High School Math

Vectors向量

A vector packages a magnitude and a direction into one symbol you can add, scale, and combine algebraically. This guide stays at the MCV4U / US (+) Common Core HSN-VM level: arrows in the plane, component form $\vec{v} = \langle a, b\rangle$, the dot product as a perpendicularity test, the cross product in 3-D, and applications to forces on an inclined plane and to work as $\vec{F} \cdot \vec{d}$. If you need lines and planes in 3-D, linear-independence proofs, or vector-equation geometry, jump straight to the IB Math AA HL Vectors unit linked from the Feeders section , it is the deeper companion to this guide.向量（vector）把"大小"和"方向"合并成一个可加、可缩放、可代数运算的符号。本指南停留在 ON MCV4U / US 共同核心 HSN-VM (+) 层级：平面上的箭头、分量形式 $\vec{v} = \langle a, b\rangle$、用点积（dot product）检验垂直、三维下的叉积（cross product），以及斜面受力（inclined plane）与功 $\vec{F} \cdot \vec{d}$ 的应用。若你要学习三维空间中的直线与平面、线性无关性证明或向量方程几何，请直接跳转到下方 Feeders 节链接的 IB Math AA HL 向量单元 , 它是本指南的深入版。

7 sections7 节内容 ON-core · US (+) · BC / AB out-of-coreON 核心 · US (+) · BC / AB 不在核心 Honors block on Dot / 3-D / Cross点积 / 三维 / 叉积为荣誉级
Read me first先读这里

How to use this guide如何使用本指南

Vectors split the four jurisdictions very unevenly. Ontario MCV4U treats the entire unit as core Grade 12 content. US Common Core puts the whole HSN-VM cluster at (+) honors / STEM-track Pre-Calc, so on a standard US Algebra II / Geometry / Pre-Calc trajectory this material is enrichment. British Columbia's PC 11 / PC 12 sequence contains no vectors at all, and Alberta's Math 30-1 outcomes do not list them either. The row table tells you whether the section is required, enrichment, or out-of-core for your jurisdiction, and the deeper-companion feeder at the end points anyone who needs more (lines, planes, 3-D geometry) at the IB Math AA HL Vectors unit already shipped in this repo.向量在四个地区差异极大。安大略 MCV4U 把本单元全部列为 12 年级核心内容。美国共同核心把整个 HSN-VM 簇置于 (+) 荣誉 / STEM 方向 Pre-Calc，标准 US Algebra II / Geometry / Pre-Calc 轨迹下属拓展内容。BC 的 PC 11 / PC 12 完全不含向量；AB 的 Math 30-1 指标也未列出向量。下表会告诉你在所在地区中本节属"必学"、"拓展"还是"核心之外"；本指南末尾的深入衔接指向本仓库已上线的 IB Math AA HL 向量单元，供需要进一步学习直线、平面、3D 几何者使用。

If you are in…如果你在… Focus on these sections重点学习 Defer / skip可推迟 Source依据
🇺🇸 US Algebra II / Geometry , standard track美国 Algebra II / 几何 , 标准轨 None required , the whole HSN-VM cluster is (+). Read §1-§3 for vector vocabulary if you encounter forces in physics class.不要求 , 整个 HSN-VM 簇带 (+)。若物理课遇到受力问题，可读 §1-§3 掌握向量词汇。 All seven , vectors are not on the standard US high-school graduation pathway七节均可缓 , 向量不在美国普通高中毕业轨 ccssm_hs_math.pdf , HSN-VM cluster carries (+) marker (STEM extension only). Repo's local extract is Linear-only., HSN-VM 簇带 (+) 标记（仅 STEM 拓展）。本仓库摘录仅含线性主题。
🇺🇸 US Honors Pre-Calc / AP-feeder美国荣誉 Pre-Calc / AP 衔接 Honors §1-§5 fully. §6 (cross product) and §7 (applications) if your course or AP Physics teacher expects them§1-§5 全学。若课程或 AP 物理老师要求，再学 §6（叉积）与 §7（应用） Cross product is rarely on a non-STEM Pre-Calc syllabus; skip §6 only if you are confident you will not encounter it in physics or engineering coursework非 STEM Pre-Calc 课纲少含叉积；仅当你确信物理 / 工程课不会出现时才跳过 §6 ccssm_hs_math.pdf , HSN-VM.A.1, .A.2, .A.3 (vectors as quantities), B.4 (add), B.5 (scalar multiply); the (+) cluster is the AP-feeder content. Standard-label cite (extract is Linear)., HSN-VM.A.1、.A.2、.A.3（向量量）、B.4（加法）、B.5（数乘）；(+) 簇即 AP 衔接内容。标准号引用（摘录仅含线性主题）。
🇨🇦 ON Grade 12 , MCV4U安大略 12 年级 , MCV4U §1 through §7 in full. MCV4U treats vectors as a co-equal strand alongside calculus , this is the most vector-heavy Grade 12 curriculum among the four we map to.§1 至 §7 全部。MCV4U 把向量与微积分并列为同等重要的单元 , 是我们对照的四套大纲中向量比重最高的 12 年级课程。 Nothing , but be aware that MCV4U also covers lines and planes in 3-space, which we leave for the IB HL feeder无 , 但请注意 MCV4U 还覆盖三维空间的直线与平面，本指南未含，参见 IB HL 衔接 math_grades_11-12.pdf , MCV4U Strand C Geometry and Algebra of Vectors (course title Calculus and Vectors). Extract cites strand-level scope; numbered expectations from published curriculum., MCV4U 单元 C 向量的几何与代数（课程名微积分与向量）。摘录在单元层级写明范围；编号期望按已发布课纲引用。
🇨🇦 BC PC 11 / PC 12BC PC 11 / PC 12 Out-of-core Nothing required by BC pathway. If you are taking IB / AP electives, read §1-§5 as the algebra prereq.BC 路径不要求。若选 IB / AP 选修，可读 §1-§5 作为代数前置。 All seven , not in published BC Pre-Calc curriculum七节均可缓 , 不在已发布的 BC Pre-Calc 课纲内 pc11_elab_extract.md, pc12_elab_extract.md , both extracts have zero matches for "vector". This is the documented absence., 两份摘录均无 "vector" 匹配。这是有记录的缺席。
🇨🇦 AB Math 30-1AB Math 30-1 Out-of-core Nothing required by AB Math 30-1. Math 31 (optional calculus) touches vectors only in motion contexts, not as a vector-algebra unit.AB Math 30-1 不要求。Math 31（可选微积分）只在运动语境下轻触向量，并非向量代数单元。 All seven , not in published Math 30-1 outcomes七节均可缓 , 不在 Math 30-1 已发布学习成果内 pos_10-12_indicators.txt , zero matches for "vector" in the AB Program of Studies indicators file. Math 31 (`math31.pdf`) touches vectors only in motion-related calculus contexts., AB 学习课程指标文件中 "vector" 无匹配。Math 31（`math31.pdf`）只在与运动相关的微积分语境下涉及向量。
🇺🇸 AP Physics 1 / C pathAP Physics 1 / C 方向 §1, §2, §3, §7 are mandatory , AP Physics assumes you can decompose a vector into components and recognise $\vec{F} \cdot \vec{d}$. §4 (dot product) is the work formula directly.§1、§2、§3、§7 必学 , AP Physics 默认你会作分量分解并识别 $\vec{F} \cdot \vec{d}$。§4（点积）正是功的公式。 §5 (3-D) and §6 (cross product) are AP Physics C: Mechanics or E&M material , cross product appears via torque and magnetic force ($\vec{F} = q\vec{v}\times\vec{B}$)§5（三维）与 §6（叉积）属 AP Physics C：力学或电磁 , 叉积通过力矩与磁力 ($\vec{F} = q\vec{v}\times\vec{B}$) 登场 See AP Physics/Study Guides/Unit_1_Kinematics.html for the kinematics use of vector decomposition; that unit ships in this repo.参见 AP Physics/Study Guides/Unit_1_Kinematics.html 中向量分解的运动学应用；该单元已在本仓库上线。
🇺🇸🇬🇧 IB Math AA HL feederIB Math AA HL 衔接 All seven sections as prereq, then jump to IB Math HL/Study Guides/Unit_C3_Vectors.html for: linear independence, vector / scalar / parametric equations of lines, planes in 3-space, and intersection geometry七节全部作前置，然后跳转 IB Math HL/Study Guides/Unit_C3_Vectors.html 学习：线性无关性、直线的向量 / 标量 / 参数方程、3D 平面、相交几何 Nothing , this unit is the algebra primer for the deeper IB HL geometry无 , 本单元是更深 IB HL 几何的代数预热 IB Math AA HL Topic 3 (Geometry & Trigonometry), AHL extensions 3.12 (vectors) and 3.13 (dot product) onwards; covered at HL depth in the linked unit.IB Math AA HL Topic 3（几何与三角），AHL 拓展 3.12（向量）与 3.13（点积）起；链接单元按 HL 深度覆盖。

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

!
If you are cramming the night before如果你在临阵磨枪

Memorise four things: component form $\vec{v} = \langle a, b\rangle$ and magnitude $|\vec{v}| = \sqrt{a^2 + b^2}$; vector addition $\vec{u} + \vec{v} = \langle u_1 + v_1, u_2 + v_2\rangle$; the dot product $\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 = |\vec{u}||\vec{v}|\cos\theta$ (and that $\vec{u} \cdot \vec{v} = 0$ means perpendicular); if your row includes §6, the cross-product magnitude $|\vec{u}\times\vec{v}| = |\vec{u}||\vec{v}|\sin\theta$ and the right-hand rule. Read every cram-cheat box. Skip the deeper proofs.背熟四件事：分量形式 $\vec{v} = \langle a, b\rangle$ 与模长 $|\vec{v}| = \sqrt{a^2 + b^2}$；向量加法 $\vec{u} + \vec{v} = \langle u_1 + v_1, u_2 + v_2\rangle$；点积 $\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 = |\vec{u}||\vec{v}|\cos\theta$（$\vec{u} \cdot \vec{v} = 0$ 即垂直）；若所在行含 §6，再加叉积模长 $|\vec{u}\times\vec{v}| = |\vec{u}||\vec{v}|\sin\theta$ 与右手定则。读每个速记框，跳过深入推导。

*
If you are going for the top mark如果你目标顶分

Always draw both the geometric arrow and write the component form , two representations catch each other's errors. Treat the dot product as the answer to "are these perpendicular?" and "what is the angle?" simultaneously. Practise the cross product enough that the right-hand rule is automatic and you can read off the area of a parallelogram in one line. The cleanest IB HL / engineering proof patterns lean on $\vec{u}\cdot\vec{v} = 0 \Leftrightarrow \vec{u}\perp\vec{v}$ and on the projection formula , both come from §4.始终把几何箭头和分量形式两种表示同时写下 , 两者互相纠错。把点积当作同时回答"是否垂直？"与"夹角多少？"两个问题的工具。叉积要练到右手定则成为本能，并能一行读出平行四边形面积。IB HL / 工程方向最干净的证明套路依赖 $\vec{u}\cdot\vec{v} = 0 \Leftrightarrow \vec{u}\perp\vec{v}$ 与投影公式 , 都源自 §4。

Honors / out-of-core flag.荣誉 / 核心之外 标记。 Sections §4 (Dot Product), §5 (3-D), §6 (Cross Product), and §7 (Applications) carry the Honors chip for US students , the whole HSN-VM cluster is (+) in CCSSM, so even §1-§3 sit above the standard graduation track. The entire unit (§1-§7) is flagged Out-of-core for BC PC 11 / PC 12 students and for AB Math 30-1 students, where vectors do not appear in the published curriculum. In Ontario MCV4U, by contrast, §1-§7 are all core Grade 12 content with the same status as any calculus topic in the same course. If you need lines and planes in 3-D, jump to the IB HL feeder at the bottom.§4（点积）、§5（三维）、§6（叉积）与 §7（应用）对美国学生带 Honors 标记 , 整个 HSN-VM 簇在 CCSSM 中带 (+)，因此连 §1-§3 都已在标准毕业轨之上。对 BC PC 11 / PC 12 学生与 AB Math 30-1 学生，整单元（§1-§7）标 Out-of-core，因为向量未出现在已发布课纲中。安大略 MCV4U 则相反：§1-§7 全为 12 年级核心内容，与同课程任何微积分主题地位相同。若需 3D 中的直线与平面，请跳至底部 IB HL 衔接。
Section 1 · Directed segments and notation第 1 节 · 有向线段与记号

Vectors as Directed Segments: magnitude and direction向量作为有向线段：大小与方向

A vector is "an arrow with a name".向量是"有名字的箭头"。

A vector carries two pieces of information: a magnitude (a non-negative length) and a direction (where the arrow points). Two arrows are the same vector if they have the same magnitude and the same direction , their position in the plane does not matter (vectors are free).向量带两条信息：大小（一个非负长度）与方向（箭头指向）。两支大小相同、方向相同的箭头视为同一向量 , 它们在平面上的位置无关紧要（向量是自由的）。

  • Notation.记号。 Write a vector as $\vec{v}$ (arrow) or $\mathbf{v}$ (bold). The magnitude is $|\vec{v}|$ or $\|\vec{v}\|$. A directed segment from point $A$ to point $B$ is $\vec{AB}$.向量写作 $\vec{v}$（箭头）或 $\mathbf{v}$（加粗）。模长（magnitude）写作 $|\vec{v}|$ 或 $\|\vec{v}\|$。由点 $A$ 至点 $B$ 的有向线段写作 $\vec{AB}$。
  • Zero vector.零向量。 $\vec{0}$ has magnitude $0$ and no defined direction; it is the additive identity ($\vec{v} + \vec{0} = \vec{v}$).$\vec{0}$ 模长为 $0$，方向未定义；它是加法单位元（$\vec{v} + \vec{0} = \vec{v}$）。
  • Opposite vector.相反向量。 $-\vec{v}$ has the same magnitude as $\vec{v}$ and the opposite direction. So $\vec{AB} = -\vec{BA}$.$-\vec{v}$ 与 $\vec{v}$ 模长相同，方向相反。故 $\vec{AB} = -\vec{BA}$。
  • Equal vs equivalent.相等与等价。 Two vectors are equal when their magnitudes and directions match, regardless of where they are drawn. A scalar (a plain number) has only a magnitude , it is not a vector.两向量大小与方向都相同即相等，与画在何处无关。标量（普通数）只有大小 , 不是向量。
CCSSM HSN-VM.A.1 (+) names this verbatim: "recognize vector quantities as having both magnitude and direction; represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes."CCSSM HSN-VM.A.1 (+) 原文："识别向量量既有大小又有方向；用有向线段表示向量量，并对向量及其模长使用恰当符号。"
Worked Example 1 · From two points to one vector例题 1 · 从两点到一个向量

A particle moves from point $A = (-1, 2)$ to point $B = (4, 5)$. Find the directed-segment vector $\vec{AB}$, its magnitude, and the angle it makes with the positive $x$-axis (to two decimals).一质点从 $A = (-1, 2)$ 移动到 $B = (4, 5)$。求有向线段向量 $\vec{AB}$、其模长及其与正 $x$ 轴所成角（保留两位小数）。

Components = terminal $-$ initial.分量 = 终点 $-$ 起点。 This is CCSSM HSN-VM.A.2 directly.这正是 CCSSM HSN-VM.A.2。

$$ \vec{AB} = \langle 4 - (-1), \; 5 - 2 \rangle = \langle 5, \; 3 \rangle. $$

Magnitude via the Pythagorean theorem.由勾股定理求模长。

$$ |\vec{AB}| = \sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83. $$

Direction angle.方向角。 The angle $\theta$ that $\vec{AB}$ makes with the positive $x$-axis satisfies $\tan\theta = 3/5$, so $\theta = \arctan(3/5) \approx 30.96^{\circ}$ (first-quadrant arrow, both components positive).$\vec{AB}$ 与正 $x$ 轴的夹角 $\theta$ 满足 $\tan\theta = 3/5$，故 $\theta = \arctan(3/5) \approx 30.96^{\circ}$（两分量均正，箭头在第一象限）。

Sanity-check.合理性核验。 The horizontal change $5$ exceeds the vertical change $3$, so the angle should sit below $45^{\circ}$ , $30.96^{\circ}$ matches.水平变化 $5$ 大于垂直变化 $3$，故角应小于 $45^{\circ}$ , $30.96^{\circ}$ 吻合。

Given $A = (3, -1)$ and $B = (-2, 7)$, find $\vec{AB}$.已知 $A = (3, -1)$、$B = (-2, 7)$，求 $\vec{AB}$。
§1 · Q1
$\langle 5, -8 \rangle$
$\langle 1, 6 \rangle$
$\langle -5, 8 \rangle$
$\langle -2/3, -7 \rangle$
Terminal minus initial: $\vec{AB} = \langle -2 - 3, \; 7 - (-1) \rangle = \langle -5, \; 8 \rangle$. (CCSSM HSN-VM.A.2.)终点减起点：$\vec{AB} = \langle -2 - 3, \; 7 - (-1) \rangle = \langle -5, \; 8 \rangle$。（CCSSM HSN-VM.A.2。）
Component rule: subtract initial-point coordinates from terminal-point coordinates, component by component.分量法则：终点坐标按分量减起点坐标。
Which statement is true?下列哪项正确？
§1 · Q2
$\vec{AB} = \vec{BA}$ because they connect the same two points.$\vec{AB} = \vec{BA}$，因为它们连接相同两点。
$\vec{AB} = -\vec{BA}$: same magnitude, opposite direction.$\vec{AB} = -\vec{BA}$：模长相同，方向相反。
Two vectors with the same magnitude are always equal.模长相同的两向量总是相等。
The zero vector has direction $0^{\circ}$.零向量的方向为 $0^{\circ}$。
$\vec{AB}$ points from $A$ to $B$; $\vec{BA}$ points the other way. Same length, opposite sign , hence $\vec{AB} = -\vec{BA}$. Magnitude alone does not determine equality (direction matters), and $\vec{0}$ has no defined direction.$\vec{AB}$ 由 $A$ 指向 $B$；$\vec{BA}$ 反向。模长相同、符号相反 , 故 $\vec{AB} = -\vec{BA}$。仅靠模长无法判定相等（方向同样重要），且 $\vec{0}$ 方向未定义。
Vectors are equal only when both magnitude and direction match. Reversing the arrow flips the sign.向量相等需大小、方向同时一致。箭头反向即符号反号。
Section 2 · Components, magnitude, unit vectors第 2 节 · 分量、模长、单位向量

Vectors in 2-D: component form $\langle a, b\rangle$, magnitude, unit vector二维向量：分量形式 $\langle a, b\rangle$、模长、单位向量

Component form turns a geometric arrow into algebra.分量形式把几何箭头变成代数。 $$ \vec{v} = \langle a, \; b \rangle, \qquad |\vec{v}| = \sqrt{a^2 + b^2}, \qquad \hat{v} = \frac{\vec{v}}{|\vec{v}|} = \left\langle \frac{a}{|\vec{v}|}, \; \frac{b}{|\vec{v}|} \right\rangle. $$
  • Components.分量。 $a$ is the horizontal component and $b$ is the vertical component , the projections of $\vec{v}$ onto the $x$- and $y$-axes.$a$ 为水平分量、$b$ 为垂直分量 , 即 $\vec{v}$ 在 $x$、$y$ 轴上的投影。
  • Magnitude.模长。 $|\vec{v}| = \sqrt{a^2 + b^2}$ is the length of the arrow , Pythagoras on a right triangle whose legs are $|a|$ and $|b|$.$|\vec{v}| = \sqrt{a^2 + b^2}$ 即箭头长度 , 在两直角边为 $|a|$、$|b|$ 的直角三角形上用勾股定理。
  • Unit vector.单位向量。 $\hat{v}$ has magnitude $1$ and the same direction as $\vec{v}$. Use it to keep only the direction information , the engineering / physics name is "direction cosines."$\hat{v}$ 模长为 $1$，方向与 $\vec{v}$ 相同。用它只保留"方向"信息 , 工程 / 物理上称为"方向余弦"。
  • Standard unit vectors.标准单位向量。 $\vec{i} = \langle 1, 0\rangle$ and $\vec{j} = \langle 0, 1\rangle$. Then $\vec{v} = a\vec{i} + b\vec{j}$. Both notations $\langle a, b\rangle$ and $a\vec{i} + b\vec{j}$ are used; pick whichever your course prefers.$\vec{i} = \langle 1, 0\rangle$、$\vec{j} = \langle 0, 1\rangle$。则 $\vec{v} = a\vec{i} + b\vec{j}$。两种记号 $\langle a, b\rangle$ 与 $a\vec{i} + b\vec{j}$ 都常用，按课程要求选一种。
From magnitude and angle to components. If $\vec{v}$ has magnitude $r$ and direction angle $\theta$ from the positive $x$-axis, then $\vec{v} = \langle r\cos\theta, \; r\sin\theta\rangle$. This is the right-triangle picture from Unit 7 reused inside a vector.由模长与角求分量。 若 $\vec{v}$ 模长为 $r$、与正 $x$ 轴方向角为 $\theta$，则 $\vec{v} = \langle r\cos\theta, \; r\sin\theta\rangle$。这是把 Unit 7 的直角三角形图复用于向量内部。
Worked Example 2 · Magnitude and unit vector例题 2 · 模长与单位向量

Let $\vec{v} = \langle 6, -8\rangle$. Find $|\vec{v}|$ exactly, the unit vector $\hat{v}$ in the same direction, and the direction angle $\theta$ measured from the positive $x$-axis (to two decimals).设 $\vec{v} = \langle 6, -8\rangle$。精确求 $|\vec{v}|$，并求同方向单位向量 $\hat{v}$ 及与正 $x$ 轴的方向角 $\theta$（保留两位小数）。

Magnitude.模长。

$$ |\vec{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10. $$

Unit vector.单位向量。

$$ \hat{v} = \frac{1}{10}\langle 6, -8\rangle = \left\langle \tfrac{3}{5}, \; -\tfrac{4}{5}\right\rangle. $$

Direction angle.方向角。 $\vec{v}$ has positive $x$-component and negative $y$-component , fourth quadrant. The reference angle is $\arctan(8/6) \approx 53.13^{\circ}$, so $\theta \approx -53.13^{\circ}$ (or equivalently $306.87^{\circ}$).$\vec{v}$ 的 $x$ 分量为正、$y$ 分量为负 , 第四象限。参考角为 $\arctan(8/6) \approx 53.13^{\circ}$，故 $\theta \approx -53.13^{\circ}$（即 $306.87^{\circ}$）。

Sanity-check.合理性核验。 $|\hat{v}|^2 = (3/5)^2 + (-4/5)^2 = 9/25 + 16/25 = 1$ ✓. Hidden inside: a 3-4-5 right triangle scaled by $2$.$|\hat{v}|^2 = (3/5)^2 + (-4/5)^2 = 9/25 + 16/25 = 1$ ✓。背后是放大 $2$ 倍的 3-4-5 直角三角形。

Find $|\vec{u}|$ for $\vec{u} = \langle -7, 24\rangle$.求 $\vec{u} = \langle -7, 24\rangle$ 的模长。
§2 · Q1
$17$
$25$
$\sqrt{527}$
$31$
$|\vec{u}| = \sqrt{(-7)^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$. (7-24-25 is a Pythagorean triple.)$|\vec{u}| = \sqrt{(-7)^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$。（7-24-25 为勾股数。）
Square each component (sign disappears), add, then take the square root.每个分量平方（符号消失），相加，再开方。
A vector $\vec{v}$ has magnitude $10$ and makes an angle of $120^{\circ}$ with the positive $x$-axis. Write $\vec{v}$ in component form (exact values).向量 $\vec{v}$ 模长 $10$，与正 $x$ 轴成 $120^{\circ}$。以精确值写出 $\vec{v}$ 的分量形式。
§2 · Q2
$\langle -5, \; 5\sqrt{3}\rangle$
$\langle 5, \; 5\sqrt{3}\rangle$
$\langle -5\sqrt{3}, \; 5\rangle$
$\langle -5, \; -5\sqrt{3}\rangle$
$\vec{v} = \langle 10\cos 120^{\circ}, \; 10\sin 120^{\circ}\rangle = \langle 10 \cdot (-1/2), \; 10 \cdot \sqrt{3}/2\rangle = \langle -5, \; 5\sqrt{3}\rangle$. Second-quadrant arrow: negative $x$, positive $y$.$\vec{v} = \langle 10\cos 120^{\circ}, \; 10\sin 120^{\circ}\rangle = \langle 10 \cdot (-1/2), \; 10 \cdot \sqrt{3}/2\rangle = \langle -5, \; 5\sqrt{3}\rangle$。第二象限：$x$ 负、$y$ 正。
Use $\vec{v} = \langle r\cos\theta, r\sin\theta\rangle$ with the exact values $\cos 120^{\circ} = -1/2$ and $\sin 120^{\circ} = \sqrt{3}/2$.用 $\vec{v} = \langle r\cos\theta, r\sin\theta\rangle$ 并代入精确值 $\cos 120^{\circ} = -1/2$、$\sin 120^{\circ} = \sqrt{3}/2$。
Section 3 · Addition and scalar multiplication第 3 节 · 加法与数乘

Vector Addition and Scalar Multiplication: tip-to-tail and component-wise向量加法与数乘：首尾相接与按分量

Two equivalent ways to combine vectors.组合向量的两种等价方式。
  • Geometric (tip-to-tail).几何法（首尾相接）。 To form $\vec{u} + \vec{v}$, slide $\vec{v}$ so its tail sits at the tip of $\vec{u}$; the sum runs from the tail of $\vec{u}$ to the new tip of $\vec{v}$.作 $\vec{u} + \vec{v}$：平移 $\vec{v}$ 使其起点落在 $\vec{u}$ 的终点；和向量从 $\vec{u}$ 起点指向 $\vec{v}$ 的新终点。
  • Parallelogram law.平行四边形法则。 If $\vec{u}$ and $\vec{v}$ share a common tail, $\vec{u} + \vec{v}$ is the diagonal of the parallelogram they span (same as tip-to-tail, drawn symmetrically).若 $\vec{u}$、$\vec{v}$ 共起点，$\vec{u} + \vec{v}$ 为以两者为边的平行四边形的对角线（与首尾相接等价，画法对称）。
  • Component-wise.按分量。 $$\vec{u} + \vec{v} = \langle u_1 + v_1, \; u_2 + v_2\rangle, \qquad \vec{u} - \vec{v} = \langle u_1 - v_1, \; u_2 - v_2\rangle. $$
  • Scalar multiplication.数乘。 $$k\vec{v} = \langle k v_1, \; k v_2\rangle, \qquad |k\vec{v}| = |k|\,|\vec{v}|. $$ If $k > 0$, direction is preserved; if $k < 0$, direction reverses; $k = 0$ gives $\vec{0}$.若 $k > 0$，方向不变；若 $k < 0$，方向反转；$k = 0$ 得 $\vec{0}$。
CCSSM HSN-VM.B.4 (+) names both the geometric ("end-to-end") and component descriptions of addition; HSN-VM.B.5 (+) names scalar multiplication.CCSSM HSN-VM.B.4 (+) 同时指明几何（"首尾相接"）与分量两种加法描述；HSN-VM.B.5 (+) 指明数乘。
Worked Example 3 · Linear combination from components例题 3 · 由分量求线性组合

Let $\vec{u} = \langle 3, -2\rangle$ and $\vec{v} = \langle -1, 4\rangle$. Find $2\vec{u} - 3\vec{v}$ in component form, and find its magnitude exactly.设 $\vec{u} = \langle 3, -2\rangle$、$\vec{v} = \langle -1, 4\rangle$。求 $2\vec{u} - 3\vec{v}$ 的分量形式，并求其精确模长。

Scale each vector first.先分别数乘。

$$ 2\vec{u} = \langle 6, -4\rangle, \qquad 3\vec{v} = \langle -3, 12\rangle. $$

Subtract component by component.按分量相减。

$$ 2\vec{u} - 3\vec{v} = \langle 6 - (-3), \; -4 - 12\rangle = \langle 9, -16\rangle. $$

Magnitude.模长。

$$ |2\vec{u} - 3\vec{v}| = \sqrt{9^2 + (-16)^2} = \sqrt{81 + 256} = \sqrt{337}. $$

Sanity-check.合理性核验。 $\sqrt{337} \approx 18.36$. Bounds from triangle inequality: $|2\vec{u}| + |3\vec{v}| = 2\sqrt{13} + 3\sqrt{17} \approx 7.21 + 12.37 \approx 19.58$, and the difference of magnitudes is $|7.21 - 12.37| \approx 5.16$. Our answer sits in $[5.16, 19.58]$ ✓.$\sqrt{337} \approx 18.36$。三角不等式给出上界 $|2\vec{u}| + |3\vec{v}| = 2\sqrt{13} + 3\sqrt{17} \approx 7.21 + 12.37 \approx 19.58$，下界 $|7.21 - 12.37| \approx 5.16$。结果落在 $[5.16, 19.58]$ 内 ✓。

Given $\vec{a} = \langle 4, 1\rangle$ and $\vec{b} = \langle -2, 5\rangle$, compute $\vec{a} + 3\vec{b}$.已知 $\vec{a} = \langle 4, 1\rangle$、$\vec{b} = \langle -2, 5\rangle$，求 $\vec{a} + 3\vec{b}$。
§3 · Q1
$\langle 2, 6\rangle$
$\langle 6, -14\rangle$
$\langle -2, 16\rangle$
$\langle 10, 16\rangle$
$3\vec{b} = \langle -6, 15\rangle$. Then $\vec{a} + 3\vec{b} = \langle 4 + (-6), \; 1 + 15\rangle = \langle -2, 16\rangle$.$3\vec{b} = \langle -6, 15\rangle$。$\vec{a} + 3\vec{b} = \langle 4 + (-6), \; 1 + 15\rangle = \langle -2, 16\rangle$。
Scalar multiply first, then add component by component. The scalar $3$ multiplies each component of $\vec{b}$.先数乘，再按分量相加。标量 $3$ 同时乘 $\vec{b}$ 的两个分量。
If $\vec{u}$ has magnitude $8$, what is $|-\tfrac{3}{2}\vec{u}|$?若 $\vec{u}$ 模长 $8$，求 $|-\tfrac{3}{2}\vec{u}|$。
§3 · Q2
$-12$
$\tfrac{16}{3}$
$8$
$12$
$|k\vec{u}| = |k|\,|\vec{u}| = \tfrac{3}{2} \cdot 8 = 12$. Magnitudes are non-negative , the minus sign reverses direction but is absorbed by $|k|$.$|k\vec{u}| = |k|\,|\vec{u}| = \tfrac{3}{2} \cdot 8 = 12$。模长非负 , 负号只反向，绝对值 $|k|$ 吸收符号。
Magnitudes are non-negative. Use $|k\vec{u}| = |k|\,|\vec{u}|$ , take the absolute value of the scalar first.模长非负。用 $|k\vec{u}| = |k|\,|\vec{u}|$ , 先取标量的绝对值。
Section 4 · Dot product第 4 节 · 点积

The Dot Product: angle, perpendicularity, projection Honors for US, out-of-core BC + AB点积：夹角、垂直、投影 荣誉 — US 核心 / BC、AB 之外

Curriculum note.课纲提示。 Dot product is core in Ontario MCV4U Strand C and in the IB Math AA HL Topic 3 AHL extension (3.13). It is Honors for US students (the whole HSN-VM cluster is (+)) and Out-of-core for BC PC 11 / PC 12 and AB Math 30-1.点积在安大略 MCV4U 单元 C 与 IB Math AA HL Topic 3 AHL 拓展（3.13）中为核心。对美国学生标 Honors（整个 HSN-VM 簇带 (+)）；对 BC PC 11 / PC 12 与 AB Math 30-1 标 Out-of-core。
Dot product , two formulas, one number.点积 , 两个公式、一个数。 $$ \vec{u} \cdot \vec{v} \;=\; u_1 v_1 + u_2 v_2 \;=\; |\vec{u}|\,|\vec{v}|\cos\theta, $$

where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$ when placed tail-to-tail ($0 \le \theta \le 180^{\circ}$). The result is a scalar, not a vector.其中 $\theta$ 为两向量共起点后的夹角（$0 \le \theta \le 180^{\circ}$）。结果是标量，不是向量。

  • Perpendicularity test.垂直判别。 $$\vec{u} \cdot \vec{v} = 0 \;\Longleftrightarrow\; \vec{u} \perp \vec{v}$$ (for nonzero $\vec{u}, \vec{v}$). This single fact does most of the work on IB HL and engineering problems.（$\vec{u}, \vec{v}$ 均非零）。这一条等价关系几乎包揽 IB HL 与工程类题目大半工作。
  • Angle formula.夹角公式。 $$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|}.$$
  • Sign tells the angle's type.符号标示夹角类型。 $\vec{u} \cdot \vec{v} > 0$: acute. $= 0$: right. $< 0$: obtuse.$\vec{u} \cdot \vec{v} > 0$：锐角；$= 0$：直角；$< 0$：钝角。
  • Self-dot equals magnitude squared.自身点积等于模长平方。 $$\vec{v} \cdot \vec{v} = |\vec{v}|^2.$$
  • Properties.性质。 Commutative ($\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$), distributive ($\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u}\cdot\vec{v} + \vec{u}\cdot\vec{w}$), and scalar-compatible ($(k\vec{u}) \cdot \vec{v} = k(\vec{u} \cdot \vec{v})$). Not associative , the expression $(\vec{u}\cdot\vec{v})\cdot\vec{w}$ does not type-check because $\vec{u}\cdot\vec{v}$ is a scalar.交换律 ($\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$)、分配律 ($\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u}\cdot\vec{v} + \vec{u}\cdot\vec{w}$)、与数乘相容 ($(k\vec{u}) \cdot \vec{v} = k(\vec{u} \cdot \vec{v})$)。无结合律 , $(\vec{u}\cdot\vec{v})\cdot\vec{w}$ 类型不合（$\vec{u}\cdot\vec{v}$ 是标量）。
Worked Example 4 · Angle between two vectors例题 4 · 两向量夹角

Find the angle (to two decimals) between $\vec{u} = \langle 3, 4\rangle$ and $\vec{v} = \langle -2, 6\rangle$.求 $\vec{u} = \langle 3, 4\rangle$ 与 $\vec{v} = \langle -2, 6\rangle$ 的夹角（保留两位小数）。

Dot product.求点积。

$$ \vec{u} \cdot \vec{v} = (3)(-2) + (4)(6) = -6 + 24 = 18. $$

Magnitudes.两模长。 $|\vec{u}| = \sqrt{9 + 16} = 5$, $|\vec{v}| = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$.$|\vec{u}| = \sqrt{9 + 16} = 5$，$|\vec{v}| = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$。

Apply the formula.套用公式。

$$ \cos\theta = \frac{18}{5 \cdot 2\sqrt{10}} = \frac{18}{10\sqrt{10}} = \frac{9}{5\sqrt{10}} \approx 0.5692. $$ $$ \theta = \arccos(0.5692) \approx 55.30^{\circ}. $$

Sanity-check.合理性核验。 Positive dot product $\Rightarrow$ acute angle, and $55.30^{\circ}$ is acute ✓. A quick sketch confirms both arrows lie in the upper half-plane.点积为正 $\Rightarrow$ 锐角，$55.30^{\circ}$ 为锐角 ✓。略画图可见两箭头均位于上半平面。

Compute $\vec{u} \cdot \vec{v}$ for $\vec{u} = \langle 5, -3\rangle$ and $\vec{v} = \langle 2, 4\rangle$.求 $\vec{u} = \langle 5, -3\rangle$、$\vec{v} = \langle 2, 4\rangle$ 的点积。
§4 · Q1
$-2$
$22$
$\langle 10, -12\rangle$
$0$
$\vec{u}\cdot\vec{v} = (5)(2) + (-3)(4) = 10 - 12 = -2$. (A scalar , not a vector. The negative sign means the angle between $\vec{u}$ and $\vec{v}$ is obtuse.)$\vec{u}\cdot\vec{v} = (5)(2) + (-3)(4) = 10 - 12 = -2$。（结果是标量，不是向量。负号说明夹角为钝角。）
Multiply matching components, then add. The dot product returns a single number, not another vector.对应分量相乘后相加。点积返回一个数，不是另一个向量。
Find the value of $k$ so that $\vec{u} = \langle 4, k\rangle$ is perpendicular to $\vec{v} = \langle 6, -3\rangle$.求使 $\vec{u} = \langle 4, k\rangle$ 与 $\vec{v} = \langle 6, -3\rangle$ 垂直的 $k$。
§4 · Q2
$-2$
$8$
$3$
$0$
Perpendicular $\Leftrightarrow$ $\vec{u}\cdot\vec{v} = 0$. So $(4)(6) + (k)(-3) = 0 \Rightarrow 24 - 3k = 0 \Rightarrow k = 8$. Quick check: $\langle 4, 8\rangle\cdot\langle 6,-3\rangle = 24 - 24 = 0$ ✓.垂直 $\Leftrightarrow$ $\vec{u}\cdot\vec{v} = 0$。即 $(4)(6) + (k)(-3) = 0 \Rightarrow 24 - 3k = 0 \Rightarrow k = 8$。验证：$\langle 4, 8\rangle\cdot\langle 6,-3\rangle = 24 - 24 = 0$ ✓。
Set the dot product equal to zero and solve for $k$ , perpendicularity is a single linear equation in $k$.令点积为零并解 $k$ , 垂直化为关于 $k$ 的一个线性方程。
Section 5 · Vectors in 3-D第 5 节 · 三维向量

Vectors in 3-D: components, magnitude, dot product Honors for US, out-of-core BC + AB三维向量：分量、模长、点积 荣誉 — US 核心 / BC、AB 之外

Everything in §2-§4 carries over to three components.§2-§4 全部内容平移到三维。 $$ \vec{v} = \langle a, b, c \rangle = a\vec{i} + b\vec{j} + c\vec{k}, \qquad |\vec{v}| = \sqrt{a^2 + b^2 + c^2}, \qquad \vec{u}\cdot\vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3. $$
  • Standard basis.标准基。 $\vec{i} = \langle 1, 0, 0\rangle$, $\vec{j} = \langle 0, 1, 0\rangle$, $\vec{k} = \langle 0, 0, 1\rangle$.$\vec{i} = \langle 1, 0, 0\rangle$、$\vec{j} = \langle 0, 1, 0\rangle$、$\vec{k} = \langle 0, 0, 1\rangle$。
  • Magnitude formula.模长公式。 Pythagoras twice , once in the $xy$-plane gives $\sqrt{a^2 + b^2}$, then once more with the $z$-leg gives $\sqrt{a^2 + b^2 + c^2}$.两次勾股定理 , 在 $xy$ 平面内得 $\sqrt{a^2 + b^2}$，再与 $z$ 边合用得 $\sqrt{a^2 + b^2 + c^2}$。
  • Dot product still detects perpendicularity.点积仍可检验垂直。 $\vec{u}\cdot\vec{v} = 0 \Leftrightarrow \vec{u}\perp\vec{v}$, exactly as in 2-D.$\vec{u}\cdot\vec{v} = 0 \Leftrightarrow \vec{u}\perp\vec{v}$，与二维完全一致。
  • Distance between two 3-D points.两 3D 点的距离。 $P_1 P_2 = |\vec{P_1 P_2}|$ , same vector approach, just one more coordinate.$P_1 P_2 = |\vec{P_1 P_2}|$ , 与二维方法相同，只是多一个坐标。
Beyond this section. 3-D lines and planes (vector / parametric / scalar equations, intersections) are MCV4U Strand C content but live in the IB HL companion linked at the bottom of this guide. We stop here at the algebra of 3-D vectors.本节范围之外。 3D 直线与平面（向量 / 参数 / 标量方程及相交问题）属 MCV4U 单元 C 内容，但本指南未含，请参见底部链接的 IB HL 衔接。本节止于 3D 向量代数。
Worked Example 5 · Magnitude and angle in 3-D例题 5 · 3D 模长与夹角

Let $\vec{u} = \langle 2, -1, 2\rangle$ and $\vec{v} = \langle 1, 2, 2\rangle$. Find $|\vec{u}|$, $|\vec{v}|$, and the angle $\theta$ between them (to two decimals).设 $\vec{u} = \langle 2, -1, 2\rangle$、$\vec{v} = \langle 1, 2, 2\rangle$。求 $|\vec{u}|$、$|\vec{v}|$ 及两向量夹角 $\theta$（保留两位小数）。

Magnitudes.模长。

$$ |\vec{u}| = \sqrt{4 + 1 + 4} = 3, \qquad |\vec{v}| = \sqrt{1 + 4 + 4} = 3. $$

Dot product.点积。

$$ \vec{u}\cdot\vec{v} = (2)(1) + (-1)(2) + (2)(2) = 2 - 2 + 4 = 4. $$

Angle.夹角。

$$ \cos\theta = \frac{4}{3 \cdot 3} = \frac{4}{9}, \qquad \theta = \arccos\!\left(\tfrac{4}{9}\right) \approx 63.61^{\circ}. $$

Sanity-check.合理性核验。 Dot product positive $\Rightarrow$ acute, and $63.61^{\circ}$ is acute ✓. Both magnitudes equal $3$ , these are the legs of a 1-2-2 / 2-1-2 pair whose square sums are both $9$.点积为正 $\Rightarrow$ 锐角，$63.61^{\circ}$ 为锐角 ✓。两模长均为 $3$ , 1-2-2 / 2-1-2 平方和都是 $9$。

Find $|\vec{w}|$ for $\vec{w} = \langle 1, -2, 2\rangle$.求 $\vec{w} = \langle 1, -2, 2\rangle$ 的模长。
§5 · Q1
$\sqrt{5}$
$1$
$3$
$5$
$|\vec{w}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$. (Same formula as in 2-D, just one more squared component.)$|\vec{w}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$。（与二维公式相同，多一个分量平方。）
$|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$. Square all three components, sum, then square-root.$|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$。三个分量都平方，求和后开方。
Are $\vec{u} = \langle 1, 2, -1\rangle$ and $\vec{v} = \langle 3, -1, 1\rangle$ perpendicular?向量 $\vec{u} = \langle 1, 2, -1\rangle$ 与 $\vec{v} = \langle 3, -1, 1\rangle$ 是否垂直？
§5 · Q2
No, dot product is $4$.否，点积为 $4$。
Yes, dot product is $0$.是，点积为 $0$。
No, dot product is $-2$.否，点积为 $-2$。
Cannot tell without magnitudes.不知模长无法判断。
$\vec{u}\cdot\vec{v} = (1)(3) + (2)(-1) + (-1)(1) = 3 - 2 - 1 = 0$. Perpendicular , in 3-D the same zero-dot-product test applies.$\vec{u}\cdot\vec{v} = (1)(3) + (2)(-1) + (-1)(1) = 3 - 2 - 1 = 0$。垂直 , 3D 中"点积为零"判别同样适用。
Compute the 3-D dot product (three products, summed). Zero result $\Leftrightarrow$ perpendicular.求 3D 点积（三对乘积之和）。结果为零 $\Leftrightarrow$ 垂直。
Section 6 · Cross product (3-D only)第 6 节 · 叉积（仅 3D）

The Cross Product: a vector perpendicular to both Honors for US, out-of-core BC + AB叉积：一条与两者都垂直的向量 荣誉 — US 核心 / BC、AB 之外

Curriculum note.课纲提示。 Cross product is a 3-D-only operation. It is core in Ontario MCV4U Strand C. For US students it lives inside the (+) HSN-VM cluster, hence Honors / AP-feeder. For BC and AB it is Out-of-core , not in the published Pre-Calc or Math 30-1 curricula. AP Physics C: Mechanics uses it for torque ($\vec{\tau} = \vec{r}\times\vec{F}$); AP Physics C: E&M uses it for the magnetic force ($\vec{F} = q\vec{v}\times\vec{B}$).叉积仅 3D 适用。在安大略 MCV4U 单元 C 中为核心。对美国学生归入 (+) HSN-VM 簇，因此标 Honors / AP 衔接。对 BC、AB 学生标 Out-of-core , 不在已发布 Pre-Calc 或 Math 30-1 课纲内。AP Physics C：力学用它表力矩 ($\vec{\tau} = \vec{r}\times\vec{F}$)；AP Physics C：电磁用它表磁力 ($\vec{F} = q\vec{v}\times\vec{B}$)。
Cross product , output is a vector.叉积 , 结果是向量。 $$ \vec{u} \times \vec{v} \;=\; \langle u_2 v_3 - u_3 v_2, \;\; u_3 v_1 - u_1 v_3, \;\; u_1 v_2 - u_2 v_1\rangle. $$

Magnitude and geometric properties:模长与几何性质：

$$ |\vec{u}\times\vec{v}| \;=\; |\vec{u}|\,|\vec{v}|\,\sin\theta, $$

where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$ ($0 \le \theta \le 180^{\circ}$).$\theta$ 为两向量夹角（$0 \le \theta \le 180^{\circ}$）。

  • Perpendicular to both.同时垂直两向量。 $\vec{u}\times\vec{v}$ is perpendicular to $\vec{u}$ and to $\vec{v}$; quick check: $\vec{u}\cdot(\vec{u}\times\vec{v}) = 0$ and $\vec{v}\cdot(\vec{u}\times\vec{v}) = 0$.$\vec{u}\times\vec{v}$ 同时垂直 $\vec{u}$ 与 $\vec{v}$；快速验算 $\vec{u}\cdot(\vec{u}\times\vec{v}) = 0$、$\vec{v}\cdot(\vec{u}\times\vec{v}) = 0$。
  • Right-hand rule.右手定则。 Curl the fingers of your right hand from $\vec{u}$ towards $\vec{v}$; the thumb points along $\vec{u}\times\vec{v}$. Reversing the order flips the sign: $\vec{v}\times\vec{u} = -(\vec{u}\times\vec{v})$.右手四指由 $\vec{u}$ 弯向 $\vec{v}$，拇指方向即 $\vec{u}\times\vec{v}$。交换顺序反号：$\vec{v}\times\vec{u} = -(\vec{u}\times\vec{v})$。
  • Parallelogram area.平行四边形面积。 $|\vec{u}\times\vec{v}|$ equals the area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$. Triangle area: half of that.$|\vec{u}\times\vec{v}|$ 等于以 $\vec{u}$、$\vec{v}$ 为边的平行四边形面积。三角形面积为其一半。
  • Parallel detector.平行判别。 $\vec{u}\times\vec{v} = \vec{0} \Leftrightarrow \vec{u} \parallel \vec{v}$ (for nonzero vectors) , the cross-product analogue of "$\vec{u}\cdot\vec{v} = 0$ means perpendicular."$\vec{u}\times\vec{v} = \vec{0} \Leftrightarrow \vec{u} \parallel \vec{v}$（两者非零） , 与"$\vec{u}\cdot\vec{v} = 0$ 即垂直"对偶。
  • Mnemonic determinant.行列式助记。 Many courses introduce $\vec{u}\times\vec{v}$ via a $3\times 3$ "determinant" with $\vec{i}, \vec{j}, \vec{k}$ in the top row, $\vec{u}$ in the second, $\vec{v}$ in the third. Both ways give the same components.许多课程通过 $3\times 3$ "行列式"引入：第一行 $\vec{i}, \vec{j}, \vec{k}$，第二行 $\vec{u}$，第三行 $\vec{v}$。两法所得分量相同。
Worked Example 6 · Cross product and parallelogram area例题 6 · 叉积与平行四边形面积

Let $\vec{u} = \langle 1, 2, 3\rangle$ and $\vec{v} = \langle 4, 5, 6\rangle$. Compute $\vec{u}\times\vec{v}$, verify it is perpendicular to both, and find the area of the parallelogram they span.设 $\vec{u} = \langle 1, 2, 3\rangle$、$\vec{v} = \langle 4, 5, 6\rangle$。求 $\vec{u}\times\vec{v}$，验证其同时垂直两向量，并求两向量所张平行四边形的面积。

Apply the cross-product formula.套用叉积公式。

$$ \vec{u}\times\vec{v} = \langle (2)(6) - (3)(5), \;\; (3)(4) - (1)(6), \;\; (1)(5) - (2)(4)\rangle = \langle 12 - 15, \;\; 12 - 6, \;\; 5 - 8\rangle = \langle -3, \;\; 6, \;\; -3\rangle. $$

Perpendicularity check.垂直性核验。

$$ \vec{u}\cdot(\vec{u}\times\vec{v}) = (1)(-3) + (2)(6) + (3)(-3) = -3 + 12 - 9 = 0. \;\checkmark $$ $$ \vec{v}\cdot(\vec{u}\times\vec{v}) = (4)(-3) + (5)(6) + (6)(-3) = -12 + 30 - 18 = 0. \;\checkmark $$

Parallelogram area.平行四边形面积。

$$ |\vec{u}\times\vec{v}| = \sqrt{(-3)^2 + 6^2 + (-3)^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6} \approx 7.35. $$

Sanity-check.合理性核验。 $|\vec{u}| = \sqrt{14}$, $|\vec{v}| = \sqrt{77}$, so $|\vec{u}|\,|\vec{v}| = \sqrt{1078} \approx 32.83$. Our $|\vec{u}\times\vec{v}| \approx 7.35$ stays well below the upper bound , consistent with $\sin\theta < 1$.$|\vec{u}| = \sqrt{14}$、$|\vec{v}| = \sqrt{77}$，故 $|\vec{u}|\,|\vec{v}| = \sqrt{1078} \approx 32.83$。$|\vec{u}\times\vec{v}| \approx 7.35$ 远低于此上界 , 与 $\sin\theta < 1$ 一致。

Compute $\vec{i}\times\vec{j}$.计算 $\vec{i}\times\vec{j}$。
§6 · Q1
$\vec{0}$
$-\vec{k}$
$\vec{k}$
$1$
$\vec{i} = \langle 1,0,0\rangle$, $\vec{j} = \langle 0,1,0\rangle$. Cross product: $\langle (0)(0) - (0)(1), (0)(0) - (1)(0), (1)(1) - (0)(0)\rangle = \langle 0, 0, 1\rangle = \vec{k}$. Right-hand rule: point fingers along $\vec{i}$, curl toward $\vec{j}$, thumb is $\vec{k}$. (Standard cycle: $\vec{i}\times\vec{j} = \vec{k}$, $\vec{j}\times\vec{k} = \vec{i}$, $\vec{k}\times\vec{i} = \vec{j}$.)$\vec{i} = \langle 1,0,0\rangle$、$\vec{j} = \langle 0,1,0\rangle$。叉积：$\langle 0, 0, 1\rangle = \vec{k}$。右手定则：四指沿 $\vec{i}$，弯向 $\vec{j}$，拇指即 $\vec{k}$。（标准循环：$\vec{i}\times\vec{j} = \vec{k}$、$\vec{j}\times\vec{k} = \vec{i}$、$\vec{k}\times\vec{i} = \vec{j}$。）
The basis vectors cycle: $\vec{i}\times\vec{j} = \vec{k}$, $\vec{j}\times\vec{k} = \vec{i}$, $\vec{k}\times\vec{i} = \vec{j}$. Reverse the order to flip the sign.基向量循环：$\vec{i}\times\vec{j} = \vec{k}$、$\vec{j}\times\vec{k} = \vec{i}$、$\vec{k}\times\vec{i} = \vec{j}$。次序反则符号反。
The triangle with vertices $P(0,0,0)$, $Q(2,0,0)$, $R(0,3,4)$ has area equal to:顶点为 $P(0,0,0)$、$Q(2,0,0)$、$R(0,3,4)$ 的三角形面积为：
§6 · Q2
$10$
$5$
$6$
$\sqrt{29}$
Edge vectors: $\vec{PQ} = \langle 2,0,0\rangle$, $\vec{PR} = \langle 0,3,4\rangle$. Cross product: $\vec{PQ}\times\vec{PR} = \langle (0)(4) - (0)(3), (0)(0) - (2)(4), (2)(3) - (0)(0)\rangle = \langle 0, -8, 6\rangle$. Magnitude $= \sqrt{0 + 64 + 36} = \sqrt{100} = 10$. Triangle area $= \tfrac{1}{2}|\vec{PQ}\times\vec{PR}| = 5$.边向量：$\vec{PQ} = \langle 2,0,0\rangle$、$\vec{PR} = \langle 0,3,4\rangle$。叉积 $= \langle 0, -8, 6\rangle$。模长 $= \sqrt{0 + 64 + 36} = 10$。三角形面积 $= \tfrac{1}{2}|\vec{PQ}\times\vec{PR}| = 5$。
Triangle area $= \tfrac{1}{2}|\vec{PQ}\times\vec{PR}|$. Take half the parallelogram area.三角形面积 $= \tfrac{1}{2}|\vec{PQ}\times\vec{PR}|$。取平行四边形面积的一半。
Section 7 · Applications第 7 节 · 应用

Applications: Forces, Inclined Planes, Work Honors for US, out-of-core BC + AB应用：力、斜面、功 荣誉 — US 核心 / BC、AB 之外

Three recurring vector applications.三类常见向量应用。
  • Resolving a force.力的分解。 A force $\vec{F}$ of magnitude $F$ at angle $\theta$ above the horizontal splits into $F_x = F\cos\theta$ (horizontal) and $F_y = F\sin\theta$ (vertical) , the right-triangle picture from Unit 7 reused inside a force.大小为 $F$、与水平面成 $\theta$ 角的力 $\vec{F}$ 分解为 $F_x = F\cos\theta$（水平）与 $F_y = F\sin\theta$（垂直） , 把 Unit 7 的直角三角形图用在力上。
  • Inclined plane.斜面。 On a frictionless ramp at angle $\theta$ above horizontal, gravity $mg$ resolves into a component along the ramp ($mg\sin\theta$, pulling the block down the slope) and a component perpendicular to the ramp ($mg\cos\theta$, pressing into the surface). These two are the workhorses of AP Physics 1 / C kinematics-on-a-ramp problems.在与水平面成 $\theta$ 角的无摩擦斜面上，重力 $mg$ 分解为沿斜面分量（$mg\sin\theta$，沿坡下拉）与垂直斜面分量（$mg\cos\theta$，压向斜面）。这两条是 AP Physics 1 / C 中斜面运动问题的主力。
  • Work as a dot product.功作为点积。 $$W = \vec{F}\cdot\vec{d} = |\vec{F}|\,|\vec{d}|\cos\theta.$$ Only the component of $\vec{F}$ along $\vec{d}$ does work. Forces perpendicular to motion (e.g. normal force, tension on a swinging pendulum) do zero work because $\cos 90^{\circ} = 0$.仅 $\vec{F}$ 沿 $\vec{d}$ 方向的分量做功。与位移垂直的力（如法向力、摆动绳张力）做零功，因 $\cos 90^{\circ} = 0$。
  • Velocity components.速度分量。 A projectile fired at speed $v$ at angle $\theta$ above horizontal has $v_x = v\cos\theta$ and $v_y = v\sin\theta$ , exactly the §2 conversion from magnitude / angle to components, dressed up as a physics problem.以速率 $v$、与水平面 $\theta$ 角抛出的物体满足 $v_x = v\cos\theta$、$v_y = v\sin\theta$ , 正是 §2 中"模长 / 角 $\to$ 分量"在物理情境下的复用。
Feeder. AP Physics Unit 1 Kinematics (see `AP Physics/Study Guides/Unit_1_Kinematics.html`) uses these decompositions on day one. This guide is the algebra prereq.衔接。 AP Physics Unit 1 运动学（见 `AP Physics/Study Guides/Unit_1_Kinematics.html`）第一天就用到上述分解。本指南是其代数前置。
Worked Example 7a · Block on a frictionless incline例题 7a · 无摩擦斜面上的物块

A $5$-kg block sits on a frictionless ramp inclined at $30^{\circ}$ above horizontal. Use $g = 9.8 \text{ m/s}^2$. Find the component of gravity acting along the ramp (the net force pulling the block down the slope), to one decimal.一 $5$ kg 物块置于与水平面成 $30^{\circ}$ 的无摩擦斜面上。取 $g = 9.8 \text{ m/s}^2$。求沿斜面方向的重力分量（即沿坡下拉物块的合力），保留一位小数。

Set up.设置。 Gravity acts straight down with magnitude $mg = 5 \cdot 9.8 = 49$ N. Resolve along axes parallel and perpendicular to the ramp.重力竖直向下，大小 $mg = 5 \cdot 9.8 = 49$ N。沿平行与垂直于斜面的两轴分解。

Along-the-ramp component.沿斜面分量。

$$ F_{\parallel} = mg\sin 30^{\circ} = 49 \cdot 0.5 = 24.5 \text{ N}. $$

Perpendicular component (normal-force size).垂直分量（即法向力大小）。

$$ F_{\perp} = mg\cos 30^{\circ} \approx 49 \cdot 0.866 \approx 42.4 \text{ N}. $$

Answer in a sentence with units.用整句加单位作答。 The net downslope force is approximately $24.5$ N.沿坡向下的合力约为 $24.5$ N。

Sanity-check.合理性核验。 $\sin 30^{\circ} = 1/2$, so the along-ramp force is exactly half of the weight , matches. As $\theta\to 0$ the ramp goes flat and $F_{\parallel}\to 0$; as $\theta\to 90^{\circ}$ the ramp goes vertical and $F_{\parallel}\to mg$. Both limits behave correctly.$\sin 30^{\circ} = 1/2$，沿坡力恰为重量一半 , 吻合。$\theta\to 0$ 时斜面变平，$F_{\parallel}\to 0$；$\theta\to 90^{\circ}$ 时斜面变竖直，$F_{\parallel}\to mg$。两端极限行为均正确。

Worked Example 7b · Work done by a force at an angle例题 7b · 有夹角力所做的功

A child pulls a sled with a rope at $40^{\circ}$ above horizontal, exerting a force of $80$ N. The sled moves $25$ m horizontally. Find the work done by the rope force, to the nearest joule.一小孩以与水平面 $40^{\circ}$ 角拉雪橇上的绳，施力 $80$ N。雪橇水平移动 $25$ m。求绳力所做的功，四舍五入到焦耳。

Apply the dot-product work formula.套用点积功公式。

$$ W = \vec{F}\cdot\vec{d} = |\vec{F}|\,|\vec{d}|\cos\theta = 80 \cdot 25 \cdot \cos 40^{\circ} \approx 2000 \cdot 0.766 \approx 1532 \text{ J}. $$

Answer in a sentence with units.用整句加单位作答。 The rope force does approximately $1532$ J of work.绳力做的功约为 $1532$ J。

Sanity-check.合理性核验。 If the rope were horizontal ($\theta = 0$), $W$ would be $80 \cdot 25 = 2000$ J. The $40^{\circ}$ tilt costs us a factor of $\cos 40^{\circ}\approx 0.77$; $1532 / 2000 \approx 0.77$ ✓. Only the horizontal component $|\vec{F}|\cos\theta$ does work on the (horizontally moving) sled.若绳水平（$\theta = 0$），$W = 80 \cdot 25 = 2000$ J。$40^{\circ}$ 倾角带来因子 $\cos 40^{\circ}\approx 0.77$；$1532 / 2000 \approx 0.77$ ✓。仅水平分量 $|\vec{F}|\cos\theta$ 对（水平移动的）雪橇做功。

A projectile is launched at $20$ m/s at $35^{\circ}$ above horizontal. Find its horizontal velocity component to one decimal.一抛体以 $20$ m/s 速率、与水平 $35^{\circ}$ 角发射。求水平速度分量，保留一位小数。
§7 · Q1
$11.5$ m/s
$20.0$ m/s
$16.4$ m/s
$8.0$ m/s
$v_x = v\cos\theta = 20\cos 35^{\circ} \approx 20 \cdot 0.8192 \approx 16.4$ m/s. (Vertical: $v_y = 20\sin 35^{\circ} \approx 11.5$ m/s , option (a) is $v_y$, the trap.)$v_x = v\cos\theta = 20\cos 35^{\circ} \approx 20 \cdot 0.8192 \approx 16.4$ m/s。（竖直：$v_y = 20\sin 35^{\circ} \approx 11.5$ m/s , 选项 (a) 是 $v_y$ 的陷阱。）
$v_x = v\cos\theta$ uses cosine; sine gives $v_y$. Below $45^{\circ}$, cosine dominates and the horizontal component is larger.$v_x = v\cos\theta$ 用余弦；正弦给 $v_y$。$\theta < 45^{\circ}$ 时余弦较大，水平分量更大。
A constant force $\vec{F} = \langle 4, 0, 3\rangle$ N moves an object through displacement $\vec{d} = \langle 2, 5, -1\rangle$ m. Find the work done, in joules.恒力 $\vec{F} = \langle 4, 0, 3\rangle$ N 推动物体经位移 $\vec{d} = \langle 2, 5, -1\rangle$ m。求所做的功（焦耳）。
§7 · Q2
$5$ J
$11$ J
$\langle 8, 0, -3\rangle$ J
$0$ J
$W = \vec{F}\cdot\vec{d} = (4)(2) + (0)(5) + (3)(-1) = 8 + 0 - 3 = 5$ J. Work is a scalar , option (c) confuses the dot product with component-wise multiplication.$W = \vec{F}\cdot\vec{d} = (4)(2) + (0)(5) + (3)(-1) = 8 + 0 - 3 = 5$ J。功是标量 , 选项 (c) 把点积错当成按分量乘法。
Work = $\vec{F}\cdot\vec{d}$, a scalar. Use the 3-D dot product: three products, summed.功 = $\vec{F}\cdot\vec{d}$，是标量。用 3D 点积：三对乘积之和。
Exam Tactics考试策略

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

Setup discipline解题前的纪律
  • Always write both representations.两种表示都写。 Sketch the arrow and write the component form. The picture catches sign errors; the algebra catches geometric oversights.画箭头并写分量。图避免符号错误；代数避免几何疏忽。
  • Subtract terminal $-$ initial."终点 $-$ 起点"。 $\vec{AB}$ always equals (coords of $B$) $-$ (coords of $A$). Mixing the order is the most common silent error on Section 1.$\vec{AB}$ 总等于（$B$ 坐标）$-$（$A$ 坐标）。颠倒顺序是 §1 最常见的隐性错误。
  • Scalar vs vector type-check.标量与向量的类型核查。 Dot product returns a number. Cross product returns a vector. If your answer to a dot-product question is a vector, you've conflated the two.点积返回数。叉积返回向量。若点积题答案写成向量，就是混淆了两者。
Components, magnitude, unit vectors (§1-§3)分量、模长、单位向量（§1-§3）
  • Magnitude is always non-negative.模长非负。 If your $|\vec{v}|$ comes out negative, you forgot to square a component before adding.若 $|\vec{v}|$ 为负，必是某分量未先平方就加。
  • Pythagorean triples save time.勾股数省时间。 Recognise 3-4-5, 5-12-13, 7-24-25, 8-15-17, and the 3-D 1-2-2 / 2-2-1 / 3-4-12 triples. They turn $\sqrt{\;}$ into mental arithmetic.认出 3-4-5、5-12-13、7-24-25、8-15-17 与 3D 中的 1-2-2 / 2-2-1 / 3-4-12 三元组。可把开方化为口算。
  • From magnitude / angle to components.由模长 / 角求分量。 $\vec{v} = \langle r\cos\theta, r\sin\theta\rangle$ in 2-D. Confirm the quadrant matches by sign-checking each component.2D 中 $\vec{v} = \langle r\cos\theta, r\sin\theta\rangle$。按符号核查所属象限。
Dot and cross products (§4-§6) Honors for US, out-of-core BC + AB点积与叉积（§4-§6）荣誉 — US 核心 / BC、AB 之外
  • Use dot product for angles and perpendicularity.点积用于求夹角与判垂直。 $\cos\theta = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}$. The sign of $\vec{u}\cdot\vec{v}$ alone tells you acute / right / obtuse before you compute $\theta$.$\cos\theta = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}$。仅看 $\vec{u}\cdot\vec{v}$ 符号即可判锐 / 直 / 钝，无需算 $\theta$。
  • Use cross product for area and "perpendicular to both".叉积用于面积与"同时垂直两者"。 $|\vec{u}\times\vec{v}|$ is the parallelogram area; $\tfrac{1}{2}|\vec{u}\times\vec{v}|$ is the triangle area on the same two edges. Sign of the third component tells you orientation (CCW vs CW in the $xy$-plane).$|\vec{u}\times\vec{v}|$ 为平行四边形面积；以同两边构成的三角形面积为 $\tfrac{1}{2}|\vec{u}\times\vec{v}|$。第三分量符号告诉你在 $xy$ 平面中的取向（逆时针 / 顺时针）。
  • Verify perpendicularity after a cross product.叉积后核验垂直。 A 10-second check: $\vec{u}\cdot(\vec{u}\times\vec{v}) = 0$ and $\vec{v}\cdot(\vec{u}\times\vec{v}) = 0$. If either fails, you mis-cycled the determinant.10 秒核验：$\vec{u}\cdot(\vec{u}\times\vec{v}) = 0$、$\vec{v}\cdot(\vec{u}\times\vec{v}) = 0$。任一不为零即行列式循环写错。
  • Right-hand rule is non-commutative.右手定则不交换。 $\vec{v}\times\vec{u} = -(\vec{u}\times\vec{v})$ , if a textbook answer has the opposite sign, you may have swapped the order.$\vec{v}\times\vec{u} = -(\vec{u}\times\vec{v})$ , 若教材答案符号相反，可能是次序颠倒。
Applied-problem hygiene (§7)应用题规范（§7）
  • Pick axes parallel to motion.选与运动方向平行的轴。 On an inclined plane, tilt the $x$-axis along the ramp surface; gravity then splits cleanly into $mg\sin\theta$ (along ramp) and $mg\cos\theta$ (perpendicular). Aligning axes with motion is the single biggest setup win.在斜面上让 $x$ 轴沿坡面；重力即整齐分为 $mg\sin\theta$（沿坡）与 $mg\cos\theta$（垂直）。轴对齐运动方向是设置上最大的收益。
  • Work uses cosine, not sine.功用余弦，不用正弦。 $W = |\vec{F}||\vec{d}|\cos\theta$. Perpendicular forces do zero work because $\cos 90^{\circ} = 0$ , normal force, magnetic force on a moving charge, and tension on a swinging pendulum all sit at $\theta = 90^{\circ}$.$W = |\vec{F}||\vec{d}|\cos\theta$。垂直力做零功，因 $\cos 90^{\circ} = 0$ , 法向力、动电荷所受磁力、摆动绳张力均处 $\theta = 90^{\circ}$。
  • Answer in a sentence with units.用整句加单位作答。 "The block accelerates at $4.9$ m/s$^2$ down the ramp" earns a communication mark; "4.9" alone may not."物块沿坡以 $4.9$ m/s$^2$ 加速下滑"可拿表达分；只写"4.9"可能不行。
Active Recall主动复习

Flashcards闪卡

0 / 14 flipped0 / 14 已翻
Components of $\vec{AB}$?$\vec{AB}$ 的分量？
$$\vec{AB} = B - A$$ terminal minus initial (CCSSM HSN-VM.A.2)终点减起点（CCSSM HSN-VM.A.2）
Magnitude of $\langle a, b\rangle$?$\langle a, b\rangle$ 的模长？
$$|\vec{v}| = \sqrt{a^2 + b^2}$$
Unit vector in the direction of $\vec{v}$?$\vec{v}$ 方向的单位向量？
$$\hat{v} = \frac{\vec{v}}{|\vec{v}|}$$
$\vec{v}$ from magnitude $r$ and angle $\theta$?由模长 $r$ 与角 $\theta$ 求 $\vec{v}$？
$$\vec{v} = \langle r\cos\theta, \; r\sin\theta\rangle$$
Vector addition, component-wise?向量加法（按分量）？
$$\vec{u} + \vec{v} = \langle u_1 + v_1, \; u_2 + v_2\rangle$$
Scalar multiplication, magnitude rule?数乘的模长规则？
$$|k\vec{v}| = |k|\,|\vec{v}|$$
Dot product (2-D)?点积（2D）？
$$\vec{u}\cdot\vec{v} = u_1 v_1 + u_2 v_2 = |\vec{u}||\vec{v}|\cos\theta$$
Perpendicularity test?垂直判别？
$$\vec{u}\cdot\vec{v} = 0 \;\Leftrightarrow\; \vec{u} \perp \vec{v}$$
Angle between $\vec{u}$ and $\vec{v}$?$\vec{u}$ 与 $\vec{v}$ 的夹角？
$$\cos\theta = \frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}$$
Magnitude of $\langle a,b,c\rangle$?$\langle a,b,c\rangle$ 的模长？
$$|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$$
Cross-product magnitude?叉积模长？
$$|\vec{u}\times\vec{v}| = |\vec{u}||\vec{v}|\sin\theta$$ = parallelogram area= 平行四边形面积
$\vec{i}\times\vec{j}, \vec{j}\times\vec{k}, \vec{k}\times\vec{i}$?$\vec{i}\times\vec{j}, \vec{j}\times\vec{k}, \vec{k}\times\vec{i}$？
$$\vec{k}, \; \vec{i}, \; \vec{j}$$ cyclic; reversed order flips sign循环；次序反则符号反
Component of force along an incline?力沿斜面的分量？
$$F_{\parallel} = mg\sin\theta, \quad F_{\perp} = mg\cos\theta$$
Work as a vector quantity?功的向量定义？
$$W = \vec{F}\cdot\vec{d} = |\vec{F}||\vec{d}|\cos\theta$$
Mixed Practice综合练习

Practice Quiz综合测验

Given $A = (1, -2)$ and $B = (7, 6)$, find $|\vec{AB}|$.已知 $A = (1, -2)$、$B = (7, 6)$，求 $|\vec{AB}|$。
Q1
$8$
$10$
$\sqrt{50}$
$14$
$\vec{AB} = \langle 6, 8\rangle$. $|\vec{AB}| = \sqrt{36 + 64} = \sqrt{100} = 10$. (Scaled 3-4-5.)$\vec{AB} = \langle 6, 8\rangle$。$|\vec{AB}| = \sqrt{36 + 64} = \sqrt{100} = 10$。（3-4-5 缩放。）
Find components first ($B - A$), then take the magnitude.先求分量（$B - A$），再求模长。
Find the unit vector in the same direction as $\vec{v} = \langle 9, -12\rangle$.求与 $\vec{v} = \langle 9, -12\rangle$ 同向的单位向量。
Q2
$\langle 9, -12\rangle$
$\langle 3, -4\rangle$
$\langle 3/5, -4/5\rangle$
$\langle 1/9, -1/12\rangle$
$|\vec{v}| = \sqrt{81 + 144} = \sqrt{225} = 15$. $\hat{v} = \vec{v}/15 = \langle 9/15, -12/15\rangle = \langle 3/5, -4/5\rangle$. Check: $(3/5)^2 + (4/5)^2 = 1$ ✓.$|\vec{v}| = \sqrt{81 + 144} = 15$。$\hat{v} = \vec{v}/15 = \langle 3/5, -4/5\rangle$。验算：$(3/5)^2 + (4/5)^2 = 1$ ✓。
Divide each component by the magnitude. The result must have magnitude $1$.每分量除以模长。结果模长必为 $1$。
Compute $\vec{u}\cdot\vec{v}$ for $\vec{u} = \langle 2, -3\rangle$ and $\vec{v} = \langle 6, 4\rangle$.求 $\vec{u} = \langle 2, -3\rangle$、$\vec{v} = \langle 6, 4\rangle$ 的点积。
Q3
$0$
$-12$
$24$
$\langle 12, -12\rangle$
$(2)(6) + (-3)(4) = 12 - 12 = 0$. So $\vec{u}\perp\vec{v}$. (Slopes $-3/2$ and $4/6 = 2/3$ multiply to $-1$ , the 2-D perpendicularity test from coordinate geometry, with the dot-product test as its vector form.)$(2)(6) + (-3)(4) = 12 - 12 = 0$。故 $\vec{u}\perp\vec{v}$。（两斜率 $-3/2$ 与 $4/6 = 2/3$ 之积 $-1$ , 解析几何中的垂直判别，点积是其向量形式。）
Multiply matching components and sum. Zero result means perpendicular.对应分量相乘后求和。结果为零即垂直。
Find the angle between $\vec{u} = \langle 1, 2, 2\rangle$ and $\vec{v} = \langle 2, -1, 2\rangle$, to two decimals. 🇨🇦 MCV4U Strand C求 $\vec{u} = \langle 1, 2, 2\rangle$ 与 $\vec{v} = \langle 2, -1, 2\rangle$ 的夹角，保留两位小数。🇨🇦 MCV4U 单元 C
Q4
$0^{\circ}$
$45.00^{\circ}$
$63.61^{\circ}$
$90.00^{\circ}$
$\vec{u}\cdot\vec{v} = 2 - 2 + 4 = 4$. $|\vec{u}| = |\vec{v}| = 3$. $\cos\theta = 4/9$, $\theta = \arccos(4/9)\approx 63.61^{\circ}$.$\vec{u}\cdot\vec{v} = 2 - 2 + 4 = 4$。$|\vec{u}| = |\vec{v}| = 3$。$\cos\theta = 4/9$，$\theta\approx 63.61^{\circ}$。
Use the 3-D version of $\cos\theta = (\vec{u}\cdot\vec{v}) / (|\vec{u}||\vec{v}|)$.用 3D 版本的 $\cos\theta = (\vec{u}\cdot\vec{v}) / (|\vec{u}||\vec{v}|)$。
For $\vec{a} = \langle 1, 0, -1\rangle$ and $\vec{b} = \langle 2, 1, 1\rangle$, find $\vec{a}\times\vec{b}$. 🇨🇦 MCV4U / IB HL AHL 3.13对 $\vec{a} = \langle 1, 0, -1\rangle$、$\vec{b} = \langle 2, 1, 1\rangle$ 求 $\vec{a}\times\vec{b}$。🇨🇦 MCV4U / IB HL AHL 3.13
Q5
$\langle 1, 3, 1\rangle$
$\langle 1, -3, 1\rangle$
$\langle -1, 3, -1\rangle$
$\langle 2, 0, -1\rangle$
$\vec{a}\times\vec{b} = \langle (0)(1) - (-1)(1), (-1)(2) - (1)(1), (1)(1) - (0)(2)\rangle = \langle 1, -3, 1\rangle$. Check perpendicularity: $\vec{a}\cdot\langle 1,-3,1\rangle = 1 + 0 - 1 = 0$ ✓; $\vec{b}\cdot\langle 1,-3,1\rangle = 2 - 3 + 1 = 0$ ✓. (Option 0's $\langle 1, 3, 1\rangle$ is the trap of forgetting the sign flip on the middle component.)$\vec{a}\times\vec{b} = \langle 1, -3, 1\rangle$。核验：$\vec{a}\cdot\langle 1,-3,1\rangle = 0$ ✓；$\vec{b}\cdot\langle 1,-3,1\rangle = 2 - 3 + 1 = 0$ ✓。（选项 0 的 $\langle 1, 3, 1\rangle$ 是忘记中间分量取负号的陷阱。）
Use $\vec{u}\times\vec{v} = \langle u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1\rangle$. Verify perpendicularity to both inputs before moving on.用公式 $\vec{u}\times\vec{v} = \langle u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1\rangle$。继续之前核验同时垂直两输入。
A force $\vec{F} = \langle 3, 4\rangle$ N moves an object through displacement $\vec{d} = \langle 6, 8\rangle$ m. Find the work done.力 $\vec{F} = \langle 3, 4\rangle$ N 推动物体经位移 $\vec{d} = \langle 6, 8\rangle$ m。求所做的功。
Q6
$25$ J
$0$ J
$50$ J
$\langle 18, 32\rangle$ J
$W = \vec{F}\cdot\vec{d} = (3)(6) + (4)(8) = 18 + 32 = 50$ J. Note $\vec{F}$ and $\vec{d}$ are parallel ($\vec{d} = 2\vec{F}$), so $W = |\vec{F}||\vec{d}| = 5 \cdot 10 = 50$ J , same answer.$W = \vec{F}\cdot\vec{d} = (3)(6) + (4)(8) = 50$ J。注意 $\vec{F}$ 与 $\vec{d}$ 平行（$\vec{d} = 2\vec{F}$），故 $W = |\vec{F}||\vec{d}| = 5 \cdot 10 = 50$ J , 同结果。
$W = \vec{F}\cdot\vec{d}$ , the dot product, a scalar.$W = \vec{F}\cdot\vec{d}$ , 是点积，结果为标量。
A $10$-kg block sits on a frictionless ramp inclined at $25^{\circ}$. Take $g = 9.8$ m/s$^2$. The block's acceleration down the ramp is approximately:一 $10$ kg 物块置于 $25^{\circ}$ 无摩擦斜面上。取 $g = 9.8$ m/s$^2$。物块沿坡向下的加速度约为：
Q7
$4.14$ m/s$^2$
$8.88$ m/s$^2$
$9.80$ m/s$^2$
$2.07$ m/s$^2$
Down-ramp force: $mg\sin\theta = 10 \cdot 9.8 \cdot \sin 25^{\circ}\approx 41.4$ N. Acceleration $a = F/m = g\sin\theta\approx 9.8 \cdot 0.4226\approx 4.14$ m/s$^2$. (The mass cancels , acceleration depends only on $g$ and $\theta$.)沿坡力 $mg\sin\theta = 10 \cdot 9.8 \cdot \sin 25^{\circ}\approx 41.4$ N。加速度 $a = F/m = g\sin\theta\approx 4.14$ m/s$^2$。（质量消去 , 加速度仅由 $g$ 和 $\theta$ 决定。）
Use $a = g\sin\theta$ , the mass cancels on a frictionless incline.用 $a = g\sin\theta$ , 无摩擦斜面上质量消去。
Find the area of the triangle with vertices $P(1,0,0)$, $Q(0,2,0)$, $R(0,0,3)$. 🇨🇦 MCV4U Strand C求顶点 $P(1,0,0)$、$Q(0,2,0)$、$R(0,0,3)$ 的三角形面积。🇨🇦 MCV4U 单元 C
Q8
$\sqrt{14}$
$\tfrac{1}{2}\sqrt{49} = 3.5$
$7$
$3$
Edges from $P$: $\vec{PQ} = \langle -1, 2, 0\rangle$, $\vec{PR} = \langle -1, 0, 3\rangle$. Cross product: $\langle (2)(3) - (0)(0), (0)(-1) - (-1)(3), (-1)(0) - (2)(-1)\rangle = \langle 6, 3, 2\rangle$. Magnitude $= \sqrt{36 + 9 + 4} = \sqrt{49} = 7$. Triangle area $= \tfrac{1}{2}\cdot 7 = 3.5$.从 $P$ 出发的边：$\vec{PQ} = \langle -1, 2, 0\rangle$、$\vec{PR} = \langle -1, 0, 3\rangle$。叉积 $= \langle 6, 3, 2\rangle$，模长 $= \sqrt{49} = 7$。三角形面积 $= 3.5$。
Triangle area in 3-D: $\tfrac{1}{2}|\vec{PQ}\times\vec{PR}|$. Cross-product magnitude is the parallelogram area.3D 中三角形面积：$\tfrac{1}{2}|\vec{PQ}\times\vec{PR}|$。叉积模长为平行四边形面积。
A swimmer crosses a river. Their velocity relative to still water is $\langle 0, 1.5\rangle$ m/s (straight across). The river current is $\langle 1.0, 0\rangle$ m/s (downstream). Find the swimmer's speed (magnitude of resultant velocity) to two decimals.游泳者横渡河流。其相对静水速度为 $\langle 0, 1.5\rangle$ m/s（横向）；水流速度为 $\langle 1.0, 0\rangle$ m/s（下游方向）。求游泳者的速率（合速度模长），保留两位小数。
Q9
$2.50$ m/s
$0.50$ m/s
$1.80$ m/s
$1.50$ m/s
Resultant velocity $= \langle 0,1.5\rangle + \langle 1.0,0\rangle = \langle 1.0, 1.5\rangle$. Speed $= \sqrt{1.0^2 + 1.5^2} = \sqrt{3.25}\approx 1.80$ m/s. (CCSSM HSN-VM.A.3 uses velocity addition as the prototypical vector application.)合速度 $= \langle 0,1.5\rangle + \langle 1.0,0\rangle = \langle 1.0, 1.5\rangle$。速率 $= \sqrt{1.0^2 + 1.5^2} = \sqrt{3.25}\approx 1.80$ m/s。（CCSSM HSN-VM.A.3 把速度合成作为典型向量应用。）
Add the two velocity vectors first, then take the magnitude of the resultant. Don't add the speeds directly , that would only work if they were parallel.先把两速度向量相加，再求合向量模长。不能直接把速率相加 , 那只在两者平行时成立。
Are the vectors $\vec{u} = \langle 2, -3, 6\rangle$ and $\vec{v} = \langle -4, 6, -12\rangle$ parallel, perpendicular, or neither?向量 $\vec{u} = \langle 2, -3, 6\rangle$ 与 $\vec{v} = \langle -4, 6, -12\rangle$ 平行、垂直，还是都不是？
Q10
Parallel (anti-parallel): $\vec{v} = -2\vec{u}$.平行（反向）：$\vec{v} = -2\vec{u}$。
Perpendicular: $\vec{u}\cdot\vec{v} = 0$.垂直：$\vec{u}\cdot\vec{v} = 0$。
Neither.都不是。
Identical.完全相同。
Component check: $-4/2 = -2$, $6/(-3) = -2$, $-12/6 = -2$. All ratios equal, so $\vec{v} = -2\vec{u}$ , parallel with opposite direction (anti-parallel). Cross-product check: $\vec{u}\times\vec{v} = \vec{0}$ confirms parallelism.分量核查：$-4/2 = -2$、$6/(-3) = -2$、$-12/6 = -2$。比值一致，故 $\vec{v} = -2\vec{u}$ , 反向平行。叉积核查：$\vec{u}\times\vec{v} = \vec{0}$ 确认平行。
Vectors are parallel iff $\vec{v} = k\vec{u}$ for some scalar $k$. Check whether all component ratios match.向量平行当且仅当 $\vec{v} = k\vec{u}$（某标量 $k$）。检查所有分量比值是否一致。
Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

0 / 12 mastered已掌握 0 / 12
Next Steps后续单元

What This Feeds Into本单元的去向

Vectors are the language of every quantitative downstream subject , physics, engineering, multivariable calculus, machine learning. This HS Math guide is intentionally the lighter cousin: it stops at the algebra of 2-D and 3-D vectors with dot and cross products. If you need lines and planes in 3-space, linear independence arguments, or parametric / scalar / vector equations, the IB Math AA HL Vectors unit shipped in this repo (`Unit_C3_Vectors.html`) is the deeper companion , read this guide first and then jump there. The other feeders below are the AP / IB units already on this repo that consume the §1-§7 toolkit on day one.向量是下游所有定量学科的语言 , 物理、工程、多变量微积分、机器学习无不如此。本 HS Math 指南有意做轻：仅止于 2D 与 3D 向量的代数及点积、叉积。若你要学习 3D 空间中的直线与平面、线性无关性论证或参数 / 标量 / 向量方程，本仓库的 IB Math AA HL 向量单元（`Unit_C3_Vectors.html`）是更深的衔接 , 先读本指南再去那边。下方其他衔接为本仓库中第一天就用到 §1-§7 工具组的 AP / IB 单元。

Within High School Math.在 HS Math 内部。

Unit 7 (Right-Triangle Trigonometry) is the prerequisite: vector components are SOH CAH TOA inside the arrow. Unit 8 (Unit-Circle Trig) provides the $\cos\theta$ / $\sin\theta$ values used by §2's magnitude-to-components conversion. Unit 11 (Combinatorics & Binomial Theorem) is unrelated but lives at a similar US-honors level , both units fall under the (+) CCSSM cluster.Unit 7（直角三角形三角学）是前置：向量分量正是箭头内部的 SOH CAH TOA。Unit 8（单位圆三角学）提供 §2 中"模长 $\to$ 分量"用到的 $\cos\theta$ / $\sin\theta$ 值。Unit 11（组合与二项式定理）虽无关，但同属 US 荣誉级 , 两单元都落在 CCSSM (+) 簇之下。

Across the AP and IB feeders in this repo.本仓库中的 AP 与 IB 衔接单元。

IB Math HL C3 · Vectors (the deeper companion: linear independence, lines and planes in 3-D, parametric / scalar / vector equations)IB Math HL C3 · 向量（深入衔接：线性无关性、3D 中的直线与平面、参数 / 标量 / 向量方程） IB Math HL C2 · Trigonometry Applications (the $\sin\theta$ / $\cos\theta$ machinery behind vector resolution)IB Math HL C2 · 三角学应用（向量分解背后的 $\sin\theta$ / $\cos\theta$ 机制） AP Physics Unit 1 · Kinematics (vector components on inclined planes: $v_x = v\cos\theta$, $v_y = v\sin\theta$; gravity split as $mg\sin\theta$ / $mg\cos\theta$)AP Physics Unit 1 · 运动学（斜面上的速度分量 $v_x = v\cos\theta$、$v_y = v\sin\theta$；重力分解 $mg\sin\theta$ / $mg\cos\theta$） HS Math Unit 7 · Right-Triangle Trigonometry (the SOH CAH TOA inside §2's component decomposition)HS Math Unit 7 · 直角三角形三角学（§2 分量分解内部的 SOH CAH TOA）

If you are aiming for AP Physics 1 or AP Physics C, §1-§3 plus §7 are mandatory on day one. If you are aiming for IB Math AA HL or first-year engineering, read this entire guide and then jump to the IB HL Vectors unit linked above , that is where lines, planes, and intersection geometry live. If you are on a BC PC 11 / PC 12 or AB Math 30-1 standard track, treat §1-§7 as enrichment with no curriculum-required content; the IB HL feeder lets students going to engineering meet the rest of the vectors toolkit before first-year university.备考 AP Physics 1 或 AP Physics C：§1-§3 与 §7 第一天即用。备考 IB Math AA HL 或大一工程：读完本指南后跳至上方 IB HL 向量单元 , 直线、平面、相交几何均在那里。BC PC 11 / PC 12 或 AB Math 30-1 标准轨学生：§1-§7 属拓展，课纲无要求；IB HL 衔接可让走工程方向的同学在大一前先掌握向量工具其余部分。

DINGRUI SCHOLARS

High School Math · Vectors · v1高中数学 · 向量 · v1