The Vector Product | 向量积

📚 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 ab 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 ab 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 = 0b = 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.

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 ab 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:ab 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 ab 构成的平行四边形的面积。由 ab 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 d₁d₂,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 (ra) · 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 (ra) · 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:

(ra) · 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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