📚 4B Three-dimensional Coordinate Geometry | 三维坐标几何
Extending coordinate geometry from the plane into three dimensions opens up a rich world of spatial reasoning, essential for tackling problems in IB Mathematics. In this article, we systematically develop the key formulas, vector techniques, and equation types required to describe points, lines, planes, and spheres in 3D space. Each concept is paired with clear algebraic representations ready for examination use.
将平面坐标几何推广到三维空间，为你打开了立体的推理世界，这是求解 IB 数学中空间问题的关键基础。本文系统地梳理描述三维空间中点、直线、平面和球面所需的核心公式、向量工具以及方程类型。每一个概念都配有可以直接用于考试的代数表达与解释。
1. The 3D Coordinate System | 三维直角坐标系
he yz-plane (x=0). This right-handed system is the standard convention in the IB syllabus.
三维空间中的点由有序三元组 (x, y, z) 确定，它们相对于三个互相垂直的轴： x 轴、y 轴和 z 轴。三条轴相交于原点 O(0,0,0)，并形成三个坐标平面： xy 平面 (z=0)、xz 平面 (y=0) 和 yz 平面 (x=0)。IB 课程采用右手系作为标准约定。
To visualise a point like P(3, −2, 4), move 3 units along the positive x-direction, 2 units opposite the y-direction, and then 4 units upward along the z-axis. Understanding the coordinates as perpendicular projections onto the axes is fundamental for distance and midpoint calculations.
为了直观地想象点 P(3, −2, 4)，先沿 x 轴正向移动 3 个单位，再沿 y 轴负向移动 2 个单位，最后沿 z 轴向上移动 4 个单位。将坐标理解为点在三条轴上的垂直投影，这是计算距离和中点的基础。
2. Distance Between Two Points | 两点间的距离公式
The distance d between two points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is the length of the straight segment joining them. It is obtained by applying the Pythagorean theorem twice in 3D, yielding the formula:
d = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]
点 A(x₁, y₁, z₁) 和点 B(x₂, y₂, z₂) 之间的距离 d 是连接它们的直线段的长度。在三维中两次应用勾股定理，就得到了距离公式：
d = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]
As a special case, the distance from the origin to a point (x, y, z) simplifies to √x² + y² + z². When solving problems, always check if a point lies on a given sphere or if a triangle is isosceles by computing side lengths.
作为一种特殊情况，原点到点 (x, y, z) 的距离可以简化为 √x² + y² + z²。解题时，常常通过计算边长来判断一个点是否在给定球面上，或三角形是否为等腰三角形。
3. Midpoint of a Segment | 线段的中点公式
The midpoint M of the line segment joining A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is found by averaging the corresponding coordinates:
M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 , (z₁ + z₂)/2 )
连接 A(x₁, y₁, z₁) 和 B(x₂, y₂, z₂) 的线段的中点 M，是将对应的坐标分别取平均值得到的：
M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 , (z₁ + z₂)/2 )
Just like in 2D, the midpoint is the average position of the endpoints. This formula is crucial when finding the centre of a sphere given the endpoints of a diameter, or when solving geometry problems that involve bisected segments. A typical examination question might ask: "Given A(1, −3, 2) and B(5, 7, −4), find the midpoint of AB and verify that it lies on a given plane."
与二维情况类似，中点就是端点位置的平均值。若已知球面一条直径的端点，可以用该公式求出球心；在涉及线段平分的几何问题中，这个公式也至关重要。典型考题可能要求：“已知 A(1, −3, 2) 和 B(5, 7, −4)，求 AB 的中点，并验证该点是否在给定平面上。”
4. Vectors in 3D Space | 空间向量基础
A vector in ℝ³ can be expressed in component form as v = ⟨v₁, v₂, v₃⟩ or as a position vector from the origin. The vector from point A to point B is given by AB = ⟨x₂−x₁, y₂−y₁, z₂−z₁⟩. The magnitude (length) of vector v = ⟨a, b, c⟩ is |v| = √a² + b² + c².
ℝ³ 中的向量可以表示为分量形式 v = ⟨v₁, v₂, v₃⟩，或者用从原点出发的位置向量表示。从点 A 到点 B 的向量是 AB = ⟨x₂−x₁, y₂−y₁, z₂−z₁⟩。向量 v = ⟨a, b, c⟩ 的模（长度）为 |v| = √a² + b² + c²。
Unit vectors in the directions of the positive x-, y-, and z-axes are denoted by i = ⟨1,0,0⟩, j = ⟨0,1,0⟩, and k = ⟨0,0,1⟩. Any vector can be written as a linear combination: v = ai + bj + ck. The scalar (dot) product of two vectors u·v = u₁v₁ + u₂v₂ + u₃v₃ is fundamental for testing perpendicularity and finding angles.
沿 x、y、z 轴正方向的单位向量分别记作 i = ⟨1,0,0⟩、j = ⟨0,1,0⟩、k = ⟨0,0,1⟩。任何一个向量都可以写成这组基向量的线性组合：v = ai + bj + ck。两个向量的数量积（点积）u·v = u₁v₁ + u₂v₂ + u₃v₃ 是判断垂直关系和计算夹角的基础。
5. Equations of a Line in 3D | 空间直线的方程
A straight line in 3D can be defined by a fixed point A(x₀, y₀, z₀) and a direction vector d = ⟨a, b, c⟩. Its vector equation is:
r = r₀ + td, t ∈ ℝ
这里 r₀ = ⟨x₀, y₀, z₀⟩。对应的参数方程为：
x = x₀ + a t, y = y₀ + b t, z = z₀ + c t
三维空间中的一条直线可以由一个已知点 A(x₀, y₀, z₀) 和一个方向向量 d = ⟨a, b, c⟩ 确定。其向量方程为：
r = r₀ + td, t ∈ ℝ
where r₀ = ⟨x₀, y₀, z₀⟩。对应的参数方程为：
x = x₀ + a t, y = y₀ + b t, z = z₀ + c t
stantly, and the symmetric equations are adjusted accordingly. Always be ready to extract a direction vector and a point from any given form.
如果 a、b、c 都不为零，就可以写出对称式（笛卡尔式）： (x − x₀)/a = (y − y₀)/b = (z − z₀)/c。若某个分量为零，比如 a=0，则 x 恒等于 x₀，对称式也需要相应调整。无论题目给出哪种形式，你都应该能够快速提取出一个方向向量和一个已知点。
6. Equations of a Plane | 平面的方程
A plane can be defined by a point P(x₀, y₀, z₀) lying on it and a normal vector n = ⟨A, B, C⟩ perpendicular to the plane. The scalar (dot) product form of the plane is:
A(x − x₀) + B(y − y₀) + C(z − z₀) = 0
平面可以由平面上的一点 P(x₀, y₀, z₀) 和一个垂直于该平面的法向量 n = ⟨A, B, C⟩ 确定。利用向量点积为零的条件，可以得到平面的方程：
A(x − x₀) + B(y − y₀) + C(z − z₀) = 0
₀ + By₀ + Cz₀). The direction vectors of lines lying in the plane are all perpendicular to n. To find the equation of a plane passing through three non-collinear points, form two vectors in the plane, take their cross product to obtain a normal vector, then apply the point-normal form.
将上式展开就得到线性的一般式：Ax + By + Cz + D = 0，其中 D = −(Ax₀ + By₀ + Cz₀)。平面上所有直线的方向向量都与法向量 n 垂直。如果要找出经过三个不共线点的平面方程，可以先在平面内构造两个向量，再求它们的叉积得到法向量，然后利用点法式得出方程。
7. Angle Between Two Planes | 两平面的夹角
The angle θ between two planes is defined as the acute angle between their normal vectors. If the normals are n₁ = ⟨A₁, B₁, C₁⟩ and n₂ = ⟨A₂, B₂, C₂⟩, then:
cos θ = |n₁·n₂| / (|n₁| |n₂|)
两个平面之间的夹角 θ 被定义为它们的法向量之间的锐角。设两平面的法向量分别为 n₁ = ⟨A₁, B₁, C₁⟩ 和 n₂ = ⟨A₂, B₂, C₂⟩，则：
cos θ = |n₁·n₂| / (|n₁| |n₂|)
Absolute value is taken to ensure θ is between 0° and 90°. If the normals are parallel, the planes are parallel; if the normals are perpendicular, the planes are perpendicular. This relationship is frequently tested when determining whether two faces of a solid are orthogonal or when finding the line of intersection.
公式中取了绝对值，以保证 θ 的范围在 0° 到 90° 之间。如果两个法向量平行，则两平面平行；如果两个法向量垂直，则两平面互相垂直。在判断立体的两个面是否垂直，或者求解交线问题时，这一关系经常被考查。
8. Distance from a Point to a Plane | 点到平面的距离
0 is measured along a perpendicular. The formula, derived from the scalar projection, is:
d = |Ax₁ + By₁ + Cz₁ + D| / √A² + B² + C²
点 P(x₁, y₁, z₁) 到平面 Ax + By + Cz + D = 0 的最短距离是沿着垂线的长度。利用标量投影可以推出公式：
d = |Ax₁ + By₁ + Cz₁ + D| / √A² + B² + C²
This distance is positive if the point is on the same side as the normal vector's direction (judged by the sign of Ax₁+By₁+Cz₁+D) and zero if the point lies on the plane. Calculating this distance is essential for finding distances between parallel planes or verifying that a sphere is tangent to a plane.
如果点位于法向量指向的一侧（通过 Ax₁+By₁+Cz₁+D 的符号判断），该距离取正值；若点在平面上，则距离为零。计算点到平面的距离，对于求两平行平面之间的距离，或者验证球面与平面相切等问题，都极其关键。
9. Intersection of a Line and a Plane | 直线与平面的交点
To find where a line given in parametric form r = r₀ + td itute the expressions for x, y, z into the plane equation. This yields a linear equation in t:
A(x₀ + a t) + B(y₀ + b t) + C(z₀ + c t) + D = 0
要求参数式直线 r = r₀ + td 与平面 Ax + By + Cz + D = 0 的交点，只需将 x、y、z 的表达式代入平面方程。这样可以得到关于 t 的一元一次方程：
A(x₀ + a t) + B(y₀ + b t) + C(z₀ + c t) + D = 0
), the line is parallel to the plane and does not intersect it. This method is straightforward and highly examinable.
解出 t 即可得到交点对应的参数值。如果该方程有唯一解，则直线在该点穿过平面；如果方程化简为恒等式 0=0，说明直线完全落在平面内；如果方程变成矛盾式（如 5=0），则直线与平面平行且不相交。这种处理方法直接明了，是考试中的高频考点。
10. Equation of a Sphere | 球面方程
A sphere with centre C(h, k, l) and radius r has the standard equation:
(x − h)² + (y − k)² + (z − l)² = r²
球心为 C(h, k, l)、半径为 r 的球面，其标准方程为：
(x − h)² + (y − k)² + (z − l)² = r²
x² + y² + z² + 2gx + 2fy + 2hz + c = 0, where the centre is (−g, −f, −h) and the radius is √g² + f² + h² − c, provided g²+f²+h²−c > 0. When given endpoints of a diameter, use the midpoint formula to find the centre and half the distance for the radius. Completing the square is the key technique to convert between forms.
展开后可得到一般式：x² + y² + z² + 2gx + 2fy + 2hz + c = 0，此时球心坐标为 (−g, −f, −h)，半径为 √g² + f² + h² − c，前提是 g²+f²+h²−c > 0。若已知直径的两个端点，可以用中点公式求球心，用距离的一半作半径。配方是两种形式互相转化的核心技巧。
11. Common Problem Solving Strategies | 常见解题策略
Many IB problems combine lines, planes, and spheres, requiring multiple steps. Here are several key strategies:
- Checking if a point lies on a line or plane: Substitute coordinates into the equations and check for consistency.
- Finding foot of perpendicular from a point to a plane: Construct a line through the point with direction equal to the plane's normal, then intersect with the plane.
- Shortest distance from a point to a line: Use vector cross product method or minimise a squared distance expression.
许多 IB 问题综合了直线、平面和球面，需要多步求解。以下是几个关键策略：
- 判断点是否在直线或平面上： 将坐标代入方程，检验是否成立。
- 求点到平面的垂足： 过该点以平面的法向量为方向构造一条直线，然后求它与平面的交点。
- 求点到直线的最短距离： 可使用向量叉积法，或者对距离平方表达式求最小值。
When dealing with two skew lines, the shortest distance can be found by projecting the vector between a point on each line onto the common perpendicular direction, which is the cross product of their direction vectors. Being systematic and writing clear steps prevents sign errors.
针对两条异面直线，可将两直线上各取一点所得向量投影到公垂线方向上（即两方向向量的叉积），从而求得最短距离。解题时写出清晰步骤、保持符号准确，可以避免低级错误。
12. Summary and Revision Tips | 总结与复习建议
Three-dimensional coordinate geometry builds directly on your vector and 2D coordinate geometry skills. To master it, ensure you can fluently switch between the vector, parametric, and Cartesian forms of a line, and between point-normal and general forms of a plane. Practise deriving sphere properties by completing the square, and always sketch a rough diagram to guide your reasoning. Memorising the distance and midpoint formulas, the scalar product condition for perpendicular vectors, and the general plane equation will cover the majority of examination questions.
三维坐标几何直接建立在你的向量知识和二维坐标几何技巧之上。要想掌握它，你必须能够熟练地在直线的向量式、参数式和对称式之间转换，在平面的点法式和一般式之间切换。多做通过配方求球心与半径的练习，并在解题时随手画出示意图辅助思考。熟记距离公式、中点公式、向量垂直的点积条件以及平面的一般方程，就能应对绝大部分的考试题目。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图