📚 3D Coordinates in A-Level Maths | A-Level 数学中的三维坐标
In Edexcel A-Level Mathematics, coordinate geometry is extended from two dimensions into three. The ability to describe points, distances, midpoints, vectors, lines and simple surfaces in 3D underpins later work on vectors, mechanics, and applied problems involving solid shapes. This revision guide presents the core formulas and methods you need for 3D coordinate questions, with worked examples and exam-focused tips.
在爱德思 A-Level 数学中，坐标几何从二维扩展到三维。描述点的位置、距离、中点、向量、直线以及简单曲面的能力，为后续向量、力学以及涉及立体图形的应用问题打下基础。本复习指南介绍三维坐标问题所必需的核心公式和方法，并配有例题与考试技巧。
1. The 3D Cartesian System | 三维笛卡尔坐标系
labelled x, y and z. The origin O = (0, 0, 0) is the point where the three axes intersect.
空间中的点写作有序三元组 (x, y, z)。三个坐标沿三条相互垂直的轴测量，通常标记为 x、y 和 z。原点 O = (0, 0, 0) 是三轴交点。
In a standard right-handed system, if the positive x-axis points to the right and the positive y-axis points upward in the plane of the page, then the positive z-axis points out of the page towards you. This orientation is consistent throughout Edexcel vector and 3D geometry questions.
在标准右手系中，如果 x 轴正方向指向右方，y 轴正方向指向页面上方，那么 z 轴正方向指向页面外朝向你。这一方向约定在爱德思向量与三维几何题中保持一致。
tion x = 0. These planes divide space into eight regions called octants; the first octant is where x, y and z are all positive.
三个坐标平面是有用的参考曲面。xy 平面的方程为 z = 0，xz 平面的方程为 y = 0，yz 平面的方程为 x = 0。这些平面把空间分成八个区域，称为卦限；第一卦限是 x、y、z 均为正的区域。
2. Reading and Plotting Points | 读取与描点
To plot P(3, 4, 2), start at O, move 3 units along the positive x-axis, then 4 units parallel to the positive y-axis, and finally 2 units parallel to the positive z-axis. Negative coordinates move in the opposite directions.
要描出点 P(3, 4, 2)，从 O 出发，沿 x 轴正方向移动 3 个单位，再平行于 y 轴正方向移动 4 个单位，最后平行于 z 轴正方向移动 2 个单位。负坐标则向相反方向移动。
For example, Q(−2, 5, −1) means 2 units back along x, 5 units forward along y, and 1 unit down along z. Always read the signs carefully before locating a point, because a single sign error changes the position completely.
例如，Q(−2, 5, −1) 表示沿 x 轴后退 2 个单位，沿 y 轴前进 5 个单位，沿 z 轴下降 1 个单位。定位点之前一定要仔细读符号，因为一个符号错误就会完全改变位置。
Points on the x-axis have the form (a, 0, 0).
在 x 轴上的点形如 (a, 0, 0)。
Points on the y-axis have the form (0, b, 0).
在 y 轴上的点形如 (0, b, 0)。
Points on the z-axis have the form (0, 0, c).
在 z 轴上的点形如 (0, 0, c)。
lane have z = 0, points in the xz-plane have y = 0, and points in the yz-plane have x = 0.
xy 平面内的点满足 z = 0，xz 平面内的点满足 y = 0，yz 平面内的点满足 x = 0。
3. Distance Between Two Points | 两点间距离
The distance between A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is found by applying Pythagoras' theorem twice. First find the distance in the xy-plane, then combine it with the vertical difference in z.
A(x₁, y₁, z₁) 与 B(x₂, y₂, z₂) 两点间的距离，可以通过两次使用勾股定理得到。先在 xy 平面内求距离，再与 z 方向的差值结合。
AB = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²)
For A(1, 2, 3) and B(4, 6, 3), the differences are 3, 4 and 0. Hence AB = √3² + 4² + 0² = √25 = 5.
对于 A(1, 2, 3) 和 B(4, 6, 3)，坐标差分别为 3、4 和 0。因此 AB = √3² + 4² + 0² = √25 = 5。
In exam questions, the distance formula is often used to prove that a triangle is isosceles, to find the radius of a sphere, or to check whether a point lies on a given sphere. Always square the differences before adding, and do not subtract the square roots.
在考试题中，距离公式常用于证明三角形为等腰三角形、求球半径，或检验某点是否在给定球面上。一定要先平方差再相加，不要先开方再相减。
4. Midpoint and Section Formula | 中点与定比分点
The midpoint M of the line segment joining A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is the average of the coordinates.
连接 A(x₁, y₁, z₁) 与 B(x₂, y₂, z₂) 的线段中点 M 是各坐标的平均值。
M = ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2)
For A(1, 2, 3) and B(4, 6, 3), M = ((1 + 4)/2, (2 + 6)/2, (3 + 3)/2) = (2.5, 4, 3).
对于 A(1, 2, 3) 和 B(4, 6, 3)，M = ((1 + 4)/2, (2 + 6)/2, (3 + 3)/2) = (2.5, 4, 3)。
ht A by n and B by m before dividing by m + n. The midpoint is the special case m = n = 1.
若 R 内分 AB，且 AR : RB = m : n，则 R = (nA + mB)/(m + n)。也就是说，用 n 乘 A、用 m 乘 B，再除以 m + n。中点是 m = n = 1 的特殊情况。
his formula extends directly from 2D to 3D and is sometimes used in mechanics problems on centres of mass.
顶点为 A、B、C 的三角形重心 G 是三个位置向量的平均：G = (A + B + C)/3。该公式可直接从二维推广到三维，有时也用于质心力学问题。
5. Vectors in 3D | 三维向量
or a = 2i − j + 3k, where i, j and k are unit vectors along the positive x, y and z axes.
三维向量可写成列向量或分量形式。例如 a = (2, −1, 3) 或 a = 2i − j + 3k，其中 i、j、k 分别是沿 x、y、z 轴正方向的单位向量。
The magnitude of a = (a₁, a₂, a₃) is its length.
a = (a₁, a₂, a₃) 的模就是它的长度。
|a| = √a₁² + a₂² + a₃²
A unit vector in the direction of a is a divided by its magnitude: â = a / |a| e useful for describing directions and for finding the perpendicular component of a force.
与 a 同方向的单位向量是 a 除以其模：â = a / |a|。单位向量常用于描述方向以及求力的垂直分量。
Vector addition: (a₁, a₂, a₃) + (b₁, b₂, b₃) = (a₁ + b₁, a₂ + b₂, a₃ + b₃).
向量加法：(a₁, a₂, a₃) + (b₁, b₂, b₃) = (a₁ + b₁, a₂ + b₂, a₃ + b₃)。
Scalar multiplication: k(a₁, a₂, a₃) = (ka₁, ka₂, ka₃).
数乘：k(a₁, a₂, a₃) = (ka₁, ka₂, ka₃)。
The vector from A to B is AB = B − A, where A and B are position vectors.
从 A 到 B 的向量为 AB = B − A，其中 A、B 是位置向量。
6. Dot Product and Angle Between Vectors | 数量积与向量夹角
b = (b₁, b₂, b₃) is a scalar given by the sum of the products of corresponding components.
a = (a₁, a₂, a₃) 与 b = (b₁, b₂, b₃) 的数量积（点积）是一个标量，等于对应分量乘积之和。
a · b = a₁b₁ + a₂b₂ + a₃b₃
It also satisfies a · b = |a||b| two vectors and 0° ≤ θ ≤ 180°. Equating the two expressions gives a way to find the angle.
它还满足 a · b = |a||b| cos θ，其中 θ 是两向量夹角，且 0° ≤ θ ≤ 180°。令两式相等即可求夹角。
cos θ = (a · b) / (|a||b|)
then the vectors are parallel. These two tests appear constantly in 3D geometry questions.
若 a · b = 0，且两向量均非零向量，则两向量垂直。若存在标量 k 使 a = k b，则两向量平行。这两个判定在三维几何题中反复出现。
b = (2, −1, 0) give a · b = 2 − 2 + 0 = 0, so they are perpendicular. Their magnitudes are |a| = 3 and |b| = √5, which confirms cos θ = 0 and θ = 90°.
例如 a = (1, 2, 2) 与 b = (2, −1, 0)，a · b = 2 − 2 + 0 = 0，故两者垂直。其模分别为 |a| = 3、|b| = √5，验证 cos θ = 0，θ = 90°。
7. Lines in 3D: Vector and Parametric Forms | 三维直线：向量式与参数式
direction d = (l, m, n), its vector equation is r = a + t d, where t is a real parameter.
三维直线由直线上一点和方向向量确定。若直线经过位置向量为 a 的点 A，方向向量为 d = (l, m, n)，则其向量方程为 r = a + t d，其中 t 为实数参数。
Writing r = (x, y, z) gives the parametric equations.
令 r = (x, y, z)，得到参数方程。
x = x₀ + t l, y = y₀ + t m, z = z₀ + t n
d B(4, 6, 3), use direction d = B − A = (3, 4, 0). The line is r = (1, 2, 3) + t(3, 4, 0).
求过 A(1, 2, 3) 与 B(4, 6, 3) 的直线，取方向 d = B − A = (3, 4, 0)。直线为 r = (1, 2, 3) + t(3, 4, 0)。
Two lines in 3D may be parallel, intersecting, or skew. Skew lines are not parallel and do not intersect. To test intersection, set the parametric expressions equal and solve for the parameters; if no solution exists and the direction vectors are not parallel, the lines are skew.
三维中两条直线可能平行、相交或异面。异面直线既不平行也不相交。判断相交时，令参数表达式相等并解参数；若无解且方向向量不平行，则两直线异面。
8. Planes and Simple Surfaces | 平面与简单曲面
rm is ax + by + cz = d, where n = (a, b, c) is a normal vector perpendicular to the plane.
三维中的平面可用 x、y、z 的线性方程描述。一般形式为 ax + by + cz = d，其中 n = (a, b, c) 是垂直于该平面的法向量。
2). Similarly, x = 1 is parallel to the yz-plane, and y = −3 is parallel to the xz-plane.
坐标平面是最简单的例子。平面 z = 2 平行于 xy 平面并通过 (0, 0, 2)。类似地，x = 1 平行于 yz 平面，y = −3 平行于 xz 平面。
| Surface | Equation | Feature |
|---|---|---|
| Plane parallel to xy-plane | z = c | Constant height |
| Plane parallel to xz-plane | y = c | Constant y-value |
| Plane parallel to yz-plane | x = c | Constant x-value |
| Sphere | (x − a)² + (y − b)² + (z − c)² = r² | Centre (a, b, c), radius r |
A plane is fixed by three non-collinear points. To find its equation, form two direction vectors from the points, find a normal vector perpendicular to both, and substitute one point to find d. In A-Level questions, you are often given the normal vector directly or asked to show that a given vector is normal to the plane.
三个不共线的点确定一个平面。求其方程时，先从点中构造两个方向向量，再求与两者都垂直的法向量，最后代入一点求 d。在 A-Level 题目中，往往直接给出法向量，或要求证明某向量是平面的法向量。
9. Spheres and Distance from a Point to a Plane | 球面与点到平面距离
the square allows you to recover the centre and radius from a general quadratic equation.
球心为 C(a, b, c)、半径为 r 的球面方程为 (x − a)² + (y − b)² + (z − c)² = r²。展开并配方，可从一般二次方程中还原球心和半径。
ten as (x − 1)² + (y − 2)² + (z − 3)² = 9, so the centre is (1, 2, 3) and the radius is 3.
例如，x² + y² + z² − 2x − 4y − 6z + 5 = 0 可改写为 (x − 1)² + (y − 2)² + (z − 3)² = 9，故球心为 (1, 2, 3)，半径为 3。
numerator is the absolute value of the plane equation evaluated at P, and the denominator is the magnitude of the normal vector.
点 P(x₀, y₀, z₀) 到平面 ax + by + cz = d 的垂直距离由下式给出。分子是平面方程在 P 点取值的绝对值，分母是法向量的模。
distance = |a x₀ + b y₀ + c z₀ − d| / √a² + b² + c²
This formula is also used to find the radius of a sphere that is tangent to a plane, or to decide whether a line intersects a sphere. If the distance from the centre to the plane is less than the radius, the plane cuts the sphere in a circle.
该公式也用于求与平面相切的球半径，或判断直线是否与球面相交。若球心到平面的距离小于半径，则平面与球面相交成一个圆。
10. Parallel, Perpendicular and Intersection | 平行、垂直与相交
Two planes are parallel if their normal vectors are parallel, that is, if one normal is a scalar multiple of the other. They are perpendicular if their normals have dot product zero.
两平面平行，当且仅当它们的法向量平行，即一个法向量是另一个的标量倍。两平面垂直，当且仅当它们的法向量数量积为零。
A line is parallel to a plane if its direction vector is perpendicular to the plane's normal vector. A line is perpendicular to a plane if its direction vector is parallel to the normal vector.
直线平行于平面，当且仅当其方向向量垂直于平面的法向量。直线垂直于平面，当且仅当其方向向量平行于法向量。
Line and plane: substitute the parametric line equations into the plane equation and solve for t.
直线与平面：把直线的参数方程代入平面方程，解出 t。
Two lines: equate the parametric forms and solve the resulting system. If there is a unique solution, the lines intersect.
两直线：令参数式相等，解方程组。若有唯一解，则两直线相交。
Two planes: their intersection is a line. Solve the two linear equations simultaneously and introduce a parameter for the free variable.
两平面：其交线是一条直线。联立两个线性方程，并对自由变量引入参数。
Angle between a line and a plane: use the angle φ between the direction vector d and the normal vector n, then the angle between the line and the plane is 90° − φ. This is a common source of confusion, so sketch the configuration before calculating.
直线与平面的夹角：先求方向向量 d 与法向量 n 的夹角 φ，则直线与平面的夹角为 90° − φ。这是常见易错点，计算前应先画图确认。
11. Worked Exam-Style Examples | 考试风格例题
Example 1. Given A(1, 2, 3) and B(4, 6, 3), find AB, the midpoint M, and a unit vector in the direction AB.
例 1. 已知 A(1, 2, 3) 与 B(4, 6, 3)，求 AB、中点 M，以及与 AB 同方向的单位向量。
AB = B − A = (3, 4, 0), so AB = √3² + 4² + 0² 5. M = ((1 + 4)/2, (2 + 6)/2, (3 + 3)/2) = (2.5, 4, 3). The unit vector is (3/5, 4/5, 0).
AB = B − A = (3, 4, 0)，故 AB = √3² + 4² + 0² = 5。M = ((1 + 4)/2, (2 + 6)/2, (3 + 3)/2) = (2.5, 4, 3)。单位向量为 (3/5, 4/5, 0)。
Example 2. Find the equation of the line through A(1, 2, 3) and B(4, 6, 3), and determine whether the point C(7, 10, 3) lies on it.
例 2. 求过 A(1, 2, 3) 与 B(4, 6, 3) 的直线方程，并判断点 C(7, 10, 3) 是否在其上。
+ 3t = 7 gives t = 2. Then y = 2 + 4(2) = 10 and z = 3 + 0(2) = 3, so C lies on the line.
直线为 r = (1, 2, 3) + t(3, 4, 0)。对于 C，比较 x：1 + 3t = 7 得 t = 2。于是 y =
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