Vectors in 3D | 三维向量

📚 Vectors in 3D | 三维向量

In A-Level Mathematics, vectors are essential tools for describing position, movement and geometrical relationships in space. When we move from two dimensions to three dimensions, vectors take on a new level of depth and usefulness, enabling us to model lines, planes and solid objects mathematically.

在 A-Level 数学中，向量是描述空间中位置、运动和几何关系的重要工具。当我们从二维平面进入三维空间后，向量变得更加立体且实用，能够帮助我们建立直线、平面和立体图形的数学模型。

1. Coordinates in 3D and Position Vectors | 三维坐标与位置向量

In three-dimensional space, we use three mutually perpendicular axes: the x-axis, y-axis and z-axis. A point P in 3D is written as (x, y, z). The origin O is (0, 0, 0). The position vector of P is the vector from O to P, denoted as p = x i + y j + z k, where i, j and k are unit vectors along the x, y and z axes respectively.

在三维空间中，我们使用三条互相垂直的轴：x 轴、y 轴和 z 轴。空间中的点 P 记作 (x, y, z)，原点 O 为 (0, 0, 0)。点 P 的位置向量就是从 O 到 P 的向量，记作 p = x i + y j + z k，其中 i、j、k 分别是沿 x 轴、y 轴和 z 轴方向的单位向量。

For example, the point A(2, -1, 3) has position vector a = 2i – j + 3k. Here the x-component is 2, the y-component is -1, and the z-component is 3.

例如，点 A(2, -1, 3) 的位置向量是 a = 2i – j + 3k，其中 x 分量为 2，y 分量为 -1，z 分量为 3。

When writing position vectors, it is crucial to keep the order of i, j, k consistent. The coefficients can be written as a column vector:
[x y z]ᵀ or in row form (x, y, z). In A-Level solutions, you may use either notation, but always define it clearly.

书写位置向量时，必须保持 i、j、k 分量的顺序一致。系数可以写成列向量 [x y z]ᵀ 或行形式 (x, y, z)。在 A-Level 答题中，两种记号都可以使用，但必须清晰地定义。

2. Magnitude and Unit Vectors | 向量的模与单位向量

The magnitude (or length) of a 3D vector a = x i + y j + z k is given by the formula:
|a| = √(x² + y² + z²)

三维向量 a = x i + y j + z k 的模（或长度）由公式给出：
|a| = √(x² + y² + z²)

This is a direct extension of Pythagoras’ theorem. For example, if a = 2i – 3j + 6k, then |a| = √(4 + 9 + 36) = √49 = 7.

这是勾股定理的直接推广。例如，若 a = 2i – 3j + 6k，则 |a| = √(4 + 9 + 36) = √49 = 7。

A unit vector is a vector with magnitude 1. To find a unit vector in the direction of a, divide a by its magnitude:
â = a / |a|

单位向量是模为 1 的向量。求 a 方向上的单位向量，只需将 a 除以它的模：
â = a / |a|

In the example above, the unit vector in the direction of a is â = (2/7)i – (3/7)j + (6/7)k. Unit vectors are especially useful when we care about direction rather than magnitude.

在上述例子中，a 方向上的单位向量为 â = (2/7)i – (3/7)j + (6/7)k。单位向量在我们只关心方向而不关心大小时特别有用。

3. Addition, Subtraction and Scalar Multiplication | 向量的加减与数乘

Vectors in 3D are added and subtracted component-wise. If a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k, then:
a + b = (a₁ + b₁)i + (a₂ + b₂)j + (a₃ + b₃)k
a – b = (a₁ – b₁)i + (a₂ – b₂)j + (a₃ – b₃)k

三维向量的加减按对应分量进行。若 a = a₁i + a₂j + a₃k，b = b₁i + b₂j + b₃k，则：
a + b = (a₁ + b₁)i + (a₂ + b₂)j + (a₃ + b₃)k
a – b = (a₁ – b₁)i + (a₂ – b₂)j + (a₃ – b₃)k

Scalar multiplication multiplies every component by the scalar:
λa = λa₁i + λa₂j + λa₃k

数乘（标量乘法）将每个分量都乘以该标量：
λa = λa₁i + λa₂j + λa₃k

Geometrically, adding two vectors corresponds to placing the tail of one at the head of the other (triangle law). The order of addition does not change the result: a + b = b + a. Subtraction a – b is equivalent to a + (-b), giving the vector from the head of a to the head of b when drawn from a common origin.

从几何上看，两个向量相加对应将一个向量的起点放在另一个向量的终点（三角形法则）。加法交换律成立：a + b = b + a。减法 a – b 等价于 a + (-b)，当二者从同一点出发时，它表示从 a 的终点指向 b 的终点的向量。

For example, if a = i – 2j + 3k and b = 4i + j – k, then 2a – b = 2(i – 2j + 3k) – (4i + j – k) = (2 – 4)i + (-4 – 1)j + (6 + 1)k = -2i – 5j + 7k.

例如，设 a = i – 2j + 3k，b = 4i + j – k，则 2a – b = 2(i – 2j + 3k) – (4i + j – k) = (2 – 4)i + (-4 – 1)j + (6 + 1)k = -2i – 5j + 7k。

4. The Vector Between Two Points | 两点间的向量

Given two points A and B with position vectors a and b, the displacement vector from A to B is:
AB = b – a

已知两点 A 和 B，其位置向量分别为 a 和 b，则从 A 到 B 的位移向量为：
AB = b – a

For example, if A = (1, 2, 3) and B = (4, -2, 5), then AB = (4 – 1)i + (-2 – 2)j + (5 – 3)k = 3i – 4j + 2k.

举个例子，若 A = (1, 2, 3)，B = (4, -2, 5)，则 AB = (4 – 1)i + (-2 – 2)j + (5 – 3)k = 3i – 4j + 2k。

The distance between A and B is simply the magnitude of AB:
|AB| = √((x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)²)

A 与 B 之间的距离就是 AB 的模：
|AB| = √((x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)²)

Notice the direction of the vector matters: AB = -BA. In exam questions, always check whether you are asked for AB or BA.

注意向量的方向很重要：AB = -BA。在考试中，一定要看清题目要求的是 AB 还是 BA。

5. The Dot Product (Scalar Product) | 点积（标量积）

The dot product of two 3D vectors a and b is defined as:
a ⋅ b = a₁b₁ + a₂b₂ + a₃b₃

两个三维向量 a 和 b 的点积定义为：
a ⋅ b = a₁b₁ + a₂b₂ + a₃b₃

Alternatively, using the angle θ between the vectors:
a ⋅ b = |a||b| cos θ

另一种表达方式用两向量夹角 θ：
a ⋅ b = |a||b| cos θ

This gives a powerful way to find angles between vectors. Rearranging:
cos θ = (a ⋅ b) / (|a||b|)

这为我们提供了求向量夹角的强大工具。改写后：
cos θ = (a ⋅ b) / (|a||b|)

For example, find the angle between a = 2i + j – k and b = i – j + 3k.

例如，求 a = 2i + j – k 与 b = i – j + 3k 之间的夹角。

First compute a ⋅ b = 2×1 + 1×(-1) + (-1)×3 = 2 – 1 – 3 = -2.

先计算 a ⋅ b = 2×1 + 1×(-1) + (-1)×3 = 2 – 1 – 3 = -2。

Then |a| = √(4 + 1 + 1) = √6, |b| = √(1 + 1 + 9) = √11.

然后 |a| = √(4 + 1 + 1) = √6，|b| = √(1 + 1 + 9) = √11。

Thus cos θ = -2 / (√6 √11) ≈ -0.246, so θ ≈ 104.2°.

因此 cos θ = -2 / (√6 √11) ≈ -0.246，所以 θ ≈ 104.2°。

Note: when the angle is obtuse, the dot product is negative. When the dot product is zero, the vectors are perpendicular (θ = 90°).

注意：当夹角为钝角时，点积为负。当点积为零时，向量互相垂直（θ = 90°）。

6. Parallel and Perpendicular Vectors | 平行与垂直向量

Two non-zero vectors a and b are parallel if one is a scalar multiple of the other: a = λb for some real λ. For example, a = 2i – 4j + 6k is parallel to b = i – 2j + 3k, because a = 2b.

两个非零向量 a 与 b 平行当且仅当其中一个可以表示为另一个的标量倍：a = λb，其中 λ 为实数。例如，a = 2i – 4j + 6k 与 b = i – 2j + 3k 平行，因为 a = 2b。

Two non-zero vectors are perpendicular if their dot product is zero: a ⋅ b = 0.

两个非零向量垂直当且仅当它们的点积为零：a ⋅ b = 0。

These properties are frequently used in coordinate geometry and mechanics problems. For instance, to show that a triangle is right-angled, we can show that the dot products of adjacent side vectors equal zero.

这些性质经常出现在解析几何和力学问题中。例如，要证明一个三角形是直角三角形，只需验证其两条邻边向量的点积为零。

In exam problems, you may be asked to find a constant k such that two vectors are perpendicular. If a = (2i + 3j + k,k) and b = (i – 2j + 3k), then a ⋅ b = 0 gives 2×1 + 3×(-2) + k×3 = 2 – 6 + 3k = -4 + 3k = 0, so k = 4/3.

在考试中，常要求你找一个常数 k，使两个向量垂直。若 a = (2i + 3j + k, k) 与 b = (i – 2j + 3k)，则由 a ⋅ b = 0 得 2×1 + 3×(-2) + k×3 = 2 – 6 + 3k = -4 + 3k = 0，所以 k = 4/3。

7. Vector Equation of a Line in 3D | 三维直线的向量方程

A line in 3D can be described by a point on the line and a direction vector. If a is the position vector of a known point on the line, and d is a non-zero direction vector, then any point r on the line satisfies:
r = a + t d, t ∈ ℝ

三维空间中的直线可以用直线上一点和方向向量来描述。若 a 是直线上已知点的位置向量，d 是非零方向向量，则直线上任一点 r 满足：
r = a + t d, t ∈ ℝ

This is called the vector equation of a line. The parameter t is a real number; each value of t gives a different point on the line.

这称为直线的向量方程。参数 t 是实数；每一个 t 的取值对应直线上一个不同的点。

For example, a line passing through A(1, 2, 3) in direction d = 4i – 5j + 6k has equation r = (i + 2j + 3k) + t(4i – 5j + 6k).

例如，过点 A(1, 2, 3) 且方向为 d = 4i – 5j + 6k 的直线方程为 r = (i + 2j + 3k) + t(4i – 5j + 6k)。

We can also write the equation in parametric form by equating components:
x = 1 + 4t, y = 2 – 5t, z = 3 + 6t.

我们还可以通过比较分量写出参数方程：
x = 1 + 4t, y = 2 – 5t, z = 3 + 6t。

The direction vector can be scaled by any non-zero constant without changing the line. For example, multiplying d by 2 gives the same line with a different parameter.

方向向量可以乘以任意非零常数而不改变直线本身。例如，将 d 乘以 2 会得到同一条直线，但参数不同。

8. Cartesian Equation of a Line | 直线的笛卡尔方程

By eliminating the parameter t from the vector equation r = a + t d, we obtain the Cartesian equation of a line. If d = l i + m j + n k, then:
(x – x₀)/l = (y – y₀)/m = (z – z₀)/n
provided l, m and n are all non-zero.

从向量方程 r = a + t d 中消去参数 t，可以得到直线的笛卡尔方程。若 d = l i + m j + n k，则：
(x – x₀)/l = (y – y₀)/m = (z – z₀)/n
其中 l、m、n 都不为零。

Here (x₀, y₀, z₀) is a point on the line. If any of l, m, n is zero, we handle it separately. For example, if l = 0, then x = x₀ for every point on the line; we write the equation without the x term.

其中 (x₀, y₀, z₀) 是直线上的一点。如果 l、m、n 中有一个为零，则需要单独处理。例如，若 l = 0，则直线上所有点的 x 坐标都等于 x₀；书写方程时就不包含含 x 的项。

Example: Convert r = i + 2j – k + t(2i – j + 3k) to Cartesian form.

例：将 r = i + 2j – k + t(2i – j + 3k) 化为笛卡尔形式。

Here x = 1 + 2t, y = 2 – t, z = -1 + 3t. So t = (x – 1)/2 = (y – 2)/(-1) = (z + 1)/3.

这里 x = 1 + 2t，y = 2 – t，z = -1 + 3t。因此 t = (x – 1)/2 = (y – 2)/(-1) = (z + 1)/3。

Write the simplified equation:
(x – 1)/2 = (2 – y)/1 = (z + 1)/3

写出简化后的方程：
(x – 1)/2 = (2 – y)/1 = (z + 1)/3

This form is very useful for checking whether a given point lies on the line, or for solving intersections.

这种形式在判断给定点是否在直线上或求交点时非常有用。

9. Intersection of Two Lines | 两条直线的交点

To find the intersection of two lines in 3D, we write each in parametric form and equate the components. If there is a unique solution for the parameters, the lines intersect; if not, they are either parallel or skew.

要求两条三维直线的交点，我们将每条直线写成参数形式并令各分量相等。如果参数存在唯一解，则两直线相交；若否则它们平行或异面（skew）。

Consider:
L₁: r = (i + j + k) + s(2i – j + k)
L₂: r = (2i + 3j – k) + t(i + 2j – k)

考虑：
L₁: r = (i + j + k) + s(2i – j + k)
L₂: r = (2i + 3j – k) + t(i + 2j – k)

Equating components gives:
x: 1 + 2s = 2 + t
y: 1 – s = 3 + 2t
z: 1 + s = -1 – t

令分量相等：
x: 1 + 2s = 2 + t
y: 1 – s = 3 + 2t
z: 1 + s = -1 – t

From the first equation, t = 2s – 1. Substitute into the y equation: 1 – s = 3 + 2(2s – 1) = 3 + 4s – 2 = 1 + 4s. Thus 1 – s = 1 + 4s ⇒ -s = 4s ⇒ s = 0. Then t = -1.

由第一个方程得 t = 2s – 1。代入 y 方程：1 – s = 3 + 2(2s – 1) = 3 + 4s – 2 = 1 + 4s。因此 1 – s = 1 + 4s ⇒ -s = 4s ⇒ s = 0。于是 t = -1。

Check the z equation with s = 0: 1 + 0 = 1, and -1 – (-1) = 0. They do not match, so the lines do not intersect. In this case, the lines are skew because their direction vectors are not parallel.

用 s = 0 检验 z 方程：左边 1 + 0 = 1，右边 -1 – (-1) = 0。两边不相等，因此两直线不相交。这种情况下，因为方向向量不平行，故两直线为异面直线。

To avoid mistakes, always verify all three component equations. A common exam trap is to find s and t from two equations but forget to check the third.

为避免出错，必须验证所有三个分量方程。考试中常见陷阱是只用两个方程解出 s 和 t，却忘记检查第三个方程。

10. Distance from a Point to a Line | 点到直线的距离

Although not always examinable in the core A-Level, the distance from a point to a line in 3D is a useful extension. For a line r = a + t d and a point P with position vector p, the distance d is given by:
d = |(p – a) × d| / |d|
where × represents the cross product (if studied in Further Mathematics).

虽然并非核心 A-Level 必考内容，但三维空间中点到直线的距离是一个有用的扩展。对于直线 r = a + t d 和位置向量为 p 的点 P，距离 d 为：
d = |(p – a) × d| / |d|
其中 × 表示叉积（若在进阶数学中学习过）。

For the standard A-Level Mathematics syllabus, you can find the distance by using projection. If the foot of the perpendicular from P to the line is F, then F is found by solving (p – r) ⋅ d = 0. Then the distance is |p – r|.

对于标准 A-Level 数学大纲，你可以通过投影来求距离。若从 P 向直线作垂线，垂足为 F，则 F 可通过 (p – r) ⋅ d = 0 求得。之后距离就是 |p – r|。

Example: For line r = (i + 2j + 3k) + t(i – j + 2k) and point P(3, 1, 2), we set r(t) and require (p – r) ⋅ d = 0. Here p – r = (3 – 1 – t)i + (1 – 2 + t)j + (2 – 3 – 2t)k = (2 – t)i + (-1 + t)j + (-1 – 2t)k. Dot with d = i – j + 2k:

例：对于直线 r = (i + 2j + 3k) + t(i – j + 2k) 和点 P(3, 1, 2)，设 r(t) 并令 (p – r) ⋅ d = 0。这里 p – r = (3 – 1 – t)i + (1 – 2 + t)j + (2 – 3 – 2t)k = (2 – t)i + (-1 + t)j + (-1 – 2t)k。与 d = i – j + 2k 作点积：

(2 – t)(1) + (-1 + t)(-1) + (-1 – 2t)(2) = 2 – t + 1 – t – 2 – 4t = 1 – 6t = 0 ⇒ t = 1/6.

(2 – t)(1) + (-1 + t)(-1) + (-1 – 2t)(2) = 2 – t + 1 – t – 2 – 4t = 1 – 6t = 0 ⇒ t = 1/6。

Then r = (7/6)i + (11/6)j + (10/3)k. The distance is the magnitude of p – r.

于是 r = (7/6)i + (11/6)j + (10/3)k。距离就是 p – r 的模。

11. Worked Exam-Style Problem | 考试型例题精讲

Let us apply several skills in one problem. Points A(1, -1, 2), B(3, 2, -1), C(0, 4, 5) are given.

让我们把多个技巧应用到一个综合例题中。已知点 A(1, -1, 2)、B(3, 2, -1)、C(0, 4, 5)。

(a) Find the vector AB and AC.

(a) 求向量 AB 与 AC。

AB = (3 – 1)i + (2 + 1)j + (-1 – 2)k = 2i + 3j – 3k.

AB = (3 – 1)i + (2 + 1)j + (-1 – 2)k = 2i + 3j – 3k。

AC = (0 – 1)i + (4 + 1)j + (5 – 2)k = -i + 5j + 3k.

AC = (0 – 1)i + (4 + 1)j + (5 – 2)k = -i + 5j + 3k。

(b) Show that angle BAC is 90°.

(b) 证明角 BAC 为 90°。

Compute AB ⋅ AC = 2(-1) + 3(5) + (-3)(3) = -2 + 15 – 9 = 4. This is not zero, so the angle is not 90°.

计算 AB ⋅ AC = 2(-1) + 3(5) + (-3)(3) = -2 + 15 – 9 = 4。这不是零，所以角 BAC 不是 90°。

Using point A and direction AB:
r = i – j + 2k + t(2i + 3j – 3k)

以点 A 和方向 AB 可得：
r = i – j + 2k + t(2i + 3j – 3k)

(d) Determine whether point D(5, 4, -4) lies on this line.

(d) 判断点 D(5, 4, -4) 是否在此直线上。

Set r = 5i + 4j – 4k. Then:
1 + 2t = 5 ⇒ t = 2
-1 + 3t = 4 ⇒ t = 5/3
2 – 3t = -4 ⇒ t = 2

令 r = 5i + 4j – 4k。则：
1 + 2t = 5 ⇒ t = 2
-1 + 3t = 4 ⇒ t = 5/3
2 – 3t = -4 ⇒ t = 2

The t values do not all agree, so D is not on the line.

t 的值不一致，因此 D 不在此直线上。

12. Common Pitfalls and Revision Tips | 常见错误与复习建议

Many students lose marks on 3D vector questions because of sign errors, missing the third component, or forgetting to verify a solution. Here we list the most common pitfalls.

许多学生在三维向量题目中失分，原因往往包括符号错误、漏掉第三个分量，或忘记验证解。我们在这里列出最常见的陷阱。

Mixing up position vectors and displacement vectors. Position vectors start at the origin; displacement vectors point from one point to another.
未区分位置向量和位移向量。位置向量从原点出发；位移向量从一点指向另一点。
Forgetting to take the square root when calculating magnitude.
计算模时忘记开方。
Using the wrong formula for the angle. Remember cos θ = (a ⋅ b)/(|a||b|), not just a ⋅ b.
用错夹角公式。记住 cos θ = (a ⋅ b)/(|a||b|)，而不是只算 a ⋅ b。
Not checking all three equations when finding intersections of lines.
求直线交点时没有验证全部三个方程。
Incorrectly handling zero components in direction vectors while converting to Cartesian form.
在转化为笛卡尔形式时错误处理方向向量中的零分量。

To revise effectively, practise classifying lines as parallel, intersecting or skew. Draw simple diagrams in 3D (or use 3D graphing software) to build intuition. Always write your vectors in a clear component ordering and check your arithmetic step by step.

为了高效复习，应练习判断两直线是平行、相交还是异面。可以在纸上粗略画三维图（或使用 3D 绘图软件）来建立直观感觉。书写向量时保持清晰的分量排列，并逐步检查计算。

Finally, remember that vector methods in 3D are consistent with the 2D cases you already know. The dot product, magnitude and line equations all extend naturally by adding the k-component. Once you feel comfortable with the notation, 3D vector questions become a matter of careful algebra.

最后，请记住三维向量的方法与已知的二维情形是一致的。点积、模和直线方程都通过添加 k 分量自然地扩展。一旦你熟悉了这些符号，三维向量问题就变成了仔细的代数和计算。

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Vectors in 3D | 三维向量