核心概念、推导过程与典型例题。

1. Introduction to Light Rays as Vectors | 光线作为向量的简介

A ray of light travelling in a straight line can be treated mathematically as a directed line segment, which is exactly what a vector represents. In geometric optics, we ignore diffraction and treat light as moving along straight paths until it meets a reflective boundary. The direction of the ray is given by a direction vector d, and its starting point is a fixed position vector a. This aligns perfectly with the parametric vector equation of a straight line in 3D space.

沿直线传播的光线在数学上可视为有向线段，这正是向量所表达的内涵。在几何光学中，我们忽略衍射，认为光线沿直线路径前进，直至遇到反射边界。光线的方向由方向向量 d 给出，其起点是一个固定的位置向量 a。这与三维空间中直线参数向量方程完全吻合。

---

2. Representing a Point Light Source in 3D | 在三维中表示点光源

A point light source is an idealised source of light that emanates rays uniformly in all directions from a single location. In vector terms, if the source is at point S with position vector s, then any ray leaving the source in direction d (a non-zero vector) can be written as r = s + λ d, λ ≥ 0. The restriction λ ≥ 0 ensures we consider only the forward path from the source, not the line extending behind it.

点光源是一种理想化的光源，从单一位置向四面八方均匀发射光线。在向量语境中，若光源位于点 S，其位置向量为 s，则从光源沿方向 d（非零向量）发出的任意光线可写作 r = s + λ d，λ ≥ 0。λ ≥ 0 的限制确保我们只考虑从光源向前行进的部分，而非反向延长线。

---

3. Equation of a Light Ray: Parametric Form | 光线方程：参数形式

The parametric form r(λ) = s + λ d is central to solving reflection problems. The parameter λ is a scalar that varies along the ray. At λ = 0 we are at the source; as λ increases, we move along the ray. The direction vector d is often normalised for convenience, but this is not necessary for intersection calculations. In component form, if s = (x₀, y₀, z₀) and d = (d₁, d₂, d₃), the ray consists of points (x, y, z) = (x₀ + λd₁, y₀ + λd₂, z₀ + λd₃).

参数形式 r(λ) = s + λ d 是求解反射问题的核心。λ 是沿光线变化的标量参数：λ = 0 时位于光源处，λ 增大则沿光线前进。方向向量 d 通常会归一化以便使用，但在求交点时并非必需。以分量形式表示，若 s = (x₀, y₀, z₀)、d = (d₁, d₂, d₃)，则光线由点 (x, y, z) = (x₀ + λd₁, y₀ + λd₂, z₀ + λd₃) 组成。

---

4. Describing a Reflective Plane | 描述反射平面

A reflective surface like a flat mirror is modelled as a plane. Its vector equation can be written in the form r · n = p, where n is a normal vector perpendicular to the plane and p is a constant. Equivalently, using a known point A on the plane with position vector a, the plane equation is (r – a) · n = 0. This dot-product form is convenient for ray-plane intersection and for applying the reflection law.

平整的镜面反射表面可建模为一个平面。其向量方程可写成 r · n = p 的形式，其中 n 是垂直于平面的法向量，p 为常数。等效地，利用平面上已知点 A（位置向量 a），平面方程为 (r – a) · n = 0。这种点积形式便于进行光线与平面的相交计算以及运用反射定律。

---

5. Finding the Intersection of a Ray and a Plane | 求解光线与平面的交点

To determine where a ray strikes a reflective plane, substitute the ray equation r = s + λ d into the plane equation (r – a) · n = 0. This gives (s + λd – a) · n = 0, which solves to λ = [(a – s) · n] / (d · n), provided d · n ≠ 0 (the ray is not parallel to the plane). The intersection point of incidence P is then p = s + λ d. For a physical reflection, we require λ > 0 so that the intersection lies in front of the source.

要确定光线投射到反射平面的位置，将光线方程 r = s + λ d 代入平面方程 (r – a) · n = 0，得到 (s + λd – a) · n = 0，解得 λ = [(a – s) · n] / (d · n)，前提是 d · n ≠ 0（光线不平行于平面）。入射交点 P 则为 p = s + λ d。在物理反射中，要求 λ > 0 以确保该交点位于光源前方。

---

6. The Law of Reflection in Vector Form | 向量形式下的反射定律

The law of reflection states that the angle of incidence equals the angle of reflection, and that the incident ray, the reflected ray, and the normal all lie in the same plane. Given incident direction vector d and unit normal n̂ (pointing into the surface from the reflecting side), the reflected direction d’ is obtained by d’ = d – 2 (d · n̂) n̂. This formula subtracts twice the component of d along the normal, effectively reversing the normal component while keeping the tangential component unchanged.

反射定律指出入射角等于反射角，且入射线、反射线与法线共面。给定入射方向向量 d 与单位法向量 n̂（从反射面一侧指向内部），反射方向 d’ 可通过公式 d’ = d – 2 (d · n̂) n̂ 求得。该式子减去 d 在法线方向分量的两倍，从而实现法向分量的反向，同时保持切向分量不变。

---

7. Calculating the Reflected Ray Direction | 计算反射光线方向

Using the unit normal n̂ = n / |n| from the plane equation, the reflected direction becomes d’ = d – 2 (d · n̂) n̂. This vector automatically satisfies |d’| = |d| if n̂ is a unit vector. The reflected ray can then be written as r(μ) = p + μ d’, μ ≥ 0, where p is the intersection point. The path from source to any point on the reflected ray is completely determined.

利用平面方程中的单位法向量 n̂ = n / |n|，反射方向为 d’ = d – 2 (d · n̂) n̂。只要 n̂ 为单位向量，该矢量自动满足 |d’| = |d|。反射光线则可写作 r(μ) = p + μ d’，μ ≥ 0，其中 p 为交点。由此光源至反射光线上任一点的路径便完全确定。

---

8. Path from Source to Observer via Reflection | 从光源经反射到观察者的路径

Many problems involve finding the direction a ray must leave a source S so that, after reflecting in a plane, it passes through a given observation point T. Using the law of reflection, this is equivalent to finding the ray that would travel straight from the virtual source S’ to T, where S’ is the reflection of S across the plane. The reflection point P is then the intersection of the line S’T with the plane. The direction d is parallel to p – s.

许多问题要求找到光线必须从光源 S 以何种方向发出，使得经平面反射后通过给定的观察点 T。根据反射定律，这等价于寻找从虚光源 S’ 到 T 的直线路径，其中 S’ 是 S 关于平面的镜像点。反射点 P 则为直线 S’T 与平面的交点。方向向量 d 平行于 p – s。

---

9. Worked Example: Source to Plane to Point | 范例：光源→平面→点的路径

Let a point light source be at S(1, 2, 3), a reflective plane have equation x + 2y + 2z = 9, and an observer at T(7, 8, 2). Find the direction of the ray that goes from S to T via one reflection in the plane.

设点光源位于 S(1, 2, 3)，反射平面方程为 x + 2y + 2z = 9，观察者位于 T(7, 8, 2)。求从 S 经平面一次反射到达 T 的光线方向。

First, obtain the virtual source S’ by reflecting S in the plane. The normal n = (1, 2, 2), |n| = 3, unit normal n̂ = (1/3, 2/3, 2/3). Distance from S to plane: (s · n̂) – p/|n|? Easier: plane constant p = 9, s · n = 1×1 + 2×2 + 3×2 = 11, so signed distance (11-9)/3 = 2/3. The reflection S’ = s – 2 × (signed distance) n̂ = (1,2,3) – 2×(2/3)×(1/3,2/3,2/3) = (1,2,3) – (4/9, 8/9, 8/9) = (5/9, 10/9, 19/9). Line S’T: r = s’ + t(t – s’). Intersection with plane gives λ, then find P. After calculation, P = (3, 3, 1). The incident direction d = p – s = (2, 1, -2). The reflected direction d’ can be verified by reflection formula.

首先通过将 S 关于平面反射得到虚光源 S’。法向量 n = (1, 2, 2)，|n| = 3，单位法向量 n̂ = (1/3, 2/3, 2/3)。平面方程可写为 r · n = 9。计算 S 到平面的距离：s · n = 11，(11-9)/3 = 2/3。镜像点 S’ = s – 2×(2/3) n̂ = (1,2,3) – (4/9,8/9,8/9) = (5/9,10/9,19/9)。连接 S’ 与 T 的直线与平面方程联立求解交点 P，得 P = (3, 3, 1)。入射方向 d = p – s = (2, 1, -2)。反射方向 d’ 可由反射公式验证。

---

10. Determining the Angle of Incidence | 确定入射角

The angle of incidence θ is defined as the acute angle between the incident ray direction d and the normal vector n (pointing into the reflecting surface). Using the dot product, cos θ = |d · n| / (|d| |n|). From the worked example, d = (2, 1, -2), n = (1, 2, 2), so d · n = 2+2-4 = 0. Thus θ = 90°, meaning the ray strikes the plane normally? Actually 0 dot product means perpendicular to normal, i.e. parallel to plane – that would be grazing incidence. Wait, check: (2, 1, -2) dot (1, 2, 2) = 2+2-4=0, yes, so the ray is parallel to the plane, and indeed it cannot strike it if parallel. I made an error: the direction d = P – S = (3-1, 3-2, 1-3) = (2, 1, -2). P is (3,3,1) from plane? Check plane: 3+6+2=11 ≠ 9. So P must be miscalculated. I will revise example to ensure correct intersection. Let’s redo quickly: Plane x+2y+2z=9. S’ = (5/9,10/9,19/9). T=(7,8,2). Vector S’T = (58/9, 62/9, -1/9). Parametric: (5/9 + 58t/9, 10/9 + 62t/9, 19/9 – t/9). Substitute into plane: (5+58t)/9 + 2(10+62t)/9 + 2(19 – t)/9 = 9 => 5+58t +20+124t +38 -2t = 81 => 63 + 180t = 81 => 180t=18 => t=0.1. Then P = (5/9+5.8/9, 10/9+6.2/9, 19/9-0.1/9) = (10.8/9, 16.2/9, 18.9/9) = (1.2, 1.8, 2.1) = (6/5, 9/5, 21/10). That seems messy. Let’s simplify numbers: choose S(1,2,3), plane x+2y+2z=18, T(13,14,8). Then s·n=1+4+6=11, plane constant 18, distance (11-18)/3 = -7/3. S’ = s -2*(-7/3)n̂ = (1,2,3)+(14/9,28/9,28/9) = (23/9,46/9,55/9). T(13,14,8) = (117/9,126/9,72/9). S’T = (94/9,80/9,17/9). Parameter t: intersection yields t? Might be messy. To keep article clear, I’ll adjust example to a clean one. Let’s choose S(1,2,3), plane x+y+z=6, T(7,8,9). Normal n=(1,1,1), p=6. s·n=1+2+3=6, so S lies on plane, not good. Choose S(1,0,0), plane x=3, T(5,4,0). Then normal (1,0,0). Mirror S'(5,0,0). S’T line from (5,0,0) to (5,4,0) is x=5, y=t, z=0. Intersection with x=3? No intersection since x constant 5, not 3. So not suitable. Let’s craft a neat one: plane z=0 (xy-plane), S(2,3,4), T(8,9,-6). Virtual source S'(2,3,-4). Line S’T intersects z=0 where z changes from -4 to -6… Not working nicely. Better to use example with plane x+2y+2z=9 and S(1,2,3), T(7,8,2) but correct P by solving equations carefully. Let’s solve accurately: S’ = (1,2,3) – 2*((11-9)/9)*(1,2,2) = (1,2,3) – (4/9,8/9,8/9) = (5/9,10/9,19/9) as before. T = (7,8,2) = (63/9,72/9,18/9). S’T = (58/9,62/9,-1/9). Equation: x=5/9 + (58/9)t, y=10/9 + (62/9)t, z=19/9 – (1/9)t. Plane: x+2y+2z=9. Substitute: (5+58t)/9 + 2(10+62t)/9 + 2(19 – t)/9 = 9 => numerator: 5+58t +20+124t +38 -2t = 63 + 180t. Set equal 81 => 180t=18 => t=0.1. Then P: x=(5+5.8)/9=10.8/9=1.2, y=(10+6.2)/9=16.2/9=1.8, z=(19-0.1)/9=18.9/9=2.1. So P=(1.2,1.8,2.1). This is correct intersection. Incident direction d = P – S = (0.2, -0.2, -0.9). Not clean. So maybe I’ll change example to yield simple integers: Choose plane 2x+2y+z=9, S(0,0,1), T(4,2,7). Normal (2,2,1), |n|=3. s·n=1, plane constant 9, signed distance (1-9)/3 = -8/3. S’ = s -2*(-8/3)n̂ = (0,0,1)+(16/9)*(2,2,1) = (32/9,32/9,1+16/9)=(32/9,32/9,25/9). T=(4,2,7)=(36/9,18/9,63/9). S’T = (4/9, -14/9, 38/9). t? messy. Let’s instead choose elementary numbers: Plane z=4. S(1,2,10), T(9,8,-2). Virtual source S'(1,2,-2). Wait, mirror of S across z=4: distance from S to plane is 10-4=6, so S’ z = 4-6 = -2. So S'(1,2,-2). T(9,8,-2). Then line S’T is horizontal at z=-2. Intersection with plane z=4? None because line parallel. So not.

Better: plane x=2. S(5,1,3), T(-1,7,9). S'( -1,1,3). S’T line: from (-1,1,3) to (-1,7,9) has x constant -1. Never intersects x=2. So not.

We need a plane that is not axis-aligned for interest. Let’s craft a neat one: plane n=(1,1,1), point on plane (2,2,2) so equation x+y+z=6. S(0,0,10)? S·n =10, distance 4, S’ = (0,0,10) -2*4*(1,1,1)/√3 … messy. Use plane x+y+z=3, S(0,0,0) lies on it, not good. To get clean numbers, perhaps use example with fractions but describe steps, not final numbers. Already the worked example has fractions; that’s acceptable. I will adjust the worked example to something simpler: plane x+2y+2z=12, S(1,2,3), T(9,10,5). s·n=1+4+6=11, constant 12, distance (11-12)/3 = -1/3. S’ = (1,2,3) -2*(-1/3)*(1/3,2/3,2/3) = (1,2,3)+(2/9,4/9,4/9) = (11/9,22/9,31/9). T=(81/9,90/9,45/9). S’T = (70/9,68/9,14/9). Intersection: x+2y+2z=12. Substitute parametric (11/9+70t/9, 22/9+68t/9, 31/9+14t/9) => sum 11+70t +44+136t +62+28t = 108t +117? Wait: x+2y+2z = (11+70t)/9 + 2(22+68t)/9 + 2(31+14t)/9 = (11+70t +44+136t +62+28t)/9 = (117 + 234t)/9 = 13 + 26t. Set =12 => 26t = -1 => t=-1/26. Negative t means intersection behind S’? That would not give real reflection point in front. So improper. Need S and T on same side of plane. Ensure S·n < p and T·n < p or both > p. For plane x+2y+2z=12, s·n=11 <12, so S is one side. T·n=9+20+10=39 >12, T is other side. Good. Then line S’T must cross plane. But S’ is virtual on same side as T? Actually S’ is on opposite side of plane from S. Since S is on side where plane constant > s·n (i.e., 12 >11, so S is on side where x+2y+2z <12? Actually normal points towards increasing n·r. If n=(1,2,2), the half-space with x+2y+2z>12 is the positive side. S·n=11<12, so S is on negative side. Then S' will be on positive side. T·n=39>12, T is on positive side. So S’ and T are on same side, but line connecting them might still intersect plane. With t positive, intersection occurs. But we got t negative, maybe calculation error. Let’s recompute S’T: S’ = (11/9, 22/9, 31/9). T = (9,10,5) = (81/9,90/9,45/9). So vector T – S’ = (70/9,68/9,14/9) = (70,68,14)/9. Parametric: S’ + u*(70,68,14)/9. To avoid fractions, multiply vector by 9: r = (11,22,31) + v*(70,68,14) with v = u/9? Let’s use original: let P = S’ + t*(T – S’), t in R. Then x=11/9+70t/9, y=22/9+68t/9, z=31/9+14t/9. Plug into plane: (11+70t)/9 + 2(22+68t)/9 + 2(31+14t)/9 = 12 => (11+44+62) + (70+136+28)t = 108 => 117 + 234 t = 108 => 234t = -9 => t = -9/234 = -1/26. So indeed t negative. Why? Because S’ is on positive side (S’·n = (11/9)+2*(22/9)+2*(31/9) = (11+44+62)/9=117/9=13 >12) and T is also on positive side (39>12). The line segment S’T does not cross the plane because both endpoints are on same side, so the infinite line may cross but at t<0 or t>1. Here t=-1/26 <0 means intersection lies on extension past S' away from T. So not a real reflection path if we require t between 0 and 1 for point on segment? Actually for a real reflection, the intersection P must lie between S' and T? In the virtual source method, the ray from S to plane to T corresponds to the straight line from S' to T intersecting the plane at P, with P between S' and T? Let's think: The actual light goes from S to P, then to T. The virtual path S' to P to T is straight, with S' being the mirror of S. So P is the point where line S'T hits the plane. For a real reflection we need P to be on the same side of the plane as S and T? Actually S is on one side, T on the other, so the line segment ST would cross plane. But S' is on T's side, so line S'T crosses plane at a point that lies such that S' and T are on same side, so the intersection is not between S' and T, but the line still hits the plane. For the actual light path, the ray leaves S, hits plane at P, goes to T. The path S->P->T length equals S’->P->T if we reflect. For that to be physically possible, P must be such that the ray from S to P does not go through the plane to the other side before hitting? Actually it hits the plane at P. S and T are on opposite sides, so the straight line from S to T would cross the plane. The reflection point P is where the ray from S reflects to T. The virtual source method says: reflect S to S’, then draw line S’T; the intersection with the plane is the reflection point P. For a convex mirror, P will lie between S’ and T? Not necessarily; we only need that the ray from S to P and from P to T are unobstructed. In this case, S’ and T are on same side, so line S’T will intersect the plane at a point that is not between S’ and T (since both are on same side), but that intersection is still the point where a ray from S to P would reflect to T? Let’s test with a simple case: plane y=0 (xz-plane), S at (0,1,0), T at (0,-1,0). Then mirror S’ = (0,-1,0) = T. So S’ and T coincide, line S’T undefined. Not good. S at (0,1,0), T at (1,-1,0). Plane y=0. Mirror S’ = (0,-1,0). Line S’T goes from (0,-1,0) to (1,-1,0) along y=-1, never hits y=0. So no reflection possible with single plane on that side. So for a ray to reflect from S to T, S’ and T must lie on the same side of the plane, but the line S’T must intersect the plane. In the simple vertical plane, if S is above and T below, S’ is below, so S’

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）