PART I · SHORT RESPONSE第一部分 · 简答题ON MCV4U + US HSN-VM (+) · 18 marksON MCV4U + US HSN-VM (+) · 18 分
Section A · Short ResponseA 节 · 简答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. Write vectors in angle-bracket component form $\langle a, b\rangle$ unless the question states otherwise. State magnitudes to two decimals or exact-radical form as the question requires. No calculator on Q1–Q3; calculator permitted on Q4–Q5.题型混合：选择题与简答题。选择题请圈出字母，并在空白处保留可验证的过程。除非题目另有说明，向量请用尖括号分量形式 $\langle a, b\rangle$ 表示。模长按题目要求保留两位小数或精确根式。Q1–Q3 不可用计算器；Q4–Q5 可用计算器。
The vector $\vec{AB}$ has initial point $A(-1, 4)$ and terminal point $B(5, -4)$. Which of the following is the component form of $\vec{AB}$, and what is its magnitude $|\vec{AB}|$?向量 $\vec{AB}$ 的起点为 $A(-1, 4)$，终点为 $B(5, -4)$。下列哪一项是 $\vec{AB}$ 的分量形式与其模长 $|\vec{AB}|$？
Let $\vec{v} = \langle -3, 4\rangle$. Which of the following gives $|\vec{v}|$ and the unit vector $\hat{v}$ in the same direction as $\vec{v}$?设 $\vec{v} = \langle -3, 4\rangle$。下列哪一项给出 $|\vec{v}|$ 及与 $\vec{v}$ 同向的单位向量 $\hat{v}$？
A wind vector $\vec{w}$ has magnitude $|\vec{w}| = 20$ km/h and points in a direction $35^{\circ}$ north of east. Take east as the positive $x$-axis and north as the positive $y$-axis.风向向量 $\vec{w}$ 的模长 $|\vec{w}| = 20$ km/h，方向为北偏东 $35^{\circ}$（即在东方向上方 $35^{\circ}$）。取东向为正 $x$ 轴，北向为正 $y$ 轴。
(a)Write $\vec{w}$ in component form $\langle a, b\rangle$, with each component rounded to two decimals (km/h).将 $\vec{w}$ 写成分量形式 $\langle a, b\rangle$，每个分量保留两位小数（km/h）。[2]
(b)Verify your answer by computing $|\vec{w}|$ from your components; confirm it rounds to $20$ km/h.用所得分量计算 $|\vec{w}|$ 进行验算，确认其四舍五入后为 $20$ km/h。[2]
Let $\vec{a} = \langle 3, -2\rangle$ and $\vec{b} = \langle 4, k\rangle$ where $k$ is a real parameter.设 $\vec{a} = \langle 3, -2\rangle$，$\vec{b} = \langle 4, k\rangle$，其中 $k$ 为实参数。
(a)Compute $\vec{a} \cdot \vec{b}$ in terms of $k$.求 $\vec{a} \cdot \vec{b}$ 关于 $k$ 的表达式。[2]
(b)Find the value of $k$ for which $\vec{a}$ and $\vec{b}$ are perpendicular. State the perpendicularity test you used.求使 $\vec{a}$ 与 $\vec{b}$ 垂直的 $k$ 值。写出所用的垂直判别式。[3]
PART II · EXTENDED RESPONSE第二部分 · 长答题AP-feeder FRQ + ON MCV4U C3 · 35 marksAP 衔接 FRQ + ON MCV4U C3 · 35 分
Section B · Extended ResponseB 节 · 长答题
Show every step of each dot- and cross-product computation. For cross products, write the $3 \times 3$ determinant expansion explicitly and watch the sign of the middle component (Sarrus / cofactor expansion both work). State angles in degrees rounded to one decimal unless the question asks for radians. No calculator on Q6–Q8; calculator permitted on Q9.点积与叉积的每一步都要写出来。叉积要明确写出 $3 \times 3$ 行列式展开，并注意中间分量的符号（Sarrus 与代数余子式展开均可）。角度若无特别说明保留到 0.1 度。Q6–Q8 不可用计算器；Q9 可用计算器。
(b)Compute $|\vec{u}|$ and $|\vec{v}|$ in exact-radical form.以精确根式计算 $|\vec{u}|$ 与 $|\vec{v}|$。[2]
(c)Use the geometric form $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos\theta$ to find the angle $\theta$ between $\vec{u}$ and $\vec{v}$, in degrees, rounded to one decimal.利用几何形式 $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos\theta$ 求 $\vec{u}$ 与 $\vec{v}$ 的夹角 $\theta$，以度为单位，保留 0.1。[3]
(d)State whether $\theta$ is acute, right, or obtuse, and link that classification back to the sign of $\vec{u} \cdot \vec{v}$.指出 $\theta$ 是锐角、直角还是钝角，并把该分类与 $\vec{u} \cdot \vec{v}$ 的符号关联起来。[1]
Q7HARDHonors🇨🇦 ONON Provincial-style§5 Vectors in 3-D · MCV4U C1-C3[9 marks]
In $\mathbb{R}^{3}$, let $\vec{p} = \langle 1, -2, 2\rangle$ and $\vec{q} = \langle 3, 0, -4\rangle$.在 $\mathbb{R}^{3}$ 中，设 $\vec{p} = \langle 1, -2, 2\rangle$，$\vec{q} = \langle 3, 0, -4\rangle$。
(a)Compute $|\vec{p}|$ and $|\vec{q}|$.计算 $|\vec{p}|$ 与 $|\vec{q}|$。[2]
(b)Find the unit vector $\hat{p}$ in the direction of $\vec{p}$.求 $\vec{p}$ 方向上的单位向量 $\hat{p}$。[2]
In $\mathbb{R}^{3}$, let $\vec{a} = \langle 2, -1, 3\rangle$ and $\vec{b} = \langle 1, 4, -2\rangle$.在 $\mathbb{R}^{3}$ 中，设 $\vec{a} = \langle 2, -1, 3\rangle$，$\vec{b} = \langle 1, 4, -2\rangle$。
(a)Compute $\vec{a} \times \vec{b}$ by writing out the $3 \times 3$ determinant expansion explicitly. Be careful with the sign of the $\hat{j}$-component, the cofactor expansion picks up an extra minus sign.通过明确写出 $3 \times 3$ 行列式展开来计算 $\vec{a} \times \vec{b}$。注意 $\hat{j}$ 分量的符号，代数余子式展开会多出一个负号。[4]
(b)Verify your answer is perpendicular to both $\vec{a}$ and $\vec{b}$ by checking $(\vec{a} \times \vec{b}) \cdot \vec{a} = 0$ and $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$.通过验证 $(\vec{a} \times \vec{b}) \cdot \vec{a} = 0$ 与 $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$，确认所得向量与 $\vec{a}$、$\vec{b}$ 均垂直。[2]
(c)Compute $|\vec{a} \times \vec{b}|$, and state its geometric meaning (area of which figure?).计算 $|\vec{a} \times \vec{b}|$，并说明其几何含义（是哪一个图形的面积？）。[2]
In $\mathbb{R}^{3}$, let $\vec{u} = \langle 1, 2, -1\rangle$ and $\vec{v} = \langle 2, 0, 1\rangle$.在 $\mathbb{R}^{3}$ 中，设 $\vec{u} = \langle 1, 2, -1\rangle$，$\vec{v} = \langle 2, 0, 1\rangle$。
(a)Compute $\vec{u} \cdot \vec{v}$ and the angle $\theta$ between $\vec{u}$ and $\vec{v}$ (in degrees, to one decimal).计算 $\vec{u} \cdot \vec{v}$ 与 $\vec{u}$、$\vec{v}$ 的夹角 $\theta$（度，保留 0.1）。[3]
(b)Compute $\vec{u} \times \vec{v}$ via the determinant expansion.用行列式展开计算 $\vec{u} \times \vec{v}$。[3]
(c)Verify the identity $|\vec{u}|^{2} |\vec{v}|^{2} = (\vec{u} \cdot \vec{v})^{2} + |\vec{u} \times \vec{v}|^{2}$ (Lagrange's identity). Show both sides numerically.验证恒等式 $|\vec{u}|^{2} |\vec{v}|^{2} = (\vec{u} \cdot \vec{v})^{2} + |\vec{u} \times \vec{v}|^{2}$（拉格朗日恒等式）。给出两侧的数值。[3]
PART III · MODELING / APPLIED第三部分 · 建模 / 应用AP Physics-feeder + ON MCV4U C2-C3 · 28 marksAP Physics 衔接 + ON MCV4U C2-C3 · 28 分
Section C · Modeling and ApplicationsC 节 · 建模与应用
Name your axes and units before writing equations. Set up the vector model, solve, and conclude with a one-sentence answer in context (with units). Calculator permitted throughout Part III. These three problems are exactly the kinds of force / work / moment computations students will see in AP Physics Unit 1 (kinematics & force vectors) and IB Math HL Topic C3.列方程前先命名坐标轴和单位。建立向量模型、求解，并用一句话给出结合情境的结论（带单位）。第三部分全程可用计算器。这三道题正是 AP Physics 第 1 单元（运动学与力向量）与 IB Math HL 主题 C3 中常见的力 / 功 / 力矩计算。
A $10$ kg crate sits on a frictionless ramp inclined at $25^{\circ}$ above the horizontal. Take gravity $g = 9.8$ m/s$^{2}$ acting straight down with magnitude $W = mg$.一个 $10$ kg 的箱子放在与水平面成 $25^{\circ}$ 的无摩擦斜面上。取重力加速度 $g = 9.8$ m/s$^{2}$ 竖直向下作用，大小为 $W = mg$。
(a)Compute the magnitude of the gravitational force $W = mg$ in newtons.计算重力大小 $W = mg$（牛顿）。[1]
(b)Resolve $W$ into a component parallel to the ramp surface (call it $W_{\parallel}$, pointing down the slope) and a component perpendicular to the ramp (call it $W_{\perp}$, pressing into the surface). Express both as multiples of $W$ with trig factors, then evaluate to the nearest tenth of a newton.将 $W$ 分解为沿斜面方向的分量（记作 $W_{\parallel}$，沿斜面向下）与垂直于斜面方向的分量（记作 $W_{\perp}$，压向斜面）。先用 $W$ 与三角函数因子表示，再求值，结果保留到 0.1 N。[4]
(c)State the minimum force, parallel to the ramp and directed up the slope, needed to hold the crate stationary.写出使箱子保持静止所需的沿斜面向上的最小作用力。[2]
(d)Sanity-check: compute $W_{\parallel}^{2} + W_{\perp}^{2}$ and confirm it equals $W^{2}$ (Pythagoras on perpendicular components).合理性检验：计算 $W_{\parallel}^{2} + W_{\perp}^{2}$，确认其等于 $W^{2}$（垂直分量上的勾股定理）。[2]
Q11MEDIUMHonors🇨🇦 ON🇺🇸 USAP-feeder FRQ§7 Work as Dot Product · HSN-VM.A.3 (+)[9 marks]
A worker pulls a sled by a rope angled $30^{\circ}$ above the horizontal. The rope tension has magnitude $|\vec{F}| = 80$ N. The sled is displaced by $\vec{d} = \langle 12, 0\rangle$ (m) along level ground (east).一工人用绳子拉雪橇，绳与水平方向成 $30^{\circ}$ 向上。绳的张力大小 $|\vec{F}| = 80$ N。雪橇沿水平地面（向东）移动 $\vec{d} = \langle 12, 0\rangle$（m）。
(a)Write the force vector $\vec{F}$ in component form $\langle F_{x}, F_{y}\rangle$, taking east as $+x$ and up as $+y$. Round each to one decimal place.取东向为正 $x$、向上为正 $y$，将力向量 $\vec{F}$ 写成分量形式 $\langle F_{x}, F_{y}\rangle$。每个分量保留 1 位小数。[3]
(b)Compute the work $W = \vec{F} \cdot \vec{d}$ using the algebraic (componentwise) form. State the units.用代数（按分量）形式计算功 $W = \vec{F} \cdot \vec{d}$，并写出单位。[3]
(c)Verify your answer using the geometric form $W = |\vec{F}|\,|\vec{d}|\cos\theta$, with $\theta = 30^{\circ}$. The two forms should agree to within rounding.取 $\theta = 30^{\circ}$，用几何形式 $W = |\vec{F}|\,|\vec{d}|\cos\theta$ 验证答案。两种形式在舍入误差内应一致。[2]
(d)Explain in one sentence why the vertical component of $\vec{F}$ does no work on the sled in this scenario.用一句话说明：在此情境中，为何 $\vec{F}$ 的竖直分量对雪橇不做功。[1]
A wrench is used to tighten a bolt at the origin. The applied force is $\vec{F} = \langle 0, 10, 0\rangle$ N. The point of application (the end of the wrench handle) is at $\vec{r} = \langle 0.30, 0, 0\rangle$ m. Torque about the origin is defined by $\vec{\tau} = \vec{r} \times \vec{F}$.用扳手紧固位于原点处的螺栓。施加力为 $\vec{F} = \langle 0, 10, 0\rangle$ N。作用点（扳手手柄末端）位于 $\vec{r} = \langle 0.30, 0, 0\rangle$ m。关于原点的力矩定义为 $\vec{\tau} = \vec{r} \times \vec{F}$。
(a)Compute $\vec{\tau} = \vec{r} \times \vec{F}$ in component form (N·m). Write out the determinant expansion explicitly.以分量形式计算 $\vec{\tau} = \vec{r} \times \vec{F}$（N·m）。明确写出行列式展开。[3]
(b)State $|\vec{\tau}|$ (the magnitude of the torque), and the direction of $\vec{\tau}$ relative to the standard right-hand coordinate axes.写出 $|\vec{\tau}|$（力矩大小）以及 $\vec{\tau}$ 相对于标准右手坐标轴的方向。[2]
(c)The mechanic now repositions the wrench so that the force is applied at $\vec{r}' = \langle 0.30, 0.10, 0\rangle$ m with the same $\vec{F} = \langle 0, 10, 0\rangle$ N. Re-compute $\vec{\tau}' = \vec{r}' \times \vec{F}$.技师调整扳手位置，使作用点变为 $\vec{r}' = \langle 0.30, 0.10, 0\rangle$ m，力仍为 $\vec{F} = \langle 0, 10, 0\rangle$ N。重新计算 $\vec{\tau}' = \vec{r}' \times \vec{F}$。[3]
(d)Compare $|\vec{\tau}|$ to $|\vec{\tau}'|$. Explain in one sentence why the $y$-component of $\vec{r}'$ does not increase the torque, and connect this to the geometric formula $|\vec{\tau}| = |\vec{r}|\,|\vec{F}|\sin\theta$.比较 $|\vec{\tau}|$ 与 $|\vec{\tau}'|$。用一句话说明为何 $\vec{r}'$ 的 $y$ 分量不会增大力矩，并将其与几何公式 $|\vec{\tau}| = |\vec{r}|\,|\vec{F}|\sin\theta$ 联系起来。[2]
🇨🇦 Ontario MCV4U (core)安大略 MCV4U （核心）Strand C Geometry and Algebra of Vectors: C1 representing vectors, C2 vector operations, C3 dot/cross product, C4 lines/planes (lines/planes covered in IB HL feeder, not this unit).C 单元 向量的几何与代数：C1 向量的表示，C2 向量运算，C3 点积/叉积，C4 直线/平面（直线/平面在 IB HL 衔接单元中讲，不在本单元）。
🇨🇦 BC PC11/PC12 + AB Math 30-1 (out-of-core)BC PC11/PC12 + AB Math 30-1 （不在核心）No vectors content in BC or AB published outcomes. Application home is IB Math AA HL Topic C3 (vectors) + AP Physics Unit 1 (kinematics & force vectors).BC 与 AB 已发布大纲中均无向量内容。应用归宿是 IB Math AA HL 主题 C3（向量）与 AP Physics 第 1 单元（运动学与力向量）。
Full 4-region Syllabus Map lives in ../Study Guides/Unit_14_Vectors.html.完整 4 区域大纲对照见 ../Study Guides/Unit_14_Vectors.html。