📚 The Vector Product | 向量积
The vector product (also called the cross product) is a binary operation on two vectors in three-dimensional space. Unlike the scalar product, which gives a scalar result, the vector product gives a vector that is perpendicular to both original vectors. This operation is essential for finding normals to planes, areas of parallelograms, and solving geometric problems in 3D Mechanics.
向量积(又称叉积)是三维空间中两个向量的二元运算。与得到标量结果的标量积不同,向量积的结果是一个向量,该向量同时垂直于原来的两个向量。这种运算对于求平面的法向量、平行四边形的面积以及解决三维力学中的几何问题至关重要。
1. Definition of the Vector Product | 向量积的定义
The vector product of two vectors a and b is denoted by a × b (read ‘a cross b’). It is defined as a vector with magnitude |a||b| sin θ, where θ is the angle between the vectors (0 ≤ θ ≤ π), and with direction given by the right-hand rule — perpendicular to both a and b.
两个向量 a 和 b 的向量积记为 a × b(读作“a 叉乘 b”)。其定义是:大小为 |a||b| sin θ 的向量,其中 θ 是两向量之间的夹角(0 ≤ θ ≤ π),方向由右手定则确定——垂直于 a 和 b 所在的平面。
|a × b| = |a||b| sin θ
If a = 0 or b = 0, the vector product is the zero vector. Note that the vector product is only defined for 3D vectors; in AQA A-Level Maths you will use vectors written in component form i, j, k or as column vectors.
如果 a = 0 或 b = 0,则向量积为零向量。请注意,向量积仅对三维向量有定义;在 AQA A-Level 数学中,你会使用分量形式 i, j, k 或列向量表示的向量。
2. Component Formula | 分量公式
Given a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k, the vector product is computed using a 3×3 determinant:
已知 a = a₁i + a₂j + a₃k,b = b₁i + b₂j + b₃k,向量积通过以下 3×3 行列式计算:
a × b = (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k
This can also be written as a determinant expansion:
这也可以写成行列式展开的形式:
| a × b = det | i j k a₁ a₂ a₃ b₁ b₂ b₃ |
A useful memory aid: for the i-component, cover the first column, compute the 2×2 determinant of the remaining entries with a positive sign; for the j-component, cover the second column, compute the determinant with a negative sign; for the k-component, cover the third column, compute with a positive sign.
一个有用的记忆技巧:求 i 分量时,遮住第一列,对剩余元素求 2×2 行列式,取正号;求 j 分量时,遮住第二列,求行列式取负号;求 k 分量时,遮住第三列,求行列式取正号。
3. Worked Example 1 | 示例 1
Let a = 2i + 3j − k and b = i − 2j + 4k. Find a × b.
设 a = 2i + 3j − k,b = i − 2j + 4k。求 a × b。
Using the component formula:
利用分量公式:
a × b = (3×4 − (−1)(−2))i − (2×4 − (−1)(1))j + (2×(−2) − 3×1)k
Calculate each component:
逐项计算各分量:
- i-component: (3×4) − ((−1)×(−2)) = 12 − 2 = 10
- i 分量: (3×4) − ((−1)×(−2)) = 12 − 2 = 10
- j-component: −[(2×4) − ((−1)×1)] = −(8 − (−1)) = −9
- j 分量: −[(2×4) − ((−1)×1)] = −(8 − (−1)) = −9
- k-component: (2×(−2)) − (3×1) = −4 − 3 = −7
- k 分量: (2×(−2)) − (3×1) = −4 − 3 = −7
Therefore a × b = 10i − 9j − 7k. This is the vector perpendicular to both a and b.
因此 a × b = 10i − 9j − 7k。这就是同时垂直于 a 和 b 的向量。
4. Properties of the Vector Product | 向量积的性质
You must know the key properties, especially the anti-commutative nature, which differs from ordinary multiplication.
你必须掌握以下关键性质,尤其是反交换性,这与普通乘法不同。
- Anti-commutative: a × b = −(b × a)
- 反交换律: a × b = −(b × a)
- Parallel vectors: If a and b are parallel (θ = 0 or π), then a × b = 0
- 平行向量: 若 a 与 b 平行(θ = 0 或 π),则 a × b = 0
- Self product: a × a = 0
- 自身叉积: a × a = 0
- Scalar multiplication: (ka) × b = a × (kb) = k(a × b) for any scalar k
- 标量倍乘: 对任意标量 k,(ka) × b = a × (kb) = k(a × b)
- Distributive: a × (b + c) = a × b + a × c
- 分配律: a × (b + c) = a × b + a × c
The unit vector (i, j, k) products are worth memorising:
单位向量(i, j, k)的叉积结果值得记忆:
| × | i | j | k |
| i | 0 | k | −j |
| j | −k | 0 | i |
| k | j | −i | 0 |
For example, i × j = k, j × k = i, and k × i = j. Reversing the order changes the sign: j × i = −k.
例如,i × j = k,j × k = i,k × i = j。交换顺序会改变符号:j × i = −k。
5. Geometric Interpretation: Area | 几何意义:面积
The magnitude |a × b| gives the area of the parallelogram formed by vectors a and b. The area of the triangle formed by a and b is half of this magnitude.
模长 |a × b| 给出由向量 a 和 b 构成的平行四边形的面积。由 a 和 b 构成的三角形面积为其一半。
Parallelogram area = |a × b|, Triangle area = ½ |a × b|
This is particularly useful when coordinates are given in 3D and finding the perpendicular height via trigonometry is inconvenient.
当给定 3D 坐标且用三角函数求垂直高度不便时,这一性质尤为实用。
6. Worked Example 2: Area of a Triangle | 示例 2:三角形面积
Points A(1, 0, 2), B(3, 2, −1) and C(0, 1, 4) form a triangle. Find its area.
点 A(1, 0, 2)、B(3, 2, −1) 和 C(0, 1, 4) 构成一个三角形。求其面积。
First form two vectors from A:
首先从 A 点构造两个向量:
- AB = (3 − 1)i + (2 − 0)j + (−1 − 2)k = 2i + 2j − 3k
- AB = (3 − 1)i + (2 − 0)j + (−1 − 2)k = 2i + 2j − 3k
- AC = (0 − 1)i + (1 − 0)j + (4 − 2)k = −i + j + 2k
- AC = (0 − 1)i + (1 − 0)j + (4 − 2)k = −i + j + 2k
Compute AB × AC:
计算 AB × AC:
= (2×2 − (−3)(1))i − (2×2 − (−3)(−1))j + (2×1 − 2×(−1))k
- i: 4 + 3 = 7
- j: −(4 − 3) = −1
- k: 2 + 2 = 4
So AB × AC = 7i − j + 4k. Its magnitude is √(7² + (−1)² + 4²) = √(49 + 1 + 16) = √66.
因此 AB × AC = 7i − j + 4k。其模长为 √(7² + (−1)² + 4²) = √(49 + 1 + 16) = √66。
Triangle area = ½ × √66 square units.
三角形面积 = ½ × √66 平方单位。
7. Normal Vector to a Plane | 平面的法向量
Given two non-parallel vectors d₁ and d₂ lying in a plane, the vector d₁ × d₂ is perpendicular to the plane, hence it is a normal vector n to the plane.
给定平面内两个不平行向量 d₁ 和 d₂,则向量 d₁ × d₂ 垂直于该平面,因此它是该平面的法向量 n。
If a plane has equation r · n = d, and point A with position vector a lies in the plane, you can find the Cartesian equation using n = d₁ × d₂, then (r − a) · n = 0.
若平面方程为 r · n = d,且点 A 的位置向量为 a 且位于该平面内,你可以用 n = d₁ × d₂ 求出笛卡尔方程,即 (r − a) · n = 0。
For AQA A-Level, you may be asked to find the equation of a plane given three points in the plane. The cross product provides the most direct method.
在 AQA A-Level 考试中,你可能会被要求根据平面上的三个点求出平面方程。向量积提供了最直接的方法。
8. Worked Example 3: Equation of a Plane | 示例 3:平面方程
Find the Cartesian equation of the plane passing through A(1, 2, 1), B(2, 0, 3) and C(0, 1, 2).
求过点 A(1, 2, 1)、B(2, 0, 3) 和 C(0, 1, 2) 的平面的笛卡尔方程。
Step 1 — find two vectors in the plane:
步骤 1——求平面内两个向量:
- AB = i − 2j + 2k
- AB = i − 2j + 2k
- AC = −i − j + k
- AC = −i − j + k
Step 2 — compute the normal vector n = AB × AC:
步骤 2——计算法向量 n = AB × AC:
= ((−2)(1) − (2)(−1))i − ((1)(1) − (2)(−1))j + ((1)(−1) − (−2)(−1))k
- i: −2 + 2 = 0
- j: −(1 + 2) = −3
- k: −1 − 2 = −3
So n = 0i − 3j − 3k, which we can simplify to n = j + k (dividing by −3). Using point A, a = i + 2j + k:
因此 n = 0i − 3j − 3k,可简化为 n = j + k(除以 −3)。使用点 A,a = i + 2j + k:
(r − a) · n = 0
Let r = xi + yj + zk. Then (x − 1)i + (y − 2)j + (z − 1)k dotted with (j + k) gives (y − 2) + (z − 1) = 0.
设 r = xi + yj + zk。则 (x − 1)i + (y − 2)j + (z − 1)k 与 (j + k) 点乘得到 (y − 2) + (z − 1) = 0。
y + z = 3
This is the required Cartesian equation of the plane.
这就是所求平面的笛卡尔方程。
9. Vector Product in Mechanics | 向量积在力学中的应用
In AQA A-Level Further Mathematics (or A-Level Maths with mechanics option), the vector product appears in the context of moments of a force about a point. The moment of a force F acting at a position vector r relative to a point P is defined as:
在 AQA A-Level 进阶数学(或 A-Level 数学的力学模块)中,向量积出现在力对点之矩的背景下。力 F 作用在相对于点 P 的位置向量 r 上时,力矩定义为:
M = r × F
The direction of M gives the axis of rotation (right-hand screw rule), and its magnitude gives the turning effect. In rigid-body problems, this vector formulation avoids the need to find perpendicular distances manually.
M 的方向给出旋转轴(右手螺旋法则),其大小给出转动效应的大小。在刚体问题中,这种向量表述避免手动求垂直距离的需要。
10. Common Mistakes and Exam Tips | 常见错误与考试提示
Students frequently lose marks on vector product questions for avoidable reasons. Here are the most common pitfalls and how to avoid them.
学生经常因可以避免的原因在向量积题目中失分。以下是最常见的陷阱及避免方法。
- Wrong sign for the j-component: The j-component in a × b is always subtracted. Remember: i positive, j negative, k positive.
- j 分量符号错误: a × b 的 j 分量始终为负。记住:i 正、j 负、k 正。
- Confusing a × b with b × a: They are opposites. Always check the order of the vectors in the question.
- 混淆 a × b 与 b × a: 它们是相反向量。始终检查题目中向量的顺序。
- Forgetting to simplify the normal vector: A normal vector can be multiplied by any non-zero scalar; simplifying (e.g., dividing by a common factor) makes the plane equation cleaner.
- 忘记化简法向量: 法向量可以乘以任何非零标量;化简(例如除以公因子)使平面方程更简洁。
- Using the scalar product formula accidentally: The scalar product uses cos θ; the vector product uses sin θ. Read the question carefully.
- 不小心使用标量积公式: 标量积使用 cos θ;向量积使用 sin θ。仔细阅读题目。
Examiners often award method marks even for incorrect calculations, so always write down the determinant or component formula clearly before substituting values.
考官通常会为即使计算错误的题目给方法分,因此务必在代入数值前清楚写出行列式或分量公式。
11. Practice Questions | 练习题
Test yourself with these AQA-style questions. Full solutions are provided below.
用以下 AQA 风格题目自测。完整解答附后。
Question 1: Given p = 3i − 2j + k and q = i + 4j − 2k, find p × q.
问题 1: 已知 p = 3i − 2j + k,q = i + 4j − 2k,求 p × q。
Question 2: The points P(2, −1, 3), Q(4, 1, 0), R(0, 2, 1) define a triangle. Find the area of triangle PQR correct to 3 significant figures.
问题 2: 点 P(2, −1, 3)、Q(4, 1, 0)、R(0, 2, 1) 确定一个三角形。求三角形 PQR 的面积(精确到 3 位有效数字)。
Question 3: Find a unit vector perpendicular to both u = i − j + 2k and v = 2i + 3j − k.
问题 3: 求同时垂直于 u = i − j + 2k 和 v = 2i + 3j − k 的单位向量。
12. Solutions to Practice Questions | 练习解答
Solution 1:
解答 1:
p × q = ((−2)(−2) − (1)(4))i − ((3)(−2) − (1)(1))j + ((3)(4) − (−2)(1))k
p × q = ((−2)(−2) − (1)(4))i − ((3)(−2) − (1)(1))j + ((3)(4) − (−2)(1))k
= (4 − 4)i − (−6 − 1)j + (12 + 2)k = 0i + 7j + 14k.
= (4 − 4)i − (−6 − 1)j + (12 + 2)k = 0i + 7j + 14k。
p × q = 7j + 14k
Solution 2: First PQ = 2i + 2j − 3k and PR = −2i + 3j − 2k. Then PQ × PR = ((2)(−2) − (−3)(3))i − ((2)(−2) − (−3)(−2))j + ((2)(3) − (2)(−2))k = (−4 + 9)i − (−4 − 6)j + (6 + 4)k = 5i + 10j + 10k.
解答 2: 首先 PQ = 2i + 2j − 3k,PR = −2i + 3j − 2k。则 PQ × PR = ((2)(−2) − (−3)(3))i − ((2)(−2) − (−3)(−2))j + ((2)(3) − (2)(−2))k = (−4 + 9)i − (−4 − 6)j + (6 + 4)k = 5i + 10j + 10k。
Magnitude = √(5² + 10² + 10²) = √225 = 15. Area = ½ × 15 = 7.5 square units.
模长 = √(5² + 10² + 10²) = √225 = 15。面积 = ½ × 15 = 7.5 平方单位。
Solution 3: u × v = ((−1)(−1) − (2)(3))i − ((1)(−1) − (2)(2))j + ((1)(3) − (−1)(2))k = (1 − 6)i − (−1 − 4)j + (3 + 2)k = −5i + 5j + 5k.
解答 3: u × v = ((−1)(−1) − (2)(3))i − ((1)(−1) − (2)(2))j + ((1)(3) − (−1)(2))k = (1 − 6)i − (−1 − 4)j + (3 + 2)k = −5i + 5j + 5k。
Its magnitude is √(25 + 25 + 25) = √75 = 5√3. Therefore the unit vector is (−1/√3)i + (1/√3)j + (1/√3)k, i.e., dividing all components by 5√3.
其模长为 √(25 + 25 + 25) = √75 = 5√3。因此单位向量为 (−1/√3)i + (1/√3)j + (1/√3)k,即所有分量除以 5√3。
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