IB Mathematics: Analysis & Approaches HL · 鼎睿学苑

Unit C: Geometry单元 C：几何

The language of shape, angle, and direction. Master 3D solids, the full trigonometric toolkit (SL & HL), and the HL vectors topic — including lines, planes, dot/cross products, and their intersections.几何是关于形状、角度和方向的语言。本单元覆盖三维几何体、完整的三角学工具箱（SL 与 HL）以及 HL 向量（vector）专题——含直线、平面、数量积（点积）（dot product）与向量积（叉积）（cross product），以及它们的交集与夹角。

35 HL Teaching Hours35 小时 HL 教学 Papers 1 · 2 · 3 3 Sub-topics · 21 Concepts3 子单元 · 21 个核心概念
Topic C1.1 · Surface Areas & VolumesTopic C1.1 · 表面积与体积

Volume & Surface Area of 3D Solids三维几何体的体积与表面积 SL 3.1

Core Idea Every composite 3D figure in IB exams decomposes into a short list of named solids: prisms, cylinders, pyramids, cones, spheres, and hemispheres. You must know their formulas cold and be able to add/subtract them to handle combined solids.
核心思想 IB 考试中每个组合三维图形都能拆解为几种命名几何体：棱柱（prism）、圆柱（cylinder）、棱锥（pyramid）、圆锥（cone）、球（sphere）与半球（hemisphere）。要把它们的公式背得滚瓜烂熟，并能通过加减拼出组合体。
Essential 3D Formulas — Memorize必记的三维公式
$$ \text{Cuboid: } V = lwh \qquad \text{Cylinder: } V = \pi r^2 h,\; SA = 2\pi r h + 2\pi r^2 $$ $$ \text{Right Pyramid: } V = \tfrac{1}{3}(\text{base area}) \cdot h $$ $$ \text{Cone: } V = \tfrac{1}{3}\pi r^2 h, \qquad SA = \pi r \ell + \pi r^2 \text{ where } \ell = \sqrt{r^2 + h^2} $$ $$ \text{Sphere: } V = \tfrac{4}{3}\pi r^3, \qquad SA = 4\pi r^2 $$ $$ \text{Hemisphere: } V = \tfrac{2}{3}\pi r^3, \qquad SA = 3\pi r^2 \text{ (including flat disc)} $$
Strategy — Composite Solids Break the figure into named pieces. Volume is additive: $V_{\text{total}} = \sum V_i$. Surface area is NOT — you must exclude faces that are "glued" together. Sketch the net if unsure.
解题策略——组合几何体 把图形拆成命名几何体。体积可以直接相加：$V_{\text{total}} = \sum V_i$。表面积（surface area）则不能直接相加——必须把"粘"在一起的面去掉。不确定时画出展开图（net）。
Exam Tip For cones, the IB formula booklet gives the curved SA as $\pi r \ell$ where $\ell$ is the slant height, not the vertical height. If the problem gives $h$, use $\ell = \sqrt{r^2 + h^2}$ first.
考试 Tip IB formula booklet 中圆锥的侧面积公式是 $\pi r \ell$，其中 $\ell$ 是斜高（slant height），不是垂直高度。如果题目只给 $h$，要先用 $\ell = \sqrt{r^2 + h^2}$ 算出斜高。
Worked Example — Composite Solid (Paper 2)例题——组合几何体 (Paper 2)

Problem: A grain silo consists of a cylinder of radius $3$ m and height $8$ m with a hemispherical cap. Find (a) its total volume and (b) the total external surface area including the circular base.

(a) Volume:

$$\begin{aligned} V_{\text{cyl}} &= \pi(3)^2(8) = 72\pi \\ V_{\text{hemi}} &= \tfrac{2}{3}\pi(3)^3 = 18\pi \\ V_{\text{total}} &= 72\pi + 18\pi = 90\pi \approx 282.7 \text{ m}^3 \end{aligned}$$

(b) Surface area: The cylinder's top is replaced by the hemisphere's flat disc — those cancel. Include: curved cylinder, curved hemisphere, and bottom disc.

$$\begin{aligned} SA &= \underbrace{2\pi(3)(8)}_{\text{curved cyl}} + \underbrace{2\pi(3)^2}_{\text{curved hemi}} + \underbrace{\pi(3)^2}_{\text{bottom}} \\ &= 48\pi + 18\pi + 9\pi = 75\pi \approx 235.6 \text{ m}^2 \end{aligned}$$

题目：一个粮仓由半径 $3$ m、高 $8$ m 的圆柱和上方的半球形顶盖组成。求 (a) 总体积，(b) 含底部圆盘在内的外表面积。

(a) 体积：

$$\begin{aligned} V_{\text{cyl}} &= \pi(3)^2(8) = 72\pi \\ V_{\text{hemi}} &= \tfrac{2}{3}\pi(3)^3 = 18\pi \\ V_{\text{total}} &= 72\pi + 18\pi = 90\pi \approx 282.7 \text{ m}^3 \end{aligned}$$

(b) 表面积：圆柱的顶面被半球的底面圆盘"封住"，两者抵消。包含：圆柱侧面、半球曲面、底部圆盘。

$$\begin{aligned} SA &= \underbrace{2\pi(3)(8)}_{\text{圆柱侧面}} + \underbrace{2\pi(3)^2}_{\text{半球曲面}} + \underbrace{\pi(3)^2}_{\text{底面}} \\ &= 48\pi + 18\pi + 9\pi = 75\pi \approx 235.6 \text{ m}^2 \end{aligned}$$
A solid cone has radius $5$ cm and slant height $13$ cm. Its curved surface area is:一个实心圆锥的半径为 $5$ cm，斜高为 $13$ cm。它的侧面积是：
C1.1
$25\pi$ cm²
$60\pi$ cm²
$65\pi$ cm²
$90\pi$ cm²
Correct! Curved SA of a cone $= \pi r \ell = \pi(5)(13) = 65\pi$ cm².
正确！圆锥侧面积 $= \pi r \ell = \pi(5)(13) = 65\pi$ cm²。
Curved SA $= \pi r \ell$. Here $r = 5$ and $\ell = 13$, so $SA = \pi(5)(13) = 65\pi$ cm².
侧面积 $= \pi r \ell$。代入 $r = 5$、$\ell = 13$，得 $SA = \pi(5)(13) = 65\pi$ cm²。
Topic C1.2 · Angles in 3DTopic C1.2 · 三维角度

Angles Between Lines & Planes直线与平面之间的夹角 SL 3.2

Two Key Angles 1. Between two intersecting lines: locate the shared vertex and form a triangle — use right-triangle trig or the cosine rule.
2. Between a line and a plane: the angle between the line and its projection onto the plane. Always found via a right triangle between the line, its foot, and the perpendicular.
两类关键夹角 1. 两条相交直线之间（angle between two intersecting lines）：找到公共顶点构造三角形，再用直角三角学或余弦定理（cosine rule）。
2. 直线与平面之间：定义为该直线与它在平面上的投影之间的角。永远通过"直线—垂足—垂线"构成的直角三角形求解。
Space Diagonal of a Cuboid长方体的体对角线
$$ d = \sqrt{l^2 + w^2 + h^2} $$
Strategy — Solving 3D Problems (1) Redraw the relevant 2D triangle in isolation. (2) Label known lengths; mark the right angle. (3) Solve with SOHCAHTOA or Pythagoras. (4) Always state the angle to 1 decimal place in degrees (or 3 s.f.) — IB standard.
解题策略——三维问题 (1) 把相关的二维三角形单独画出来。 (2) 标出已知长度，标记直角。 (3) 用 SOHCAHTOA 或三维毕达哥拉斯（3D Pythagoras）求解。 (4) 角度统一保留到小数点后 1 位（度）或 3 位有效数字——IB 标准格式。
Worked Example — Angle Between Diagonal and Base例题——体对角线与底面的夹角

Problem: A rectangular box has dimensions $6 \times 4 \times 3$ cm. Find the angle between the space diagonal and the base.

The base diagonal $b = \sqrt{6^2 + 4^2} = \sqrt{52}$.

The space diagonal, the height, and the base diagonal form a right triangle with the right angle at the base. So:

$$\tan \theta = \frac{\text{height}}{\text{base diagonal}} = \frac{3}{\sqrt{52}}$$ $$\theta = \arctan\left(\frac{3}{\sqrt{52}}\right) \approx 22.6°$$

题目：一个长方体的尺寸为 $6 \times 4 \times 3$ cm。求体对角线与底面之间的夹角。

底面对角线 $b = \sqrt{6^2 + 4^2} = \sqrt{52}$。

体对角线、高、底面对角线构成一个直角三角形，直角位于底面。于是：

$$\tan \theta = \frac{\text{高}}{\text{底面对角线}} = \frac{3}{\sqrt{52}}$$ $$\theta = \arctan\left(\frac{3}{\sqrt{52}}\right) \approx 22.6°$$
In a cube of side $a$, the space diagonal has length:边长为 $a$ 的立方体，其体对角线长度为：
C1.2
$a\sqrt{2}$
$a\sqrt{3}$
$2a$
$3a$
Correct! $d = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$.
正确！$d = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$。
Use $d = \sqrt{l^2 + w^2 + h^2}$ with $l = w = h = a$: $d = \sqrt{3a^2} = a\sqrt{3}$.
代入 $d = \sqrt{l^2 + w^2 + h^2}$，$l = w = h = a$：$d = \sqrt{3a^2} = a\sqrt{3}$。
Topic C1.3 · The CircleTopic C1.3 · 圆

Radian Measure弧度制 SL 3.4

Definition One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. Radians are dimensionless — a ratio of two lengths.
定义 一弧度（radian）是圆心角所对的弧长等于半径时的角度。弧度无量纲——它就是两个长度的比。
Degree ↔ Radian Conversion度 ↔ 弧度换算
$$ \pi \text{ rad} = 180° \qquad \Longrightarrow \qquad \text{deg} = \text{rad} \cdot \frac{180}{\pi}, \quad \text{rad} = \text{deg} \cdot \frac{\pi}{180} $$
Must-Know Exact Values (in radians)
Deg0°30°45°60°90°180°270°360°
Rad$0$$\tfrac{\pi}{6}$$\tfrac{\pi}{4}$$\tfrac{\pi}{3}$$\tfrac{\pi}{2}$$\pi$$\tfrac{3\pi}{2}$$2\pi$
必背精确值（弧度）
度0°30°45°60°90°180°270°360°
弧度$0$$\tfrac{\pi}{6}$$\tfrac{\pi}{4}$$\tfrac{\pi}{3}$$\tfrac{\pi}{2}$$\pi$$\tfrac{3\pi}{2}$$2\pi$
Common Trap All HL calculus (derivatives of $\sin x$, $\cos x$, etc.) assumes $x$ is in radians. Check your calculator mode on Paper 2/3 — "degree mode" is the single biggest source of silent errors in this unit.
常见陷阱 所有 HL 微积分内容（如 $\sin x$、$\cos x$ 的求导）都默认 $x$ 用弧度。Paper 2/3 务必检查计算器模式——"度模式"是本单元最大的无声错误来源。
Topic C1.4 · Arcs & SectorsTopic C1.4 · 弧与扇形

Arc Length & Sector Area弧长与扇形面积 SL 3.4

Arc & Sector (θ in radians)弧长与扇形（θ 用弧度）
$$ s = r\theta \qquad A_{\text{sector}} = \tfrac{1}{2}r^2\theta $$
Segment Area A segment is the region between a chord and the arc. Its area equals sector area minus the triangle formed by the two radii and the chord: $$ A_{\text{segment}} = \tfrac{1}{2}r^2(\theta - \sin\theta) $$
弓形面积 弓形（segment）是弦与弧之间的区域。其面积等于扇形面积（sector area）减去由两条半径与弦围成的三角形面积： $$ A_{\text{segment}} = \tfrac{1}{2}r^2(\theta - \sin\theta) $$
Worked Example — Segment Area例题——弓形面积

Problem: A circle has radius $10$ cm. Two radii meet at an angle of $\tfrac{2\pi}{3}$ rad. Find (a) the arc length, (b) the sector area, and (c) the minor segment area.

(a) Arc: $s = r\theta = 10 \cdot \tfrac{2\pi}{3} = \tfrac{20\pi}{3} \approx 20.9$ cm.

(b) Sector: $A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(100)\cdot\tfrac{2\pi}{3} = \tfrac{100\pi}{3} \approx 104.7$ cm².

(c) Segment:

$$A = \tfrac{1}{2}r^2(\theta - \sin\theta) = \tfrac{1}{2}(100)\left(\tfrac{2\pi}{3} - \sin\tfrac{2\pi}{3}\right) = 50\left(\tfrac{2\pi}{3} - \tfrac{\sqrt{3}}{2}\right) \approx 61.4 \text{ cm}^2$$

题目：圆的半径为 $10$ cm，两条半径之间的夹角为 $\tfrac{2\pi}{3}$ 弧度。求 (a) 弧长（arc length），(b) 扇形面积，(c) 劣弓形的面积。

(a) 弧长：$s = r\theta = 10 \cdot \tfrac{2\pi}{3} = \tfrac{20\pi}{3} \approx 20.9$ cm。

(b) 扇形：$A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(100)\cdot\tfrac{2\pi}{3} = \tfrac{100\pi}{3} \approx 104.7$ cm²。

(c) 弓形：

$$A = \tfrac{1}{2}r^2(\theta - \sin\theta) = \tfrac{1}{2}(100)\left(\tfrac{2\pi}{3} - \sin\tfrac{2\pi}{3}\right) = 50\left(\tfrac{2\pi}{3} - \tfrac{\sqrt{3}}{2}\right) \approx 61.4 \text{ cm}^2$$
A sector has radius $6$ and subtends an angle of $\tfrac{\pi}{4}$ rad. Its area is:扇形的半径为 $6$，圆心角为 $\tfrac{\pi}{4}$ 弧度。其面积为：
C1.4
$\tfrac{9\pi}{2}$
$9\pi$
$\tfrac{3\pi}{2}$
$\tfrac{\pi}{4}$
Correct! $A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(36)(\tfrac{\pi}{4}) = \tfrac{9\pi}{2}$.
正确！$A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(36)(\tfrac{\pi}{4}) = \tfrac{9\pi}{2}$。
Apply $A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(6)^2(\tfrac{\pi}{4}) = \tfrac{9\pi}{2}$. Always check: is $\theta$ in radians?
套公式 $A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(6)^2(\tfrac{\pi}{4}) = \tfrac{9\pi}{2}$。先检查：$\theta$ 是不是用弧度？
Topic C2.1 · Right-Triangle TrigTopic C2.1 · 直角三角形三角学

Sine, Cosine, Tangent in Right Triangles直角三角形中的正弦、余弦、正切 SL 3.3

SOHCAHTOA In any right triangle, for an acute angle $\theta$: $$\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}}$$
SOHCAHTOA 在任何直角三角形中，对锐角 $\theta$，正弦（sine）、余弦（cosine）、正切（tangent）定义为： $$\sin\theta = \frac{\text{对边}}{\text{斜边}}, \quad \cos\theta = \frac{\text{邻边}}{\text{斜边}}, \quad \tan\theta = \frac{\text{对边}}{\text{邻边}}$$
Pythagoras' Theorem毕达哥拉斯定理
$$ a^2 + b^2 = c^2 \quad (c \text{ is the hypotenuse}) $$
When to Use Right-Triangle Trig Use SOHCAHTOA whenever a triangle is — or can be split into — right-angled pieces. If no right angle is present, default to sine/cosine rule (C2.2).
何时使用直角三角学 当三角形本身是直角三角形，或可以拆分为若干个直角三角形时，使用 SOHCAHTOA。如果题目中没有直角，则改用正弦定理（sine rule）/ 余弦定理（C2.2）。
Topic C2.2 · Non-Right TrianglesTopic C2.2 · 非直角三角形

Sine Rule, Cosine Rule & Triangle Area正弦定理、余弦定理与三角形面积 SL 3.3

Sine Rule正弦定理
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$
Cosine Rule余弦定理
$$ c^2 = a^2 + b^2 - 2ab\cos C \qquad \Longleftrightarrow \qquad \cos C = \frac{a^2 + b^2 - c^2}{2ab} $$
Area of a Triangle三角形面积公式
$$ A = \tfrac{1}{2} ab \sin C $$
Decision Tree Sine rule: you know an angle and its opposite side (and one more piece).
Cosine rule (for a side): you know two sides and the included angle (SAS).
Cosine rule (for an angle): you know all three sides (SSS).
Area: two sides and their included angle.
选择哪条定理 正弦定理（sine rule）：已知一个角和它的对边（再加一条信息）。
余弦定理（cosine rule）求边：已知两边与它们的夹角（SAS）。
余弦定理求角：已知三条边（SSS）。
三角形面积公式（area of triangle）：已知两边及其夹角。
The Ambiguous Case (SSA) When using the sine rule to find an angle, if you're given SSA, there may be two valid triangles. After computing $\arcsin(x) = \alpha$, also consider $180° - \alpha$. Test both against the triangle sum ($< 180°$). IB frequently awards marks for identifying both cases.
SSA 模糊情形 用正弦定理求角时，如果给的是 SSA（两边一对角），可能存在两个合法三角形。算出 $\arcsin(x) = \alpha$ 后，还要考虑 $180° - \alpha$；把两个候选分别检验三角形内角和（$< 180°$）。IB 经常给同时识别出两种情况的考生加分。
Worked Example — Cosine Rule (SAS)例题——余弦定理 (SAS)

Problem: In triangle $ABC$, $a = 7$, $b = 10$, and $C = 42°$. Find side $c$ and the area.

$$\begin{aligned} c^2 &= 7^2 + 10^2 - 2(7)(10)\cos 42° \\ &= 49 + 100 - 140\cos 42° \\ &\approx 149 - 104.03 = 44.97 \\ c &\approx 6.71 \end{aligned}$$

Area: $A = \tfrac{1}{2}(7)(10)\sin 42° \approx 23.4$.

题目：在三角形 $ABC$ 中，$a = 7$、$b = 10$、$C = 42°$。求边 $c$ 与三角形面积。

$$\begin{aligned} c^2 &= 7^2 + 10^2 - 2(7)(10)\cos 42° \\ &= 49 + 100 - 140\cos 42° \\ &\approx 149 - 104.03 = 44.97 \\ c &\approx 6.71 \end{aligned}$$

面积：$A = \tfrac{1}{2}(7)(10)\sin 42° \approx 23.4$。

In $\triangle ABC$, $a = 8$, $b = 5$, and $C = 60°$. The area is:$\triangle ABC$ 中，$a = 8$、$b = 5$、$C = 60°$。三角形面积为：
C2.2
$10$
$20$
$5\sqrt{3}$
$10\sqrt{3}$
Correct! $A = \tfrac{1}{2}(8)(5)\sin 60° = 20 \cdot \tfrac{\sqrt{3}}{2} = 10\sqrt{3}$.
正确！$A = \tfrac{1}{2}(8)(5)\sin 60° = 20 \cdot \tfrac{\sqrt{3}}{2} = 10\sqrt{3}$。
Use $A = \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(8)(5)\sin 60° = 20 \cdot \tfrac{\sqrt{3}}{2} = 10\sqrt{3}$.
用 $A = \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(8)(5)\sin 60° = 20 \cdot \tfrac{\sqrt{3}}{2} = 10\sqrt{3}$。
Topic C2.3 · ApplicationsTopic C2.3 · 应用

Bearings & Angles of Elevation/Depression方位角与仰角/俯角 SL 3.3

Bearings A bearing is measured clockwise from North and written as a 3-digit number (e.g. $045°$, $120°$, $330°$). Always draw a North line at each point before reading or computing a bearing.
方位角 方位角（bearing）从正北方向顺时针量出，写成三位数（例如 $045°$、$120°$、$330°$）。读或算方位角之前，永远先在每个观测点画出一条北向参考线。
Elevation & Depression The angle of elevation is measured upward from the horizontal; the angle of depression is measured downward. They are always acute, and by alternate angles they equal each other between two points looking at each other.
仰角与俯角 仰角（angle of elevation）从水平向上量；俯角（angle of depression）从水平向下量。它们都是锐角；两个互相对望的观测点，仰角与俯角依据内错角相等。
Exam Tip Always construct a labelled diagram from the word problem before calculating. Mark all given angles, lengths, and the North line. IB awards marks for a correct diagram alone on many 6+ mark problems.
考试 Tip 遇到文字题先画一张带标注的图再动笔。标出所有已知角度、长度和正北线。在 6 分以上的题目中，IB 经常仅凭一张正确的草图就给分。
Topic C2.4 · The Unit CircleTopic C2.4 · 单位圆

Unit Circle Definitions & Exact Values单位圆定义与精确值 SL 3.5

Unit Circle Definition For a point $P$ on the unit circle ($x^2 + y^2 = 1$) at angle $\theta$ measured counter-clockwise from the positive $x$-axis: $$ \cos\theta = x, \qquad \sin\theta = y, \qquad \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{y}{x} \;(x \ne 0) $$
单位圆定义 设点 $P$ 在单位圆（unit circle）$x^2 + y^2 = 1$ 上，$\theta$ 从正 $x$ 轴逆时针量起。则： $$ \cos\theta = x, \qquad \sin\theta = y, \qquad \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{y}{x} \;(x \ne 0) $$
Exact Values — Memorize必背精确值
$\theta$$0$$\tfrac{\pi}{6}$$\tfrac{\pi}{4}$$\tfrac{\pi}{3}$$\tfrac{\pi}{2}$
$\sin\theta$$0$$\tfrac{1}{2}$$\tfrac{\sqrt{2}}{2}$$\tfrac{\sqrt{3}}{2}$$1$
$\cos\theta$$1$$\tfrac{\sqrt{3}}{2}$$\tfrac{\sqrt{2}}{2}$$\tfrac{1}{2}$$0$
$\tan\theta$$0$$\tfrac{\sqrt{3}}{3}$$1$$\sqrt{3}$undef.
Signs by Quadrant — "All Students Take Calculus" Q1 ($0 \to \tfrac{\pi}{2}$): all positive. Q2 ($\tfrac{\pi}{2} \to \pi$): sine positive. Q3 ($\pi \to \tfrac{3\pi}{2}$): tangent positive. Q4 ($\tfrac{3\pi}{2} \to 2\pi$): cosine positive.
各象限的符号——ASTC 口诀 口诀 "All Students Take Calculus"（全、正、切、余）：Q1（$0 \to \tfrac{\pi}{2}$）全正。Q2（$\tfrac{\pi}{2} \to \pi$）正弦为正。Q3（$\pi \to \tfrac{3\pi}{2}$）正切为正。Q4（$\tfrac{3\pi}{2} \to 2\pi$）余弦为正。
Topic C2.5 · IdentitiesTopic C2.5 · 恒等式

Pythagorean & Double-Angle Identities平方和恒等式与二倍角公式 SL 3.6

Pythagorean Identity (Fundamental)平方和恒等式（基本式）
$$ \sin^2\theta + \cos^2\theta = 1 $$
Double-Angle Formulas二倍角公式
$$ \sin 2\theta = 2\sin\theta\cos\theta $$ $$ \cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta $$ $$ \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} \quad (\text{HL}) $$
When to Use Which Form of $\cos 2\theta$ The three forms differ only algebraically — pick the one that creates a common variable with the rest of your expression. If the rest uses $\sin$, choose $1 - 2\sin^2\theta$; if $\cos$, choose $2\cos^2\theta - 1$.
$\cos 2\theta$ 三种形式怎么选 三个形式只是代数上不同——选与表达式其余部分共用一个变量的那一个。若其余部分用 $\sin$，选 $1 - 2\sin^2\theta$；若用 $\cos$，选 $2\cos^2\theta - 1$。这就是二倍角（double-angle）的实战选择策略。
Top Identity Trap $\sin(A + B) \ne \sin A + \sin B$ and $\sin 2\theta \ne 2\sin\theta$. These must be expanded using the actual identity — a non-negotiable Paper 1 expectation.
头号恒等式陷阱 $\sin(A + B) \ne \sin A + \sin B$，$\sin 2\theta \ne 2\sin\theta$。必须用真正的恒等式展开——Paper 1 不容妥协的要求。
Worked Example — Proving an Identity例题——证明恒等式

Prove: $\dfrac{1 - \cos 2\theta}{\sin 2\theta} = \tan\theta$

Start from LHS and use double-angle forms that let $\sin\theta$ and $\cos\theta$ cancel:

$$\begin{aligned} \text{LHS} &= \frac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta} \\[4pt] &= \frac{2\sin^2\theta}{2\sin\theta\cos\theta} \\[4pt] &= \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS} \quad \blacksquare \end{aligned}$$

求证：$\dfrac{1 - \cos 2\theta}{\sin 2\theta} = \tan\theta$

从 LHS 出发，选择能让 $\sin\theta$ 与 $\cos\theta$ 相消的二倍角形式：

$$\begin{aligned} \text{LHS} &= \frac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta} \\[4pt] &= \frac{2\sin^2\theta}{2\sin\theta\cos\theta} \\[4pt] &= \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS} \quad \blacksquare \end{aligned}$$
If $\sin\theta = \tfrac{3}{5}$ and $\theta$ is in Q1, then $\cos 2\theta$ equals:若 $\sin\theta = \tfrac{3}{5}$ 且 $\theta$ 在第一象限，则 $\cos 2\theta = $
C2.5
$\tfrac{24}{25}$
$-\tfrac{24}{25}$
$\tfrac{7}{25}$
$-\tfrac{7}{25}$
Correct! $\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2(\tfrac{9}{25}) = 1 - \tfrac{18}{25} = \tfrac{7}{25}$.
正确！$\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2(\tfrac{9}{25}) = 1 - \tfrac{18}{25} = \tfrac{7}{25}$。
Use $\cos 2\theta = 1 - 2\sin^2\theta$. With $\sin\theta = \tfrac{3}{5}$: $1 - 2(\tfrac{9}{25}) = \tfrac{7}{25}$.
套 $\cos 2\theta = 1 - 2\sin^2\theta$。代入 $\sin\theta = \tfrac{3}{5}$：$1 - 2(\tfrac{9}{25}) = \tfrac{7}{25}$。
Topic C2.6 · Circular FunctionsTopic C2.6 · 圆函数

Graphs of Sin, Cos & Tan — Transformations正弦、余弦、正切函数的图像与变换 SL 3.7

Core Model Any sinusoidal curve can be written as $$ f(x) = a\sin(b(x - c)) + d \qquad \text{or} \qquad f(x) = a\cos(b(x-c)) + d $$ where $|a|$ is the amplitude, $\tfrac{2\pi}{|b|}$ is the period, $c$ is the horizontal shift, and $d$ is the vertical shift (principal axis $y = d$).
核心模型 任何正弦/余弦型曲线都可以写成 $$ f(x) = a\sin(b(x - c)) + d \qquad \text{或} \qquad f(x) = a\cos(b(x-c)) + d $$ 其中 $|a|$ 是振幅（amplitude），$\tfrac{2\pi}{|b|}$ 是周期（period），$c$ 是水平平移（也叫phase shift，相位），$d$ 是垂直平移（中轴线 $y = d$）。
Key Features关键特征
$$ \text{Max} = d + |a|, \quad \text{Min} = d - |a|, \quad \text{Period} = \frac{2\pi}{|b|} $$ $$ \text{For } \tan x: \text{ period } = \pi, \text{ VAs at } x = \tfrac{\pi}{2} + k\pi $$
Modelling Real-World Periodic Behaviour Height of a Ferris-wheel rider, tide depth, daylight hours, pendulum motion — all fit $y = a\sin(b(t - c)) + d$. Determine $a, d$ from max/min; $b$ from the period; $c$ from a known feature (e.g. time of max).
建模现实中的周期性现象 摩天轮乘客的高度、潮汐深度、日照时长、单摆运动——都符合 $y = a\sin(b(t - c)) + d$ 的形状。$a$ 与 $d$ 由最大/最小值确定；$b$ 由周期确定；$c$ 由某个已知特征（如取最大值的时刻）确定。
Worked Example — Fitting a Sinusoidal Model例题——拟合正弦型模型

Problem: The depth of water in a harbour varies between $2$ m (at low tide, $t = 4$ h) and $12$ m (at high tide). One full cycle takes $12$ hours. Model the depth $D(t)$.

Centre line: $d = \tfrac{12+2}{2} = 7$. Amplitude: $a = \tfrac{12-2}{2} = 5$.

Period $T = 12 \Rightarrow b = \tfrac{2\pi}{12} = \tfrac{\pi}{6}$.

Low tide (minimum) of a negative-cosine model is at $t = c$, so shift $c = 4$:

$$D(t) = -5\cos\!\left(\tfrac{\pi}{6}(t - 4)\right) + 7$$

题目：某港口水深在低潮 $2$ m（发生在 $t = 4$ h）与高潮 $12$ m 之间变化，一个完整周期为 $12$ 小时。建立水深 $D(t)$ 的模型。

中轴线：$d = \tfrac{12+2}{2} = 7$。振幅：$a = \tfrac{12-2}{2} = 5$。

周期 $T = 12 \Rightarrow b = \tfrac{2\pi}{12} = \tfrac{\pi}{6}$。

负余弦模型的最小值在 $t = c$ 处，故 $c = 4$：

$$D(t) = -5\cos\!\left(\tfrac{\pi}{6}(t - 4)\right) + 7$$
The function $f(x) = 3\sin(2x) + 4$ has amplitude and period:函数 $f(x) = 3\sin(2x) + 4$ 的振幅与周期为：
C2.6
amp $3$, period $2\pi$振幅 $3$，周期 $2\pi$
amp $3$, period $\pi$振幅 $3$，周期 $\pi$
amp $4$, period $\pi$振幅 $4$，周期 $\pi$
amp $6$, period $\tfrac{\pi}{2}$振幅 $6$，周期 $\tfrac{\pi}{2}$
Correct! Amplitude $= |a| = 3$; period $= \tfrac{2\pi}{|b|} = \tfrac{2\pi}{2} = \pi$. The $+4$ is a vertical shift, not amplitude.
正确！振幅 $= |a| = 3$；周期 $= \tfrac{2\pi}{|b|} = \tfrac{2\pi}{2} = \pi$。"$+4$" 是垂直平移，不是振幅。
Amplitude is $|a| = 3$. Period is $\tfrac{2\pi}{|b|} = \tfrac{2\pi}{2} = \pi$. The $+4$ shifts vertically but doesn't change amplitude.
振幅是 $|a| = 3$。周期是 $\tfrac{2\pi}{|b|} = \tfrac{2\pi}{2} = \pi$。"$+4$" 把图像整体上移，不改变振幅。
Topic C2.7 · Trig EquationsTopic C2.7 · 三角方程

Solving Trigonometric Equations解三角方程 SL 3.8

General Strategy 1. Isolate the trig function.
2. Find the principal value with $\arcsin$, $\arccos$, or $\arctan$.
3. Use the unit circle (or symmetry of the graph) to find all solutions in the stated interval.
4. If the argument is $bx + c$, solve over the scaled interval first, then back-substitute.
通用解题步骤 1. 把三角函数孤立到等号一侧。
2. 用 $\arcsin$、$\arccos$ 或 $\arctan$ 求主值。
3. 利用单位圆（或图像的对称性）找出所给区间内的全部解。
4. 如果三角函数的参数是 $bx + c$，先在缩放后的区间内求解，再回代到 $x$。
Solution Families (over all $\mathbb{R}$)通解族（在 $\mathbb{R}$ 上）
$$ \sin\theta = k: \quad \theta = \arcsin k + 2k\pi \;\text{ or }\; \theta = \pi - \arcsin k + 2k\pi $$ $$ \cos\theta = k: \quad \theta = \pm\arccos k + 2k\pi $$ $$ \tan\theta = k: \quad \theta = \arctan k + k\pi $$
Quadratic-in-Trig Equations Equations like $2\sin^2\theta - \sin\theta - 1 = 0$ are quadratics in $\sin\theta$. Let $u = \sin\theta$: $2u^2 - u - 1 = 0 \Rightarrow u = 1$ or $u = -\tfrac{1}{2}$. Then solve each separately. Check that each $u$-value is in $[-1, 1]$ — if not, discard.
三角函数的"二次型"方程 形如 $2\sin^2\theta - \sin\theta - 1 = 0$ 的方程本质是关于 $\sin\theta$ 的二次方程。令 $u = \sin\theta$：$2u^2 - u - 1 = 0 \Rightarrow u = 1$ 或 $u = -\tfrac{1}{2}$，再分别求解。务必检查 $u$ 是否落在 $[-1, 1]$ 内——超出范围的值要舍去。
Worked Example — All Solutions in $[0, 2\pi]$例题——求 $[0, 2\pi]$ 上的全部解

Solve: $2\cos(2x) = 1$ for $x \in [0, 2\pi]$.

Step 1: $\cos(2x) = \tfrac{1}{2}$.

Step 2: Let $u = 2x$. Since $x \in [0, 2\pi]$, we need $u \in [0, 4\pi]$.

Step 3: In $[0, 2\pi]$, $\cos u = \tfrac{1}{2}$ at $u = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$. Adding $2\pi$ for the $[2\pi, 4\pi]$ copy:

$$u = \tfrac{\pi}{3}, \tfrac{5\pi}{3}, \tfrac{7\pi}{3}, \tfrac{11\pi}{3}$$

Step 4: $x = \tfrac{u}{2} = \tfrac{\pi}{6}, \tfrac{5\pi}{6}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$. Four solutions.

解方程：$2\cos(2x) = 1$，$x \in [0, 2\pi]$。

第 1 步：$\cos(2x) = \tfrac{1}{2}$。

第 2 步：令 $u = 2x$。因 $x \in [0, 2\pi]$，故 $u \in [0, 4\pi]$。

第 3 步：在 $[0, 2\pi]$ 内 $\cos u = \tfrac{1}{2}$ 的解为 $u = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$。加上 $2\pi$ 得到 $[2\pi, 4\pi]$ 的另一组：

$$u = \tfrac{\pi}{3}, \tfrac{5\pi}{3}, \tfrac{7\pi}{3}, \tfrac{11\pi}{3}$$

第 4 步：$x = \tfrac{u}{2} = \tfrac{\pi}{6}, \tfrac{5\pi}{6}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$。共 4 个解。

How many solutions does $\sin(2x) = \tfrac{1}{2}$ have on $[0, 2\pi]$?$\sin(2x) = \tfrac{1}{2}$ 在 $[0, 2\pi]$ 上有几个解？
C2.7
$2$
$3$
$4$
$6$
Correct! Let $u = 2x \in [0, 4\pi]$. $\sin u = \tfrac{1}{2}$ gives $2$ solutions per $2\pi$-cycle, so $4$ total over $[0, 4\pi]$.
正确！令 $u = 2x \in [0, 4\pi]$。$\sin u = \tfrac{1}{2}$ 每个 $2\pi$ 周期内有 $2$ 个解，共 $4$ 个。
With $b = 2$ the argument runs through $[0, 4\pi]$ (two periods). Each period has $2$ solutions of $\sin u = \tfrac{1}{2}$, giving $4$.
$b = 2$ 时，参数取值范围为 $[0, 4\pi]$（两个完整周期）。每个周期内 $\sin u = \tfrac{1}{2}$ 有 $2$ 个解，合计 $4$ 个。
Topic C2.8 · HL OnlyTopic C2.8 · 仅 HL

Reciprocal & Inverse Trigonometric Functions倒数三角函数与反三角函数 AHL 3.9

Reciprocal Ratios倒数三角函数
$$ \sec\theta = \frac{1}{\cos\theta}, \quad \csc\theta = \frac{1}{\sin\theta}, \quad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta} $$
Extended Pythagorean Identities推广的平方和恒等式
$$ 1 + \tan^2\theta = \sec^2\theta $$ $$ 1 + \cot^2\theta = \csc^2\theta $$
Inverse Trig Functions — Principal Ranges Because $\sin$, $\cos$, $\tan$ are not one-to-one, inverses need restricted domains: $$ \arcsin x : [-1,1] \to [-\tfrac{\pi}{2}, \tfrac{\pi}{2}] $$ $$ \arccos x : [-1,1] \to [0, \pi] $$ $$ \arctan x : \mathbb{R} \to (-\tfrac{\pi}{2}, \tfrac{\pi}{2}) $$
反三角函数——主值范围 因为 $\sin$、$\cos$、$\tan$ 不是一一对应（one-to-one），反函数需要限定定义域： $$ \arcsin x : [-1,1] \to [-\tfrac{\pi}{2}, \tfrac{\pi}{2}] $$ $$ \arccos x : [-1,1] \to [0, \pi] $$ $$ \arctan x : \mathbb{R} \to (-\tfrac{\pi}{2}, \tfrac{\pi}{2}) $$
Common HL Trap $\arcsin(\sin x) \ne x$ in general — it equals $x$ only when $x$ is inside the principal range. Example: $\arcsin(\sin(\tfrac{3\pi}{4})) = \arcsin(\tfrac{\sqrt{2}}{2}) = \tfrac{\pi}{4}$, not $\tfrac{3\pi}{4}$.
HL 常见陷阱 一般情况下 $\arcsin(\sin x) \ne x$——只有当 $x$ 落在主值范围内时才相等。例：$\arcsin(\sin(\tfrac{3\pi}{4})) = \arcsin(\tfrac{\sqrt{2}}{2}) = \tfrac{\pi}{4}$，不是 $\tfrac{3\pi}{4}$。
Worked Example — Using Extended Pythagorean ID例题——使用推广的平方和恒等式

Solve: $\sec^2 x - 3\tan x - 1 = 0$ for $x \in [0, 2\pi]$.

Replace $\sec^2 x$ with $1 + \tan^2 x$ to get a quadratic in $\tan x$:

$$\begin{aligned} (1 + \tan^2 x) - 3\tan x - 1 &= 0 \\ \tan^2 x - 3\tan x &= 0 \\ \tan x (\tan x - 3) &= 0 \end{aligned}$$

Case 1: $\tan x = 0 \Rightarrow x = 0, \pi, 2\pi$.

Case 2: $\tan x = 3 \Rightarrow x = \arctan(3) \approx 1.249$ or $x \approx 1.249 + \pi \approx 4.391$.

解方程：$\sec^2 x - 3\tan x - 1 = 0$，$x \in [0, 2\pi]$。

把 $\sec^2 x$ 替换为 $1 + \tan^2 x$，化为关于 $\tan x$ 的二次方程：

$$\begin{aligned} (1 + \tan^2 x) - 3\tan x - 1 &= 0 \\ \tan^2 x - 3\tan x &= 0 \\ \tan x (\tan x - 3) &= 0 \end{aligned}$$

情形 1：$\tan x = 0 \Rightarrow x = 0, \pi, 2\pi$。

情形 2：$\tan x = 3 \Rightarrow x = \arctan(3) \approx 1.249$ 或 $x \approx 1.249 + \pi \approx 4.391$。

Topic C2.9 · HL OnlyTopic C2.9 · 仅 HL

Compound Angle Identities复合角恒等式 AHL 3.10

Compound Angle Formulas复合角公式
$$ \sin(A \pm B) = \sin A\cos B \pm \cos A\sin B $$ $$ \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B $$ $$ \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B} $$
Mnemonic Sine expands as "same-cross-same" (sin cos + cos sin). Cosine expands as "cos-cos minus sin-sin" — the sign flips between the compound and the expansion. Double-angle formulas come from setting $A = B$ here.
口诀 复合角（compound-angle）公式：正弦"sin cos + cos sin"（同名乘异名）；余弦"cos cos − sin sin"——左右两边的符号正好相反。把 $A = B$ 代入这两条就得到二倍角公式。
Exam Strategy — Exact Values at Unusual Angles To find $\sin 15°$ or $\cos\tfrac{5\pi}{12}$ in exact form, express the angle as a sum/difference of familiar angles (e.g. $15° = 45° - 30°$) and apply a compound formula.
考试策略——不常见角度的精确值 求 $\sin 15°$ 或 $\cos\tfrac{5\pi}{12}$ 等的精确形式时，把角拆成熟悉角的和或差（例如 $15° = 45° - 30°$），再用复合角公式展开。
Worked Example — Exact $\sin 75°$例题——精确求 $\sin 75°$

Find: $\sin 75°$ in exact form.

Write $75° = 45° + 30°$:

$$\begin{aligned} \sin 75° &= \sin(45° + 30°) \\ &= \sin 45°\cos 30° + \cos 45°\sin 30° \\ &= \tfrac{\sqrt{2}}{2}\cdot\tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{2}}{2}\cdot\tfrac{1}{2} \\ &= \tfrac{\sqrt{6}}{4} + \tfrac{\sqrt{2}}{4} = \tfrac{\sqrt{6}+\sqrt{2}}{4} \end{aligned}$$

求：$\sin 75°$ 的精确值。

将 $75° = 45° + 30°$：

$$\begin{aligned} \sin 75° &= \sin(45° + 30°) \\ &= \sin 45°\cos 30° + \cos 45°\sin 30° \\ &= \tfrac{\sqrt{2}}{2}\cdot\tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{2}}{2}\cdot\tfrac{1}{2} \\ &= \tfrac{\sqrt{6}}{4} + \tfrac{\sqrt{2}}{4} = \tfrac{\sqrt{6}+\sqrt{2}}{4} \end{aligned}$$
The exact value of $\cos(A + B)$ when $\sin A = \tfrac{3}{5}$, $\cos A > 0$, $\cos B = \tfrac{5}{13}$, $\sin B > 0$ is:已知 $\sin A = \tfrac{3}{5}$，$\cos A > 0$，$\cos B = \tfrac{5}{13}$，$\sin B > 0$，则 $\cos(A + B)$ 的精确值为：
C2.9
$\tfrac{-16}{65}$
$\tfrac{56}{65}$
$\tfrac{16}{65}$
$\tfrac{33}{65}$
Correct! $\cos A = \tfrac{4}{5}$, $\sin B = \tfrac{12}{13}$. $\cos(A+B) = \cos A\cos B - \sin A\sin B = \tfrac{4}{5}\cdot\tfrac{5}{13} - \tfrac{3}{5}\cdot\tfrac{12}{13} = \tfrac{20-36}{65} = -\tfrac{16}{65}$.
正确！$\cos A = \tfrac{4}{5}$，$\sin B = \tfrac{12}{13}$。$\cos(A+B) = \cos A\cos B - \sin A\sin B = \tfrac{4}{5}\cdot\tfrac{5}{13} - \tfrac{3}{5}\cdot\tfrac{12}{13} = \tfrac{20-36}{65} = -\tfrac{16}{65}$。
First build the complete triple: $\cos A = \tfrac{4}{5}$, $\sin B = \tfrac{12}{13}$. Then $\cos(A+B) = \tfrac{4}{5}\cdot\tfrac{5}{13} - \tfrac{3}{5}\cdot\tfrac{12}{13} = -\tfrac{16}{65}$.
先补齐所有三角函数值：$\cos A = \tfrac{4}{5}$，$\sin B = \tfrac{12}{13}$。再代入 $\cos(A+B) = \tfrac{4}{5}\cdot\tfrac{5}{13} - \tfrac{3}{5}\cdot\tfrac{12}{13} = -\tfrac{16}{65}$。
Topic C2.10 · HL OnlyTopic C2.10 · 仅 HL

Symmetry Relations of Trig Functions三角函数的对称关系 AHL 3.11

Supplementary Angle Relations $$ \sin(\pi - \theta) = \sin\theta, \qquad \cos(\pi - \theta) = -\cos\theta, \qquad \tan(\pi - \theta) = -\tan\theta $$ These come from the reflection symmetry of the unit circle across the $y$-axis and are the source of the ambiguous case in the sine rule (C2.2).
补角关系 $$ \sin(\pi - \theta) = \sin\theta, \qquad \cos(\pi - \theta) = -\cos\theta, \qquad \tan(\pi - \theta) = -\tan\theta $$ 这些关系来自单位圆关于 $y$ 轴的反射对称，也正是 C2.2 正弦定理（sine rule）模糊情形的根源。
Other Key Symmetries其他关键对称关系
$$ \sin(-\theta) = -\sin\theta \quad (\text{odd}) $$ $$ \cos(-\theta) = \cos\theta \quad (\text{even}) $$ $$ \sin(\tfrac{\pi}{2} - \theta) = \cos\theta, \qquad \cos(\tfrac{\pi}{2} - \theta) = \sin\theta $$ $$ \sin(\theta + \pi) = -\sin\theta, \qquad \cos(\theta + \pi) = -\cos\theta $$
Graphical Meaning $\sin x$ is an odd function — its graph has rotational symmetry about the origin. $\cos x$ is even — reflective symmetry about the $y$-axis. Shifting $\sin$ left by $\tfrac{\pi}{2}$ gives $\cos$: $\sin(x + \tfrac{\pi}{2}) = \cos x$.
图像意义 $\sin x$ 是奇函数——图像关于原点中心对称。$\cos x$ 是偶函数——图像关于 $y$ 轴对称。把 $\sin$ 图像向左平移 $\tfrac{\pi}{2}$ 就得到 $\cos$：$\sin(x + \tfrac{\pi}{2}) = \cos x$。
Topic C3.1 · HL OnlyTopic C3.1 · 仅 HL

Vectors — Concepts & Algebra向量——基本概念与代数运算 AHL 3.12

Definition A vector has both magnitude and direction. It can be represented as a directed line segment, a column matrix, or a sum of base vectors. Two vectors are equal if they have the same magnitude and direction — position does not matter.
定义 向量（vector）同时具有大小（magnitude，也叫"模"）和方向。它可以用有向线段、列矩阵或基向量的线性组合来表示。两个向量相等当且仅当它们的大小和方向都相同——所处位置不重要。
Component Forms分量表达式
$$ \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} $$ $$ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} $$ $$ \hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} \quad (\text{unit vector in direction of } \mathbf{v}) $$
Position vs. Displacement Vectors The position vector of point $A$ is $\overrightarrow{OA} = \mathbf{a}$ — from the origin to $A$.
The displacement from $A$ to $B$ is $$ \overrightarrow{AB} = \mathbf{b} - \mathbf{a} $$ This is the single most-used fact in HL vector problems. If a question asks for "the vector from $A$ to $B$", write $\mathbf{b} - \mathbf{a}$.
位置向量与位移向量 点 $A$ 的位置向量（position vector）是 $\overrightarrow{OA} = \mathbf{a}$，即从原点指向 $A$ 的向量。
从 $A$ 到 $B$ 的位移为 $$ \overrightarrow{AB} = \mathbf{b} - \mathbf{a} $$ 这是 HL 向量题用得最多的一条公式。题目说"从 $A$ 到 $B$ 的向量"，就写 $\mathbf{b} - \mathbf{a}$。
Algebraic Operations代数运算
$$ \text{Addition: } \mathbf{a} + \mathbf{b} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix} $$ $$ \text{Scalar multiple: } k\mathbf{v} = \begin{pmatrix} kv_1 \\ kv_2 \\ kv_3 \end{pmatrix}, \quad |k\mathbf{v}| = |k|\cdot|\mathbf{v}| $$ $$ \text{Parallel: } \mathbf{u} \parallel \mathbf{v} \iff \mathbf{u} = k\mathbf{v} \text{ for some } k \in \mathbb{R} $$
Strategy — Geometric Proofs Many HL exam questions ask to prove properties (e.g. "prove the midpoints of a quadrilateral form a parallelogram") using vectors. Express every point as a position vector from a chosen origin, build displacement vectors, and show two are equal (or one is a scalar multiple of another).
解题策略——几何证明 HL 考题里常有"用向量证明某几何性质"的题目（如"证明四边形各边中点构成平行四边形"）。做法是：选定一个原点，把每个点都写成位置向量；构造位移向量；证明两个位移向量相等（或一个是另一个的标量倍）。
Worked Example — Unit Vector & Displacement例题——单位向量与位移

Problem: Given $A(1, 2, -1)$ and $B(4, 6, 11)$, find (a) $\overrightarrow{AB}$, (b) $|\overrightarrow{AB}|$, (c) the unit vector in the direction of $\overrightarrow{AB}$.

(a) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4-1 \\ 6-2 \\ 11-(-1) \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 12 \end{pmatrix}$.

(b) $|\overrightarrow{AB}| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

(c) $\hat{\mathbf{u}} = \tfrac{1}{13}\begin{pmatrix} 3 \\ 4 \\ 12 \end{pmatrix}$.

题目：已知 $A(1, 2, -1)$ 与 $B(4, 6, 11)$，求 (a) $\overrightarrow{AB}$，(b) $|\overrightarrow{AB}|$，(c) 沿 $\overrightarrow{AB}$ 方向的单位向量（unit vector）。

(a) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4-1 \\ 6-2 \\ 11-(-1) \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 12 \end{pmatrix}$。

(b) $|\overrightarrow{AB}| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$。

(c) $\hat{\mathbf{u}} = \tfrac{1}{13}\begin{pmatrix} 3 \\ 4 \\ 12 \end{pmatrix}$。

Given $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 5 \\ 1 \\ -1 \end{pmatrix}$, the vector $\overrightarrow{AB}$ is:已知 $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ 与 $\mathbf{b} = \begin{pmatrix} 5 \\ 1 \\ -1 \end{pmatrix}$，则向量 $\overrightarrow{AB}$ 为：
C3.1
$\begin{pmatrix} 7 \\ 0 \\ 2 \end{pmatrix}$
$\begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}$
$\begin{pmatrix} -3 \\ -2 \\ 4 \end{pmatrix}$
$\begin{pmatrix} 10 \\ -1 \\ -3 \end{pmatrix}$
Correct! $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (5-2, 1-(-1), -1-3) = (3, 2, -4)$.
正确！$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (5-2, 1-(-1), -1-3) = (3, 2, -4)$。
The displacement $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (subtract, don't add). Here: $(5-2, 1-(-1), -1-3) = (3, 2, -4)$.
位移 $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$（是相减，不是相加）。本题：$(5-2, 1-(-1), -1-3) = (3, 2, -4)$。
Topic C3.2 · HL OnlyTopic C3.2 · 仅 HL

The Scalar (Dot) Product数量积（点积） AHL 3.13

Two Equivalent Formulas The scalar product of $\mathbf{a}$ and $\mathbf{b}$ is a number (not a vector): $$ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \quad (\text{algebraic}) $$ $$ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \quad (\text{geometric}) $$ Equating these gives the angle-finding formula below.
两种等价定义 $\mathbf{a}$ 与 $\mathbf{b}$ 的数量积（点积）（dot product）是一个数（不是向量）： $$ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \quad (\text{代数定义}) $$ $$ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \quad (\text{几何定义}) $$ 令两式相等就得到下面的"求夹角"公式。
Angle Between Two Vectors两向量夹角
$$ \cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} $$
Perpendicular & Parallel — Two Quick Tests $$ \mathbf{a} \perp \mathbf{b} \iff \mathbf{a} \cdot \mathbf{b} = 0 $$ $$ \mathbf{a} \parallel \mathbf{b} \iff \mathbf{a} \times \mathbf{b} = \mathbf{0} \iff \mathbf{a} = k\mathbf{b} $$ The dot product vanishes ⇔ perpendicular. Don't confuse with the cross product test for parallelism.
垂直与平行的两条快速判定 $$ \mathbf{a} \perp \mathbf{b} \iff \mathbf{a} \cdot \mathbf{b} = 0 $$ $$ \mathbf{a} \parallel \mathbf{b} \iff \mathbf{a} \times \mathbf{b} = \mathbf{0} \iff \mathbf{a} = k\mathbf{b} $$ 点积为零 ⇔ 垂直；不要与平行的判定（用向量积/叉积，cross product）混淆。
Properties of the Dot Product点积的性质
$$ \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \quad \text{(commutative)} $$ $$ \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} \quad \text{(distributive)} $$ $$ \mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 $$
Worked Example — Angle Between Vectors例题——两向量之间的夹角

Problem: Find the angle between $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix}$.

$$\mathbf{a}\cdot\mathbf{b} = (1)(3) + (2)(0) + (2)(-4) = 3 + 0 - 8 = -5$$ $$|\mathbf{a}| = \sqrt{1 + 4 + 4} = 3, \quad |\mathbf{b}| = \sqrt{9 + 0 + 16} = 5$$ $$\cos\theta = \frac{-5}{3 \cdot 5} = -\tfrac{1}{3}$$ $$\theta = \arccos(-\tfrac{1}{3}) \approx 109.5°$$

Because $\cos\theta < 0$, the angle is obtuse — a sanity check on the sign of the dot product.

题目：求 $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ 与 $\mathbf{b} = \begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix}$ 之间的夹角。

$$\mathbf{a}\cdot\mathbf{b} = (1)(3) + (2)(0) + (2)(-4) = 3 + 0 - 8 = -5$$ $$|\mathbf{a}| = \sqrt{1 + 4 + 4} = 3, \quad |\mathbf{b}| = \sqrt{9 + 0 + 16} = 5$$ $$\cos\theta = \frac{-5}{3 \cdot 5} = -\tfrac{1}{3}$$ $$\theta = \arccos(-\tfrac{1}{3}) \approx 109.5°$$

因为 $\cos\theta < 0$，夹角为钝角——也正好对应点积为负，符号自洽。

If $\mathbf{u} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 1 \\ k \\ -1 \end{pmatrix}$ are perpendicular, then $k$ equals:若 $\mathbf{u} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ 与 $\mathbf{v} = \begin{pmatrix} 1 \\ k \\ -1 \end{pmatrix}$ 垂直，则 $k = $
C3.2
$1$
$5$
$-1$
$-5$
Correct! $\mathbf{u}\cdot\mathbf{v} = 2 - k - 3 = -1 - k = 0 \Rightarrow k = -1$.
正确！$\mathbf{u}\cdot\mathbf{v} = 2 - k - 3 = -1 - k = 0 \Rightarrow k = -1$。
Set $\mathbf{u}\cdot\mathbf{v} = 0$: $(2)(1) + (-1)(k) + (3)(-1) = 2 - k - 3 = -1 - k = 0$, so $k = -1$.
令 $\mathbf{u}\cdot\mathbf{v} = 0$：$(2)(1) + (-1)(k) + (3)(-1) = 2 - k - 3 = -1 - k = 0$，故 $k = -1$。
Topic C3.3 · HL OnlyTopic C3.3 · 仅 HL

Vector Equation of a Line直线的向量方程 AHL 3.14

Three Forms of a Line in 3D Given a point $A(\mathbf{a})$ on the line and a direction vector $\mathbf{d}$:

Vector form: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}, \quad \lambda \in \mathbb{R}$
Parametric form: $x = a_1 + \lambda d_1, \; y = a_2 + \lambda d_2, \; z = a_3 + \lambda d_3$
Cartesian (symmetric) form: $\dfrac{x - a_1}{d_1} = \dfrac{y - a_2}{d_2} = \dfrac{z - a_3}{d_3}$
三维直线的三种表达 给定直线上一点 $A(\mathbf{a})$ 与方向向量（direction vector）$\mathbf{d}$：

向量形式（vector equation of a line）：$\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}, \quad \lambda \in \mathbb{R}$
参数形式（parametric form）：$x = a_1 + \lambda d_1, \; y = a_2 + \lambda d_2, \; z = a_3 + \lambda d_3$
笛卡尔（对称）形式：$\dfrac{x - a_1}{d_1} = \dfrac{y - a_2}{d_2} = \dfrac{z - a_3}{d_3}$
Angle Between Two Lines两直线夹角
$$ \cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|} $$ Absolute value ensures the acute angle.绝对值保证结果是锐角。
Finding the Line Through Two Points Given $A(\mathbf{a})$ and $B(\mathbf{b})$: use position vector $\mathbf{a}$ and direction $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$: $$ \mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a}) $$ Setting $\lambda = 0$ recovers $A$; $\lambda = 1$ recovers $B$; $\lambda = \tfrac{1}{2}$ gives the midpoint.
过两点的直线 已知 $A(\mathbf{a})$ 与 $B(\mathbf{b})$：取位置向量 $\mathbf{a}$，方向向量 $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$： $$ \mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a}) $$ $\lambda = 0$ 回到 $A$；$\lambda = 1$ 回到 $B$；$\lambda = \tfrac{1}{2}$ 给出中点。
Kinematics Interpretation If $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$ models a particle's position, then $\mathbf{v}$ is its velocity and $|\mathbf{v}|$ is its speed. Two particles collide if their $\mathbf{r}(t)$ values are equal for the same value of $t$. They pass through the same point (at different times) if there exist $t_1, t_2$ with $\mathbf{r}_1(t_1) = \mathbf{r}_2(t_2)$ — a common Paper 3 context.
运动学解读 若 $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$ 描述某质点的位置，则 $\mathbf{v}$ 是速度，$|\mathbf{v}|$ 是速率。两个质点"碰撞"当且仅当在同一 $t$ 值下 $\mathbf{r}(t)$ 相等。两条轨迹只是"经过同一点（在不同时刻）"则要求存在 $t_1, t_2$ 使 $\mathbf{r}_1(t_1) = \mathbf{r}_2(t_2)$——Paper 3 常见题型。
Worked Example — Line Through Two Points例题——过两点的直线

Problem: Find a vector equation of the line through $A(2, -1, 4)$ and $B(5, 3, -2)$.

Direction vector: $\overrightarrow{AB} = \begin{pmatrix} 3 \\ 4 \\ -6 \end{pmatrix}$.

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

Cartesian form:

$$\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 4}{-6}$$

题目：求过 $A(2, -1, 4)$ 与 $B(5, 3, -2)$ 的直线的向量方程。

方向向量：$\overrightarrow{AB} = \begin{pmatrix} 3 \\ 4 \\ -6 \end{pmatrix}$。

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

笛卡尔形式：

$$\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 4}{-6}$$
Topic C3.4 · HL OnlyTopic C3.4 · 仅 HL

Lines in Space — Four Relationships空间中的直线——四种关系 AHL 3.15

Classifying Two Lines $L_1, L_2$ Every pair of lines in 3D is exactly one of:
1. Coincident — same line, different parametrizations.
2. Parallel — same direction, but no shared point.
3. Intersecting — different directions and meet at one point.
4. Skew — different directions and never meet. (Only possible in 3D.)
两条直线 $L_1, L_2$ 的分类 三维空间中任意一对直线恰好属于以下一类：
1. 重合——同一条直线，只是参数化不同。
2. 平行——方向相同但没有公共点。
3. 相交——方向不同且交于一点。
4. 异面（skew lines）——方向不同且永不相交（只在 3D 中存在）。
Decision Algorithm判定流程
Step 1 — Direction test: Is $\mathbf{d}_1 \parallel \mathbf{d}_2$ (one is scalar multiple of other)?
   • Yes: Check if a point of $L_1$ lies on $L_2$. If yes → coincident; if no → parallel.
   • No: Go to Step 2.
Step 2 — Intersection test: Set the parametric equations equal. Solve the resulting system (use 2 of 3 equations, then verify with the 3rd).
   • Consistent: intersecting — solve for the point.
   • Inconsistent: skew.
第 1 步——方向判定：$\mathbf{d}_1 \parallel \mathbf{d}_2$ 吗（一个是另一个的标量倍）？
   • 是：验证 $L_1$ 上某点是否也在 $L_2$ 上。在 → 重合；不在 → 平行。
   • 否：进入第 2 步。
第 2 步——相交判定：令两组参数方程对应分量相等，求解（取 3 式中的 2 式联立，再用第 3 式验证）。
   • 方程组相容：相交——求出交点。
   • 方程组矛盾：异面（skew）。
Common Trap — Different Parameters When testing for intersection, you must use different parameter names for the two lines (e.g. $\lambda$ for $L_1$, $\mu$ for $L_2$). Using $\lambda$ for both assumes the meeting point corresponds to the same parameter value — it usually doesn't.
常见陷阱——参数必须不同 检验是否相交时，必须给两条直线用不同的参数名（如 $L_1$ 用 $\lambda$，$L_2$ 用 $\mu$）。两条都用 $\lambda$ 等于假设交点对应同一个参数值——通常并不成立。
Worked Example — Intersecting or Skew?例题——相交还是异面？

Problem: Decide whether these lines intersect, and if so, find the point:

$$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}, \quad L_2: \mathbf{r} = \begin{pmatrix} 3 \\ -1 \\ 5 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$$

Directions $(2,1,-1)$ and $(1,-1,2)$ are not multiples → not parallel.

Set components equal:

$$1 + 2\lambda = 3 + \mu \quad (x)$$ $$\lambda = -1 - \mu \quad (y)$$ $$2 - \lambda = 5 + 2\mu \quad (z)$$

From $(y)$: $\lambda = -1 - \mu$. Substitute into $(x)$: $1 + 2(-1 - \mu) = 3 + \mu \Rightarrow -1 - 2\mu = 3 + \mu \Rightarrow \mu = -\tfrac{4}{3}$. Then $\lambda = \tfrac{1}{3}$.

Check in $(z)$: LHS $= 2 - \tfrac{1}{3} = \tfrac{5}{3}$; RHS $= 5 + 2(-\tfrac{4}{3}) = \tfrac{7}{3}$. Not equal.

The system is inconsistent → lines are skew.

题目：判断下列两条直线是否相交；如果相交，求交点：

$$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}, \quad L_2: \mathbf{r} = \begin{pmatrix} 3 \\ -1 \\ 5 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$$

方向 $(2,1,-1)$ 与 $(1,-1,2)$ 不互为标量倍 → 不平行。

令对应分量相等：

$$1 + 2\lambda = 3 + \mu \quad (x)$$ $$\lambda = -1 - \mu \quad (y)$$ $$2 - \lambda = 5 + 2\mu \quad (z)$$

由 $(y)$：$\lambda = -1 - \mu$。代入 $(x)$：$1 + 2(-1 - \mu) = 3 + \mu \Rightarrow -1 - 2\mu = 3 + \mu \Rightarrow \mu = -\tfrac{4}{3}$。从而 $\lambda = \tfrac{1}{3}$。

代入 $(z)$ 验证：LHS $= 2 - \tfrac{1}{3} = \tfrac{5}{3}$；RHS $= 5 + 2(-\tfrac{4}{3}) = \tfrac{7}{3}$。不相等。

方程组矛盾 → 两直线异面（skew）。

Two lines have direction vectors $(1, 2, -1)$ and $(3, 6, -3)$ and share no common points. They are:两条直线的方向向量分别为 $(1, 2, -1)$ 与 $(3, 6, -3)$，且没有公共点。它们：
C3.4
Intersecting相交
Coincident重合
Parallel平行
Skew异面
Correct! $(3,6,-3) = 3(1,2,-1)$, so the directions are parallel. With no shared point, the lines are parallel (not coincident).
正确！$(3,6,-3) = 3(1,2,-1)$，方向平行。又因没有公共点，所以平行（而非重合）。
The second direction is $3\times$ the first — so the directions are parallel. If they share no point, the lines are parallel (if they did share a point, they'd be coincident).
第二个方向是第一个的 $3$ 倍——方向平行。若没有公共点，结论是平行（如果有公共点，则为重合）。
Topic C3.5 · HL OnlyTopic C3.5 · 仅 HL

The Vector (Cross) Product向量积（叉积） AHL 3.16

Definition Unlike the dot product, the cross product $\mathbf{a} \times \mathbf{b}$ is a vector. It is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, with magnitude equal to the area of the parallelogram they span, and direction given by the right-hand rule.
定义 与点积不同，向量积/叉积 $\mathbf{a} \times \mathbf{b}$（vector product / cross product）结果是一个向量。它同时与 $\mathbf{a}$、$\mathbf{b}$ 垂直，其大小（magnitude）等于两向量张成的平行四边形面积，方向由右手定则给出。
Computation (Determinant Form)计算公式（行列式形式）
$$ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \begin{pmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{pmatrix} $$
Geometric Interpretations几何解释
$$ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta = \text{area of parallelogram} $$ $$ \text{Area of triangle} = \tfrac{1}{2}|\mathbf{a} \times \mathbf{b}| $$
Key Properties $$ \mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a}) \quad (\text{anticommutative}) $$ $$ \mathbf{a} \times \mathbf{a} = \mathbf{0} $$ $$ \mathbf{a} \times \mathbf{b} = \mathbf{0} \iff \mathbf{a} \parallel \mathbf{b} $$ $$ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0, \quad (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0 $$
关键性质 $$ \mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a}) \quad (\text{反交换}) $$ $$ \mathbf{a} \times \mathbf{a} = \mathbf{0} $$ $$ \mathbf{a} \times \mathbf{b} = \mathbf{0} \iff \mathbf{a} \parallel \mathbf{b} $$ $$ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0, \quad (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0 $$
Common Use — Normal to a Plane To find a vector perpendicular to a plane containing $\mathbf{u}$ and $\mathbf{v}$, compute $\mathbf{n} = \mathbf{u} \times \mathbf{v}$. This is the starting point for the scalar/normal form of a plane equation (C3.6).
常见应用——求平面的法向量 要求一个与含有 $\mathbf{u}$、$\mathbf{v}$ 的平面垂直的向量，直接计算 $\mathbf{n} = \mathbf{u} \times \mathbf{v}$ 即可。这就是 C3.6 平面"数量/法向式"方程的起点；得到的就是该平面的法向量（normal vector）。
Worked Example — Area of a Triangle in 3D例题——三维空间中三角形的面积

Problem: Find the area of the triangle with vertices $A(1, 0, 2)$, $B(3, 2, 1)$, $C(0, 4, 3)$.

Build two edge vectors from $A$:

$$\overrightarrow{AB} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}, \quad \overrightarrow{AC} = \begin{pmatrix} -1 \\ 4 \\ 1 \end{pmatrix}$$

Compute the cross product:

$$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} (2)(1) - (-1)(4) \\ (-1)(-1) - (2)(1) \\ (2)(4) - (2)(-1) \end{pmatrix} = \begin{pmatrix} 6 \\ -1 \\ 10 \end{pmatrix}$$

Magnitude:

$$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{36 + 1 + 100} = \sqrt{137}$$

Area $= \tfrac{1}{2}\sqrt{137} \approx 5.85$ units².

题目：求顶点为 $A(1, 0, 2)$、$B(3, 2, 1)$、$C(0, 4, 3)$ 的三角形面积。

以 $A$ 为起点构造两条边向量：

$$\overrightarrow{AB} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}, \quad \overrightarrow{AC} = \begin{pmatrix} -1 \\ 4 \\ 1 \end{pmatrix}$$

求叉积：

$$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} (2)(1) - (-1)(4) \\ (-1)(-1) - (2)(1) \\ (2)(4) - (2)(-1) \end{pmatrix} = \begin{pmatrix} 6 \\ -1 \\ 10 \end{pmatrix}$$

取模：

$$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{36 + 1 + 100} = \sqrt{137}$$

面积 $= \tfrac{1}{2}\sqrt{137} \approx 5.85$ 平方单位。

For any non-zero vectors $\mathbf{a}, \mathbf{b}$, the expression $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a}$ equals:对任意非零向量 $\mathbf{a}, \mathbf{b}$，表达式 $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a}$ 等于：
C3.5
$0$
$|\mathbf{a}|^2|\mathbf{b}|$
$\mathbf{a}\cdot\mathbf{b}$
$|\mathbf{a}\times\mathbf{b}|$
Correct! $\mathbf{a}\times\mathbf{b}$ is perpendicular to $\mathbf{a}$ by construction, so their dot product is zero.
正确！由定义 $\mathbf{a}\times\mathbf{b}$ 与 $\mathbf{a}$ 垂直，所以两者点积为零。
The cross product $\mathbf{a}\times\mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$. Since perpendicular vectors have dot product zero, $(\mathbf{a}\times\mathbf{b})\cdot\mathbf{a} = 0$.
叉积 $\mathbf{a}\times\mathbf{b}$ 同时与 $\mathbf{a}$、$\mathbf{b}$ 垂直。垂直向量的点积为零，故 $(\mathbf{a}\times\mathbf{b})\cdot\mathbf{a} = 0$。
Topic C3.6 · HL OnlyTopic C3.6 · 仅 HL

Equations of a Plane平面的方程 AHL 3.17

Three Forms A plane is determined by a point $A(\mathbf{a})$ and a normal vector $\mathbf{n}$, or by a point and two non-parallel directions $\mathbf{b}, \mathbf{c}$:

Parametric (vector) form: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$
Scalar (normal) form: $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$
Cartesian form: $n_1 x + n_2 y + n_3 z = d$ (where $d = \mathbf{a} \cdot \mathbf{n}$)
三种表达形式 一个平面由一点 $A(\mathbf{a})$ 加上一条法向量（normal vector）$\mathbf{n}$ 确定，或由一点加两条不平行的方向向量 $\mathbf{b}, \mathbf{c}$ 确定：

参数（向量）形式（vector equation of a plane）：$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$
数量（法向）形式：$\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$
笛卡尔形式：$n_1 x + n_2 y + n_3 z = d$（其中 $d = \mathbf{a} \cdot \mathbf{n}$）
Converting Between Forms Given parametric form, find $\mathbf{n} = \mathbf{b} \times \mathbf{c}$ to get the normal, then plug $A$ into the scalar form. Given three points $A, B, C$ on a plane, take $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$.
不同形式之间的转换 给出参数形式时，用 $\mathbf{n} = \mathbf{b} \times \mathbf{c}$ 求法向量，再把 $A$ 代入数量形式即可。给出平面上三点 $A, B, C$ 时，取 $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$。
Distance from a Point to a Plane点到平面的距离
Given plane $n_1 x + n_2 y + n_3 z = d$ and point $P(p_1, p_2, p_3)$:
给定平面 $n_1 x + n_2 y + n_3 z = d$ 与点 $P(p_1, p_2, p_3)$：
$$ \text{dist} = \frac{|n_1 p_1 + n_2 p_2 + n_3 p_3 - d|}{\sqrt{n_1^2 + n_2^2 + n_3^2}} $$
Three Ways to Get a Plane — Cheat Sheet Given 3 points: $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$, then use $A$.
Given a line and a point not on it: direction of line is one $\mathbf{b}$, displacement from point to a point on the line is $\mathbf{c}$; take $\mathbf{n} = \mathbf{b} \times \mathbf{c}$.
Given two intersecting lines: $\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2$; use their intersection point.
确定平面的三种情景——速查表 给定 3 点：$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$，再用 $A$。
给定一直线与不在线上的一点：直线方向是 $\mathbf{b}$，点到直线某点的位移是 $\mathbf{c}$；取 $\mathbf{n} = \mathbf{b} \times \mathbf{c}$。
给定两条相交直线：$\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2$；用它们的交点。
Worked Example — Plane Through Three Points例题——过三点的平面

Problem: Find the Cartesian equation of the plane through $A(1, 2, 3)$, $B(2, 0, 1)$, $C(0, 3, 2)$.

$$\overrightarrow{AB} = \begin{pmatrix} 1 \\ -2 \\ -2 \end{pmatrix}, \quad \overrightarrow{AC} = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}$$ $$\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix} (-2)(-1)-(-2)(1) \\ (-2)(-1)-(1)(-1) \\ (1)(1)-(-2)(-1) \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ -1 \end{pmatrix}$$

Using $A(1,2,3)$: $d = 4(1) + 3(2) + (-1)(3) = 4 + 6 - 3 = 7$.

Equation: $\boxed{4x + 3y - z = 7}$.

题目：求过 $A(1, 2, 3)$、$B(2, 0, 1)$、$C(0, 3, 2)$ 的平面的笛卡尔方程。

$$\overrightarrow{AB} = \begin{pmatrix} 1 \\ -2 \\ -2 \end{pmatrix}, \quad \overrightarrow{AC} = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}$$ $$\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix} (-2)(-1)-(-2)(1) \\ (-2)(-1)-(1)(-1) \\ (1)(1)-(-2)(-1) \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ -1 \end{pmatrix}$$

代入 $A(1,2,3)$：$d = 4(1) + 3(2) + (-1)(3) = 4 + 6 - 3 = 7$。

方程：$\boxed{4x + 3y - z = 7}$。

The plane $2x - y + 3z = 6$ has normal vector:平面 $2x - y + 3z = 6$ 的法向量为：
C3.6
$(6, 6, 6)$
$(2, -1, 3)$
$(3, -1, 2)$
$(2, 1, 3)$
Correct! In Cartesian form $n_1 x + n_2 y + n_3 z = d$, the coefficients $(n_1, n_2, n_3)$ are the normal vector.
正确！在笛卡尔形式 $n_1 x + n_2 y + n_3 z = d$ 中，系数 $(n_1, n_2, n_3)$ 就是法向量。
The normal is read directly off the coefficients of $x, y, z$. Here: $(2, -1, 3)$. Watch the sign: the $y$-coefficient is $-1$, not $+1$.
法向量直接从 $x, y, z$ 的系数读出。本题：$(2, -1, 3)$。注意符号：$y$ 的系数是 $-1$，不是 $+1$。
Topic C3.7 · HL OnlyTopic C3.7 · 仅 HL

Intersections & Angles Involving Planes涉及平面的交集与夹角 AHL 3.18

Line Meets a Plane Substitute the parametric equations of the line into the Cartesian equation of the plane. Solve for $\lambda$:
• Unique $\lambda$: line crosses plane at one point.
• No solution (e.g. $0 = 5$): line is parallel to the plane, not on it.
• Identity (e.g. $0 = 0$): line lies entirely in the plane.
直线与平面相交 把直线的参数方程代入平面的笛卡尔方程，对 $\lambda$ 求解：
• $\lambda$ 唯一：直线穿过平面，交点唯一。
• 无解（如 $0 = 5$）：直线与平面平行，且不在平面上。
• 恒等式（如 $0 = 0$）：直线完全位于该平面内。
Angle Between Two Planes两平面夹角
$$ \cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} $$
Angle Between a Line and a Plane直线与平面的夹角
$$ \sin\theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} $$ Note the $\sin$ (not $\cos$): the angle you want is complementary to the angle between $\mathbf{d}$ and $\mathbf{n}$.注意是 $\sin$（不是 $\cos$）：要求的角和 $\mathbf{d}$ 与 $\mathbf{n}$ 之间的夹角互余。
Three Planes — Three Cases Solving the $3 \times 3$ system gives:
Unique solution: the three planes meet at a single point.
Infinite solutions (a line): the planes share a common line (like pages of a book).
No solution: at least two are parallel, or they form a "triangular prism" — no common point.
Use row reduction on the GDC or by hand to classify.
三个平面——三类情形 解 $3 \times 3$ 线性方程组，得到：
唯一解：三平面交于一点。
无穷解（一条直线）：三平面共有一条直线（像翻开的书页）。
无解：至少两个平面平行，或三平面构成"三棱柱"——没有公共点。
在 GDC 上或手动做行简化（row reduction）进行分类。
Sign Convention for Angles Angles between lines/planes/planes are always reported as acute — hence the absolute value in every formula above. An examiner will mark down for $109°$ when $71°$ was meant.
夹角的符号约定 直线之间、平面之间、直线与平面之间的夹角（angle between）一律按锐角报告——这就是上面所有公式都取绝对值的原因。把 $71°$ 写成 $109°$ 会被扣分。
Worked Example — Line Meets Plane例题——直线与平面相交

Problem: Find the point where line $L: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ meets plane $\pi: x + y + z = 6$.

Parametric: $x = 1 + 2\lambda$, $y = 2 - \lambda$, $z = 3\lambda$. Substitute:

$$\begin{aligned} (1 + 2\lambda) + (2 - \lambda) + (3\lambda) &= 6 \\ 3 + 4\lambda &= 6 \\ \lambda &= \tfrac{3}{4} \end{aligned}$$

Point of intersection: $(1 + \tfrac{3}{2}, 2 - \tfrac{3}{4}, \tfrac{9}{4}) = (\tfrac{5}{2}, \tfrac{5}{4}, \tfrac{9}{4})$.

题目：求直线 $L: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ 与平面 $\pi: x + y + z = 6$ 的交点。

参数方程：$x = 1 + 2\lambda$，$y = 2 - \lambda$，$z = 3\lambda$。代入：

$$\begin{aligned} (1 + 2\lambda) + (2 - \lambda) + (3\lambda) &= 6 \\ 3 + 4\lambda &= 6 \\ \lambda &= \tfrac{3}{4} \end{aligned}$$

交点：$(1 + \tfrac{3}{2}, 2 - \tfrac{3}{4}, \tfrac{9}{4}) = (\tfrac{5}{2}, \tfrac{5}{4}, \tfrac{9}{4})$。

The planes $x + 2y - z = 4$ and $2x + 4y - 2z = 9$ are:平面 $x + 2y - z = 4$ 与 $2x + 4y - 2z = 9$ 的关系是：
C3.7
Coincident重合
Intersecting in a line交于一条直线
Perpendicular互相垂直
Parallel (distinct)平行且不重合
Correct! The second equation is $2\times$ the first on the left, but the right-hand side is $9 \ne 2(4) = 8$. Parallel normals, different constants → parallel distinct planes.
正确！第二式左边是第一式的 $2$ 倍，但右边 $9 \ne 2(4) = 8$。法向量平行而常数不匹配 → 两平面平行且不重合。
Normals are $(1, 2, -1)$ and $(2, 4, -2) = 2(1, 2, -1)$, so parallel. But $9 \ne 2(4)$, so the planes don't coincide — they are parallel and distinct.
两法向量为 $(1, 2, -1)$ 与 $(2, 4, -2) = 2(1, 2, -1)$，平行。但 $9 \ne 2(4)$，所以平面不重合——两平面平行且不重合。
Review复习

Exam Strategy & Common Pitfalls考试策略与常见陷阱

M
Memorize必背
  • Volume & SA formulas for sphere, cone, cylinder, pyramid, hemisphere
  • Arc length $s = r\theta$ and sector area $A = \tfrac{1}{2}r^2\theta$ (radians)
  • Sine rule, cosine rule, triangle area $\tfrac{1}{2}ab\sin C$
  • Exact values of $\sin, \cos, \tan$ at $0, \tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}, \tfrac{\pi}{2}$
  • $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$
  • All three forms of $\cos 2\theta$ and compound angle formulas
  • Dot product: $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$
  • Cross product formula and $|\mathbf{a}\times\mathbf{b}|$ = area of parallelogram
  • Vector line: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$; plane: $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$
  • 球、圆锥、圆柱、棱锥、半球的体积与表面积公式
  • 弧长 $s = r\theta$ 与扇形面积 $A = \tfrac{1}{2}r^2\theta$（弧度制）
  • 正弦定理、余弦定理、三角形面积 $\tfrac{1}{2}ab\sin C$
  • $\sin, \cos, \tan$ 在 $0, \tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}, \tfrac{\pi}{2}$ 处的精确值
  • $\sin^2\theta + \cos^2\theta = 1$，$1 + \tan^2\theta = \sec^2\theta$
  • $\cos 2\theta$ 的三种形式 + 复合角公式
  • 点积：$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$
  • 叉积公式以及 $|\mathbf{a}\times\mathbf{b}| = $ 平行四边形面积
  • 直线向量方程 $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$；平面 $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$
U
Understand必懂
  • When to use sine vs. cosine rule (match the given information)
  • Ambiguous case of the sine rule — two valid triangles
  • How amplitude/period/phase shift appear in a sinusoidal model
  • Unit circle → signs of trig functions in each quadrant (ASTC)
  • $\arcsin(\sin x) \ne x$ outside the principal range
  • Geometric meaning of dot/cross product (scalar vs. vector)
  • The four line-pair cases (parallel, coincident, intersecting, skew)
  • Why the line-plane angle formula uses $\sin$, not $\cos$
  • How to classify three-plane systems from their equations
  • 何时用正弦定理、何时用余弦定理（按已知信息匹配）
  • 正弦定理的模糊情形（SSA）——可能两个解
  • 振幅、周期、相位（amplitude / period / phase shift）在正弦型模型中的含义
  • 单位圆 → 各象限三角函数的符号（ASTC 口诀）
  • $\arcsin(\sin x) \ne x$ 在主值范围外
  • 点积、叉积的几何含义（标量 vs 向量）
  • 两直线四种关系（平行、重合、相交、异面）
  • 为什么"直线-平面夹角"公式用 $\sin$，不是 $\cos$
  • 如何根据方程判断三平面系统的形态

Common Pitfalls常见陷阱

Top Student Errors 1. Calculator in degree mode on Paper 2/3 — silent, catastrophic.
2. Forgetting the second solution in the ambiguous case of the sine rule.
3. Treating $\sin(A+B)$ as $\sin A + \sin B$ on Paper 1.
4. Using the same parameter for both lines when checking intersection.
5. Forgetting the absolute value when computing acute angles between vectors, lines, or planes.
6. Confusing $\mathbf{a}\cdot\mathbf{b}$ (scalar) with $\mathbf{a}\times\mathbf{b}$ (vector). Only the dot product tests perpendicularity.
7. Writing "parallel" whenever direction vectors are proportional, without checking whether the lines actually coincide.
8. Reading the normal of a plane off the Cartesian equation with the wrong sign — the $y$-coefficient is $-1$, not $+1$, in $x - y + z = 4$.
学生最常犯的错 1. Paper 2/3 计算器停在"度"模式——无声但毁灭性。
2. 正弦定理模糊情形漏掉第二个解。
3. Paper 1 把 $\sin(A+B)$ 当作 $\sin A + \sin B$。
4. 检验两直线相交时给两条直线用同一个参数。
5. 算向量、直线、平面之间的锐角时忘了加绝对值。
6. 混淆 $\mathbf{a}\cdot\mathbf{b}$（标量）与 $\mathbf{a}\times\mathbf{b}$（向量）。只有点积可以判定垂直。
7. 方向向量成比例就写"平行"，没检查是否实际重合。
8. 从笛卡尔方程读法向量时符号错误——在 $x - y + z = 4$ 中 $y$ 的系数是 $-1$，不是 $+1$。
Paper-by-Paper Focus Paper 1 (no calc): exact values, algebraic trig manipulation, identity proofs, vector arithmetic, plane equations — everything must come out clean in fractions/surds.
Paper 2 (calc): triangle solving, bearings, 3D problems, sinusoidal modelling, intersections — use the GDC for arithmetic, but show method.
Paper 3 (extended): typically includes an investigative vectors or trigonometry modelling problem. Communicate reasoning clearly — method marks are the majority.
按试卷各自的考察重心 Paper 1（无计算器）：精确值、代数化简、恒等式证明、向量算术、平面方程——一切要以分数和根式形式工整得出。
Paper 2（有计算器）：解三角形、方位角、三维问题、正弦型建模、交集——算术交给 GDC，但要写出方法。
Paper 3（拓展卷）：通常包含一道探究式的向量或三角建模题。把推理过程讲清楚——方法分占大头。
Review复习

Flashcards闪卡

0 / 12 flipped已翻 0 / 12
Arc length in radians?弧长（弧度制）？
$s = r\theta$
Sector area in radians?扇形面积（弧度制）？
$A = \tfrac{1}{2}r^2\theta$
Cosine rule for side $c$?余弦定理求边 $c$？
$c^2 = a^2 + b^2 - 2ab\cos C$
Area of a triangle (SAS)?三角形面积 (SAS)？
$A = \tfrac{1}{2}ab\sin C$
Three forms of $\cos 2\theta$?$\cos 2\theta$ 的三种形式？
$\cos^2\theta - \sin^2\theta$
$= 2\cos^2\theta - 1$
$= 1 - 2\sin^2\theta$
$\sin(A + B) = \;?$$\sin(A + B) = \;?$
$\sin A\cos B + \cos A\sin B$
Extended Pythagorean identity (sec)?推广的平方和恒等式 (sec)？
$1 + \tan^2\theta = \sec^2\theta$
$\mathbf{a}\cdot\mathbf{b}$ in component form?$\mathbf{a}\cdot\mathbf{b}$ 的分量形式？
$a_1 b_1 + a_2 b_2 + a_3 b_3$
Test for $\mathbf{a}\perp\mathbf{b}$?$\mathbf{a}\perp\mathbf{b}$ 的判定？
$\mathbf{a}\cdot\mathbf{b} = 0$
Area of triangle from vectors?由向量求三角形面积？
$\tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$
Vector equation of a line?直线的向量方程？
$\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$
Angle between line & plane?直线与平面的夹角？
$\sin\theta = \dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$
Assessment测验

Unit C — Practice Quiz单元 C 练习测验

Test your understanding. Your score updates in real time at the top of the page.检验掌握情况。分数会在页面顶部实时更新。

1. A sector has radius $8$ cm and arc length $6\pi$ cm. The angle it subtends in radians is:1. 扇形半径 $8$ cm，弧长 $6\pi$ cm。它所对的圆心角（弧度）为：
Q1
$\tfrac{\pi}{4}$
$\tfrac{3\pi}{8}$
$\tfrac{3\pi}{4}$
$\tfrac{4\pi}{3}$
Correct! $\theta = \tfrac{s}{r} = \tfrac{6\pi}{8} = \tfrac{3\pi}{4}$.
正确！$\theta = \tfrac{s}{r} = \tfrac{6\pi}{8} = \tfrac{3\pi}{4}$。
Solve $s = r\theta$ for $\theta$: $\theta = \tfrac{s}{r} = \tfrac{6\pi}{8} = \tfrac{3\pi}{4}$ rad.
由 $s = r\theta$ 解 $\theta$：$\theta = \tfrac{s}{r} = \tfrac{6\pi}{8} = \tfrac{3\pi}{4}$ 弧度。
2. In $\triangle ABC$, $a = 7$, $b = 9$, $c = 12$. The largest angle is:2. $\triangle ABC$ 中，$a = 7$、$b = 9$、$c = 12$。最大的角是：
Q2
Angle $A$, about $36°$角 $A$，约 $36°$
Angle $C$, about $87°$角 $C$，约 $87°$
Angle $B$, about $58°$角 $B$，约 $58°$
Angle $C$, exactly $90°$角 $C$，恰为 $90°$
Correct! Largest side is $c = 12$, so largest angle is $C$. $\cos C = \tfrac{49 + 81 - 144}{2(7)(9)} = \tfrac{-14}{126} \approx -0.111 \Rightarrow C \approx 86.4°$.
正确！最长边是 $c = 12$，所以最大角是 $C$。$\cos C = \tfrac{49 + 81 - 144}{2(7)(9)} = \tfrac{-14}{126} \approx -0.111 \Rightarrow C \approx 86.4°$。
The largest angle is opposite the largest side ($c = 12$). Use $\cos C = \tfrac{a^2+b^2-c^2}{2ab} = \tfrac{-14}{126}$, giving $C \approx 86.4°$ (closest to $87°$).
最大角对最长边（$c = 12$）。用 $\cos C = \tfrac{a^2+b^2-c^2}{2ab} = \tfrac{-14}{126}$，得 $C \approx 86.4°$（最接近 $87°$）。
3. Simplify $\dfrac{\sin 2\theta}{1 + \cos 2\theta}$:3. 化简 $\dfrac{\sin 2\theta}{1 + \cos 2\theta}$：
Q3
$\tan\theta$
$\cot\theta$
$\tan 2\theta$
$\sin\theta$
Correct! $\tfrac{2\sin\theta\cos\theta}{1 + 2\cos^2\theta - 1} = \tfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \tan\theta$.
正确！$\tfrac{2\sin\theta\cos\theta}{1 + 2\cos^2\theta - 1} = \tfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \tan\theta$。
Expand: $\sin 2\theta = 2\sin\theta\cos\theta$ and use $\cos 2\theta = 2\cos^2\theta - 1$: $\tfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \tan\theta$.
展开：$\sin 2\theta = 2\sin\theta\cos\theta$，并用 $\cos 2\theta = 2\cos^2\theta - 1$：$\tfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \tan\theta$。
4. How many solutions does $2\cos x = 1$ have on $[0, 2\pi]$?4. $2\cos x = 1$ 在 $[0, 2\pi]$ 上有几个解？
Q4
$0$
$1$
$3$
$2$
Correct! $\cos x = \tfrac{1}{2}$ at $x = \tfrac{\pi}{3}$ and $x = \tfrac{5\pi}{3}$ in $[0, 2\pi]$ — two solutions.
正确！$\cos x = \tfrac{1}{2}$ 在 $[0, 2\pi]$ 上的解为 $x = \tfrac{\pi}{3}$ 与 $x = \tfrac{5\pi}{3}$——共 2 个。
$\cos x = \tfrac{1}{2}$ at $x = \tfrac{\pi}{3}$ (Q1) and $x = \tfrac{5\pi}{3}$ (Q4) over one full cycle — two solutions.
在一个完整周期内 $\cos x = \tfrac{1}{2}$ 的解为 $x = \tfrac{\pi}{3}$（Q1）与 $x = \tfrac{5\pi}{3}$（Q4），共 2 个。
5. Given $\mathbf{u} = (3, -2, 1)$ and $\mathbf{v} = (1, 4, -2)$, $\mathbf{u}\cdot\mathbf{v}$ equals:5. 给定 $\mathbf{u} = (3, -2, 1)$、$\mathbf{v} = (1, 4, -2)$，则 $\mathbf{u}\cdot\mathbf{v} = $
Q5
$3$
$9$
$-7$
$5$
Correct! $(3)(1) + (-2)(4) + (1)(-2) = 3 - 8 - 2 = -7$.
正确！$(3)(1) + (-2)(4) + (1)(-2) = 3 - 8 - 2 = -7$。
$\mathbf{u}\cdot\mathbf{v} = (3)(1) + (-2)(4) + (1)(-2) = 3 - 8 - 2 = -7$. Watch the signs carefully.
$\mathbf{u}\cdot\mathbf{v} = (3)(1) + (-2)(4) + (1)(-2) = 3 - 8 - 2 = -7$。注意符号。
6. The magnitude of $\mathbf{a}\times\mathbf{b}$ where $\mathbf{a} = (2, 0, 0)$ and $\mathbf{b} = (0, 3, 0)$ is:6. 设 $\mathbf{a} = (2, 0, 0)$、$\mathbf{b} = (0, 3, 0)$，则 $|\mathbf{a}\times\mathbf{b}| = $
Q6
$0$
$6$
$\sqrt{13}$
$5$
Correct! $\mathbf{a}\times\mathbf{b} = (0, 0, 6)$, so its magnitude is $6$. This is also the area of the $2\times 3$ rectangle they span.
正确！$\mathbf{a}\times\mathbf{b} = (0, 0, 6)$，其模为 $6$，也正好等于两向量张成的 $2\times 3$ 矩形面积。
Compute $\mathbf{a}\times\mathbf{b} = (0, 0, 6)$. Its magnitude is $6$ — equal to the area of the $2\times 3$ parallelogram (rectangle here) spanned by the two vectors.
计算 $\mathbf{a}\times\mathbf{b} = (0, 0, 6)$，其模为 $6$——等于两向量张成的 $2\times 3$ 平行四边形（此处即矩形）的面积。
7. The lines $L_1: \mathbf{r} = (1,0,2) + \lambda(1,1,0)$ and $L_2: \mathbf{r} = (2,1,3) + \mu(2,2,0)$ are:7. 直线 $L_1: \mathbf{r} = (1,0,2) + \lambda(1,1,0)$ 与 $L_2: \mathbf{r} = (2,1,3) + \mu(2,2,0)$ 的关系是：
Q7
Intersecting相交
Skew异面
Parallel (distinct)平行且不重合
Coincident重合
Correct! $(2,2,0) = 2(1,1,0)$ so the directions are parallel. Check whether $(2,1,3)$ is on $L_1$: need $1+\lambda=2, \lambda=1, 2=3$. The last equation fails, so the point is not on $L_1$. Lines are parallel and distinct.
正确！$(2,2,0) = 2(1,1,0)$，方向平行。检查 $(2,1,3)$ 是否在 $L_1$ 上：需要 $1+\lambda=2, \lambda=1, 2=3$。最后一条不成立，所以该点不在 $L_1$ 上。两直线平行且不重合。
Directions are parallel since $(2,2,0) = 2(1,1,0)$. But the point $(2,1,3)$ is not on $L_1$ (the $z$-coordinate forces a contradiction). So the lines are parallel and distinct, not coincident or skew.
$(2,2,0) = 2(1,1,0)$ 所以方向平行。但点 $(2,1,3)$ 不在 $L_1$ 上（$z$ 分量导致矛盾），所以两直线平行且不重合，不是异面或重合。
8. A plane has equation $\mathbf{r}\cdot(2, -1, 2) = 6$. The distance from the origin to the plane is:8. 平面方程为 $\mathbf{r}\cdot(2, -1, 2) = 6$。原点到该平面的距离为：
Q8
$6$
$2$
$\tfrac{6}{\sqrt{5}}$
$3$
Correct! $\text{dist} = \tfrac{|2(0) + (-1)(0) + 2(0) - 6|}{\sqrt{4+1+4}} = \tfrac{6}{3} = 2$.
正确！$\text{dist} = \tfrac{|2(0) + (-1)(0) + 2(0) - 6|}{\sqrt{4+1+4}} = \tfrac{6}{3} = 2$。
Use the point-to-plane formula with $(p_1,p_2,p_3) = (0,0,0)$: $\text{dist} = \tfrac{|0 - 6|}{\sqrt{4+1+4}} = \tfrac{6}{3} = 2$.
代入点到平面距离公式，$(p_1,p_2,p_3) = (0,0,0)$：$\text{dist} = \tfrac{|0 - 6|}{\sqrt{4+1+4}} = \tfrac{6}{3} = 2$。
Self-Assessment自我评估

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