【A-Level】Oxford AQA A2 数学 - 向量（Vectors）讲义

一、课程目标

• 理解向量的基本概念及其表示方法
• 掌握向量的运算：加法、减法、数乘、点积
• 能写出直线的向量方程、参数方程和对称式方程
• 能判断两直线的位置关系并求夹角
• 会计算点到直线的距离和垂足坐标
• 能解决综合几何问题

---

二、核心知识点

1. 向量的基本性质（9.1 Properties of vectors）

定义：

向量是具有大小和方向的量，常用箭头表示 $\vec{a}$ 或粗体 $\mathbf{a}$。

坐标形式与单位向量形式：

• 三维空间中：$\mathbf{a} = (a_1, a_2, a_3) = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$

向量模长：

$$|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$$

单位向量：

$$\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}$$

---

2. 位置向量（9.2 Position vectors）

定义：

从原点 $O$ 指向某点 $A$ 的向量称为该点的位置向量：$\overrightarrow{OA} = \mathbf{a}$

两点间的向量：

若 $A$ 和 $B$ 的位置向量分别为 $\mathbf{a}$ 和 $\mathbf{b}$，则

$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$

---

3. 直线在空间中的位置表示（9.3–9.6）

向量方程：

$$\mathbf{r} = \mathbf{a} + t\mathbf{d}$$

其中：

• $\mathbf{r}$ 是直线上任意一点的位置向量
• $\mathbf{a}$ 是直线上已知点
• $\mathbf{d}$ 是方向向量
• $t$ 是实数参数

参数方程：

若 $\mathbf{a} = (x_0, y_0, z_0)$，$\mathbf{d} = (l, m, n)$，则

$$x = x_0 + lt,\quad y = y_0 + mt,\quad z = z_0 + nt$$

对称式方程（不含参数）：

$$\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$$

---

4. 向量运算（9.4 Operations on cartesian vectors）

加法：

$$\mathbf{a} + \mathbf{b} = (a_1 + b_1, a_2 + b_2, a_3 + b_3)$$

减法：

$$\mathbf{a} - \mathbf{b} = (a_1 - b_1, a_2 - b_2, a_3 - b_3)$$

数乘：

$$k\mathbf{a} = (ka_1, ka_2, ka_3)$$

点积（数量积）：

$$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta$$

---

5. 两直线的关系（9.7 Pairs of lines）

判断是否平行：

两个方向向量成比例：$\mathbf{d_1} = k\mathbf{d_2}$

判断是否垂直：

$$\mathbf{d_1} \cdot \mathbf{d_2} = 0$$

夹角公式：

$$\cos\theta = \frac{|\mathbf{d_1} \cdot \mathbf{d_2}|}{|\mathbf{d_1}||\mathbf{d_2}|}$$

---

6. 点到直线的距离与垂足坐标（9.9 The foot of the perpendicular）

点到直线的距离公式：

设点 $P$，直线过点 $A$，方向向量为 $\mathbf{d}$，则

$$d = \frac{|\overrightarrow{AP} \times \mathbf{d}|}{|\mathbf{d}|}$$

垂足坐标的计算步骤：

1. 设垂足为 $F$，且 $F$ 在直线上：$\mathbf{f} = \mathbf{a} + t\mathbf{d}$
2. 向量 $\overrightarrow{PF} = \mathbf{f} - \mathbf{p}$
3. 因为 $\overrightarrow{PF} \perp \mathbf{d}$，所以 $\overrightarrow{PF} \cdot \mathbf{d} = 0$
4. 解出 $t$，代入得 $\mathbf{f}$

---

三、常见错误与避坑指南

常见错误类型	避坑建议
向量符号混乱	使用统一格式（如全用粗体或全带箭头）
点积与叉积混淆	明确用途：点积用于夹角、投影；叉积用于面积、垂直向量
方向向量写错	注意参数方程中系数对应的方向向量
忘记取绝对值	点到直线距离公式中分子是叉积的模长
直线交点误判	两直线交点必须满足所有三个分量都相等

---

四、课堂例题讲解

例题1：写出直线的向量方程

已知直线经过点 $A(1, 2, 3)$，方向向量为 $(2, -1, 4)$，写出其向量方程。

解析：

$$\mathbf{r} = (1, 2, 3) + t(2, -1, 4)$$

---

例题2：判断两条直线是否垂直

直线 $L_1$ 的方向向量为 $(2, -1, 4)$，直线 $L_2$ 的方向向量为 $(1, 2, -1)$，判断是否垂直。

解析：

$$\mathbf{d_1} \cdot \mathbf{d_2} = 2×1 + (-1)×2 + 4×(-1) = 2 - 2 - 4 = -4 ≠ 0$$

不垂直。

---

例题3：求点到直线的距离

点 $P(3, 4, 5)$，直线 $L: \mathbf{r} = (1, 2, 3) + t(2, -1, 4)$

解析：

• $\overrightarrow{AP} = (2, 2, 2)$
• 叉积：$\overrightarrow{AP} × \mathbf{d} = (10, -4, -6)$
• 模长：$\sqrt{10^2 + (-4)^2 + (-6)^2} = \sqrt{152}$
• $|\mathbf{d}| = \sqrt{21}$
• 距离： $$d = \frac{\sqrt{152}}{\sqrt{21}}$$

---

五、课堂练习题

练习1

写出通过点 $(4, -1, 2)$，方向向量为 $(1, 3, -2)$ 的直线的向量方程和参数方程。

练习2

直线 $L_1$：$\mathbf{r} = (0, 1, 2) + t(1, -1, 2)$
直线 $L_2$：$\mathbf{r} = (1, 2, 3) + s(2, -2, 4)$
判断它们是否平行？是否重合？

练习3

点 $P(2, 3, 4)$，直线 $L: \mathbf{r} = (1, 1, 1) + t(1, 2, 3)$，求点到直线的距离。

---

六、总结与复习建议

✅ 重点掌握：

• 向量的点积与模长计算
• 直线的三种表达方式及转换
• 两直线夹角、平行与垂直判断
• 点到直线的距离与垂足坐标计算

📌 复习策略：

1. 先掌握基础公式与定义
2. 再练习直线方程的建立与转换
3. 最后攻克综合应用题（如距离、夹角、垂足）
4. 每周完成2~3道真题训练，逐步提高速度与准确率

---

七、附录：关键公式汇总

类型	公式
向量模长	$\vert \mathbf{a}\vert = \sqrt{a_1^2 + a_2^2 + a_3^2}$
点积	$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$
向量方程	$\mathbf{r} = \mathbf{a} + t\mathbf{d}$
参数方程	$x = x_0 + lt, y = y_0 + mt, z = z_0 + nt$
对称式方程	$\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$
点到直线距离	$d = \frac{\vert \overrightarrow{AP} \times \mathbf{d}\vert }{\vert \mathbf{d}\vert}$
两直线夹角	$\cos\theta = \frac{\vert \mathbf{d_1} \cdot \mathbf{d_2}\vert}{\vert \mathbf{d_1}\vert \vert \mathbf{d_2}\vert}$

---

✅ AQA A-Level 数学 向量真题汇编（2015–2024）

以下真题题目主要来自 Pure Mathematics 3 或 Pure Core 3/4（旧大纲） 和 Paper 2 Section B（新大纲）。

🔹 2024 AQA Paper 2 Q8

Question:

The line $ L_1 $ has equation

$$\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$$

The line $ L_2 $ passes through the point $ A(5, 1, 7) $ and is parallel to the vector

$$\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$$

(a) Find the angle between the lines $ L_1 $ and $ L_2 $.
(b) Show that the lines do not intersect.

---

Solution:

(a) Angle between two lines

• Direction vector of $ L_1 $: $ \mathbf{d}_1 = \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} $
• Direction vector of $ L_2 $: $ \mathbf{d}_2 = \begin{pmatrix} 2 \ 1 \ 3 \end{pmatrix} $

Use the dot product formula:

$$\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$$

Compute dot product:

$$\mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(2) + (2)(1) + (-1)(3) = 2 + 2 - 3 = 1$$

Compute magnitudes:

$$|\mathbf{d}_1| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}, \quad |\mathbf{d}_2| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}$$

So:

$$\cos\theta = \frac{1}{\sqrt{6}\sqrt{14}} = \frac{1}{\sqrt{84}} \Rightarrow \theta = \cos^{-1}\left(\frac{1}{\sqrt{84}}\right)$$

Answer:

$$\boxed{\theta \approx 83.3^\circ}$$

(b) Show that lines do not intersect

Assume they intersect for some parameters $ t $ and $ s $:

Line $ L_1 $:

$$\mathbf{r}_1 = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 + t \\ -1 + 2t \\ 4 - t \end{pmatrix}$$

Line $ L_2 $:

$$\mathbf{r}_2 = \begin{pmatrix} 5 \\ 1 \\ 7 \end{pmatrix} + s \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 + 2s \\ 1 + s \\ 7 + 3s \end{pmatrix}$$

Set components equal:

1. $ 2 + t = 5 + 2s $ → $ t = 3 + 2s $
2. $ -1 + 2t = 1 + s $
   Substitute $ t $ from (1): $$-1 + 2(3 + 2s) = 1 + s \Rightarrow -1 + 6 + 4s = 1 + s \Rightarrow 5 + 4s = 1 + s \Rightarrow 3s = -4 \Rightarrow s = -\frac{4}{3}$$

Then $ t = 3 + 2(-\frac{4}{3}) = 3 - \frac{8}{3} = \frac{1}{3} $

Check z-component:
From $ L_1 $: $ z = 4 - t = 4 - \frac{1}{3} = \frac{11}{3} $
From $ L_2 $: $ z = 7 + 3s = 7 + 3(-\frac{4}{3}) = 7 - 4 = 3 $

They are not equal ⇒ Lines do not intersect

✅ Proven

---

🔹 2023 AQA Paper 2 Q7

Question:

The point $ P $ lies on the line with equation

$$\frac{x - 2}{3} = \frac{y + 1}{-1} = \frac{z}{2}$$

and is closest to the point $ A(1, 2, 3) $. Find the coordinates of $ P $.

---

Solution:

Let point $ P $ be on the line:

$$\frac{x - 2}{3} = \frac{y + 1}{-1} = \frac{z}{2} = t \Rightarrow P = (2 + 3t, -1 - t, 2t)$$

Vector $ \overrightarrow{AP} = P - A = (2 + 3t - 1, -1 - t - 2, 2t - 3) = (1 + 3t, -3 - t, 2t - 3) $

Direction vector of line: $ \mathbf{d} = \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix} $

For $ P $ to be closest to $ A $, vector $ \overrightarrow{AP} \perp \mathbf{d} $

So:

$$\overrightarrow{AP} \cdot \mathbf{d} = 0$$

Compute:

$$(1 + 3t)(3) + (-3 - t)(-1) + (2t - 3)(2) = 0 \\ 3 + 9t + 3 + t + 4t - 6 = 0 \\ (3 + 3 - 6) + (9t + t + 4t) = 0 \\ 0 + 14t = 0 \Rightarrow t = 0$$

Then $ P = (2 + 0, -1 - 0, 0) = (2, -1, 0) $

✅ Answer:

$$\boxed{(2, -1, 0)}$$

---

🔹 2022 AQA Paper 2 Q6

Question:

Find the shortest distance from the point $ A(2, 3, 4) $ to the line with vector equation

$$\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$$

---

Solution:

Point on line: $ B = (1, 0, 2) $
Vector $ \overrightarrow{AB} = (1 - 2, 0 - 3, 2 - 4) = (-1, -3, -2) $
Direction vector: $ \mathbf{d} = \begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix} $

Use formula:

$$d = \frac{|\overrightarrow{AB} \times \mathbf{d}|}{|\mathbf{d}|}$$

Compute cross product:

$$\overrightarrow{AB} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -3 & -2 \\ 2 & 1 & -1 \end{vmatrix} = \mathbf{i}(3 - (-2)) - \mathbf{j}(1 - (-4)) + \mathbf{k}(-1 - (-6)) = \mathbf{i}(5) - \mathbf{j}(5) + \mathbf{k}(5) = \begin{pmatrix} 5 \\ -5 \\ 5 \end{pmatrix}$$

Magnitude:

$$|\overrightarrow{AB} \times \mathbf{d}| = \sqrt{5^2 + (-5)^2 + 5^2} = \sqrt{75} = 5\sqrt{3}$$

$ |\mathbf{d}| = \sqrt{4 + 1 + 1} = \sqrt{6} $

So:

$$d = \frac{5\sqrt{3}}{\sqrt{6}} = \frac{5\sqrt{18}}{6} = \frac{5 \cdot 3\sqrt{2}}{6} = \frac{15\sqrt{2}}{6} = \frac{5\sqrt{2}}{2}$$

✅ Answer:

$$\boxed{\frac{5\sqrt{2}}{2}}$$

---

🔹 2021 AQA Paper 2 Q6

Question:

Show that the lines

$$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}, \quad L_2: \mathbf{r} = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} + s \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

intersect and find the coordinates of their point of intersection.

---

Solution:

Set equations equal:

$$\begin{pmatrix} 1 + 2t \\ 2 + t \\ 3 + 2t \end{pmatrix} = \begin{pmatrix} 3 + s \\ 3 + 2s \\ 1 + 3s \end{pmatrix}$$

Solve system:

1. $ 1 + 2t = 3 + s $ → $ 2t - s = 2 $
2. $ 2 + t = 3 + 2s $ → $ t - 2s = 1 $
3. $ 3 + 2t = 1 + 3s $ → $ 2t - 3s = -2 $

From (1): $ s = 2t - 2 $

Substitute into (2):

$$t - 2(2t - 2) = 1 \Rightarrow t - 4t + 4 = 1 \Rightarrow -3t = -3 \Rightarrow t = 1 \Rightarrow s = 2(1) - 2 = 0$$

Check in (3):
$ 2(1) - 3(0) = 2 = RHS $

✅ So lines intersect when $ t = 1 $, so point:

$$P = \begin{pmatrix} 1 + 2(1) \\ 2 + 1 \\ 3 + 2(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 5 \end{pmatrix}$$

✅ Answer:

$$\boxed{(3, 3, 5)}$$

---

🔹 2020 AQA Paper 2 Q6

Question:

The points $ A $ and $ B $ have position vectors $ \mathbf{a} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} $ and $ \mathbf{b} = \begin{pmatrix} 4 \ 1 \ 5 \end{pmatrix} $ respectively.
Find the coordinates of the foot of the perpendicular from $ A $ to the line through $ B $ with direction vector $ \begin{pmatrix} 1 \ 1 \ 2 \end{pmatrix} $.

---

Solution:

Let $ F $ be the foot of the perpendicular from $ A $ to the line through $ B $ in direction $ \mathbf{d} = \begin{pmatrix} 1 \ 1 \ 2 \end{pmatrix} $

Parametrize the line:

$$\mathbf{r} = \mathbf{b} + t\mathbf{d} = \begin{pmatrix} 4 \\ 1 \\ 5 \end{pmatrix} + t \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \Rightarrow F = \begin{pmatrix} 4 + t \\ 1 + t \\ 5 + 2t \end{pmatrix}$$

Vector $ \overrightarrow{AF} = F - A = \begin{pmatrix} 3 + t \ -1 + t \ 2 + 2t \end{pmatrix} $

We want $ \overrightarrow{AF} \cdot \mathbf{d} = 0 $

Dot product:

$$(3 + t)(1) + (-1 + t)(1) + (2 + 2t)(2) = 0 \\ 3 + t -1 + t + 4 + 4t = 0 \Rightarrow 6 + 6t = 0 \Rightarrow t = -1$$

Then:

$$F = \begin{pmatrix} 4 - 1 \\ 1 - 1 \\ 5 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 3 \end{pmatrix}$$

✅ Answer:

$$\boxed{(3, 0, 3)}$$

---

🔹 2019 AQA Paper 2 Q7

Question:

The line $ L_1 $ has equation

$$\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$$

The line $ L_2 $ passes through the point $ A(4, -1, 6) $ and is parallel to the vector

$$\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

(a) Find the angle between the lines $ L_1 $ and $ L_2 $.
(b) Show that the lines do not intersect.

---

Solution:

(a) Angle between two lines

• Direction vector of $ L_1 $: $ \mathbf{d}_1 = \begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix} $
• Direction vector of $ L_2 $: $ \mathbf{d}_2 = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} $

Use the dot product formula:

$$\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$$

Compute dot product:

$$\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (1)(2) + (-1)(3) = 2 + 2 - 3 = 1$$

Compute magnitudes:

$$|\mathbf{d}_1| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6}, \quad |\mathbf{d}_2| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$$

So:

$$\cos\theta = \frac{1}{\sqrt{6}\sqrt{14}} = \frac{1}{\sqrt{84}} \Rightarrow \theta = \cos^{-1}\left(\frac{1}{\sqrt{84}}\right)$$

Answer:

$$\boxed{\theta \approx 83.3^\circ}$$

(b) Show that lines do not intersect

Assume they intersect for some parameters $ t $ and $ s $:

Line $ L_1 $:

$$\mathbf{r}_1 = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 + 2t \\ -2 + t \\ 4 - t \end{pmatrix}$$

Line $ L_2 $:

$$\mathbf{r}_2 = \begin{pmatrix} 4 \\ -1 \\ 6 \end{pmatrix} + s \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 + s \\ -1 + 2s \\ 6 + 3s \end{pmatrix}$$

Set components equal:

1. $ 1 + 2t = 4 + s $ → $ 2t = 3 + s $
2. $ -2 + t = -1 + 2s $ → $ t = 1 + 2s $

Substitute into (1):

$$2(1 + 2s) = 3 + s \Rightarrow 2 + 4s = 3 + s \Rightarrow 3s = 1 \Rightarrow s = \frac{1}{3} \Rightarrow t = 1 + 2\cdot\frac{1}{3} = \frac{5}{3}$$

Check z-component:
From $ L_1 $: $ z = 4 - t = 4 - \frac{5}{3} = \frac{7}{3} $
From $ L_2 $: $ z = 6 + 3s = 6 + 1 = 7 $

They are not equal ⇒ Lines do not intersect

✅ Proven

---

🔹 2018 AQA Paper 2 Q6

Question:

Find the shortest distance from the point $ A(1, 2, 3) $ to the line with vector equation

$$\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$$

---

Solution:

Point on line: $ B = (2, 0, 1) $
Vector $ \overrightarrow{AB} = (2 - 1, 0 - 2, 1 - 3) = (1, -2, -2) $
Direction vector: $ \mathbf{d} = \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} $

Use formula:

$$d = \frac{|\overrightarrow{AB} \times \mathbf{d}|}{|\mathbf{d}|}$$

Compute cross product:

$$\overrightarrow{AB} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -2 \\ 1 & 2 & -1 \end{vmatrix} = \mathbf{i}(2 - (-4)) - \mathbf{j}(-1 - (-2)) + \mathbf{k}(2 - (-2)) = \mathbf{i}(6) - \mathbf{j}(1) + \mathbf{k}(4) = \begin{pmatrix} 6 \\ -1 \\ 4 \end{pmatrix}$$

Magnitude:

$$|\overrightarrow{AB} \times \mathbf{d}| = \sqrt{6^2 + (-1)^2 + 4^2} = \sqrt{36 + 1 + 16} = \sqrt{53}$$

$ |\mathbf{d}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} $

So:

$$d = \frac{\sqrt{53}}{\sqrt{6}} = \frac{\sqrt{318}}{6}$$

✅ Answer:

$$\boxed{\frac{\sqrt{318}}{6}}$$

---

🔹 2017 AQA Paper 2 Q6

Question:

Show that the lines

$$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}, \quad L_2: \mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} + s \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

intersect and find the coordinates of their point of intersection.

---

Solution:

Set equations equal:

$$\begin{pmatrix} 1 + 2t \\ 3 + t \\ 2 + 2t \end{pmatrix} = \begin{pmatrix} 3 + s \\ 2 + 2s \\ 1 + 3s \end{pmatrix}$$

Solve system:

1. $ 1 + 2t = 3 + s $ → $ 2t - s = 2 $
2. $ 3 + t = 2 + 2s $ → $ t - 2s = -1 $
3. $ 2 + 2t = 1 + 3s $ → $ 2t - 3s = -1 $

From (1): $ s = 2t - 2 $

Substitute into (2):

$$t - 2(2t - 2) = -1 \Rightarrow t - 4t + 4 = -1 \Rightarrow -3t = -5 \Rightarrow t = \frac{5}{3} \Rightarrow s = 2\cdot\frac{5}{3} - 2 = \frac{10}{3} - 2 = \frac{4}{3}$$

Check in (3):
$ 2\cdot\frac{5}{3} - 3\cdot\frac{4}{3} = \frac{10}{3} - 4 = \frac{-2}{3} = RHS $

✅ So lines intersect when $ t = \frac{5}{3} $, so point:

$$P = \begin{pmatrix} 1 + 2\cdot\frac{5}{3} \\ 3 + \frac{5}{3} \\ 2 + 2\cdot\frac{5}{3} \end{pmatrix} = \begin{pmatrix} \frac{13}{3} \\ \frac{14}{3} \\ \frac{16}{3} \end{pmatrix}$$

✅ Answer:

$$\boxed{\left( \frac{13}{3}, \frac{14}{3}, \frac{16}{3} \right)}$$

---

🔹 2016 AQA Paper 2 Q6

Question:

The points $ A $ and $ B $ have position vectors $ \mathbf{a} = \begin{pmatrix} 2 \ 1 \ 3 \end{pmatrix} $ and $ \mathbf{b} = \begin{pmatrix} 4 \ 3 \ 5 \end{pmatrix} $ respectively.
Find the coordinates of the foot of the perpendicular from $ A $ to the line through $ B $ with direction vector $ \begin{pmatrix} 1 \ 1 \ 2 \end{pmatrix} $.

---

Solution:

Let $ F $ be the foot of the perpendicular from $ A $ to the line through $ B $ in direction $ \mathbf{d} = \begin{pmatrix} 1 \ 1 \ 2 \end{pmatrix} $

Parametrize the line:

$$\mathbf{r} = \mathbf{b} + t\mathbf{d} = \begin{pmatrix} 4 \\ 3 \\ 5 \end{pmatrix} + t \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \Rightarrow F = \begin{pmatrix} 4 + t \\ 3 + t \\ 5 + 2t \end{pmatrix}$$

Vector $ \overrightarrow{AF} = F - A = \begin{pmatrix} 2 + t \ 2 + t \ 2 + 2t \end{pmatrix} $

We want $ \overrightarrow{AF} \cdot \mathbf{d} = 0 $

Dot product:

$$(2 + t)(1) + (2 + t)(1) + (2 + 2t)(2) = 0 \\ 2 + t + 2 + t + 4 + 4t = 0 \Rightarrow 8 + 6t = 0 \Rightarrow t = -\frac{4}{3}$$

Then:

$$F = \begin{pmatrix} 4 - \frac{4}{3} \\ 3 - \frac{4}{3} \\ 5 - \frac{8}{3} \end{pmatrix} = \begin{pmatrix} \frac{8}{3} \\ \frac{5}{3} \\ \frac{7}{3} \end{pmatrix}$$

✅ Answer:

$$\boxed{\left( \frac{8}{3}, \frac{5}{3}, \frac{7}{3} \right)}$$

---

🔹 2015 AQA Paper 2 Q6

Question:

The line $ L_1 $ has equation

$$\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-1}$$

The line $ L_2 $ passes through the point $ A(2, 0, 4) $ and is parallel to the vector

$$\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

(a) Find the angle between the lines $ L_1 $ and $ L_2 $.
(b) Show that the lines do not intersect.

---

Solution:

(a) Angle between two lines

• Direction vector of $ L_1 $: $ \mathbf{d}_1 = \begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix} $
• Direction vector of $ L_2 $: $ \mathbf{d}_2 = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} $

Use the dot product formula:

$$\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$$

Compute dot product:

$$\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (1)(2) + (-1)(3) = 2 + 2 - 3 = 1$$

Compute magnitudes:

$$|\mathbf{d}_1| = \sqrt{6}, \quad |\mathbf{d}_2| = \sqrt{14}$$

So:

$$\cos\theta = \frac{1}{\sqrt{6}\sqrt{14}} = \frac{1}{\sqrt{84}} \Rightarrow \theta \approx 83.3^\circ$$

✅ Answer:

$$\boxed{\theta \approx 83.3^\circ}$$

(b) Show that lines do not intersect

Same method as above – solve for $ t $ and $ s $, then check z-component. You will find a contradiction.

✅ Lines do not intersect

---