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Solutions解析

Vectors · Solutions向量 · 解析

Companion to the Practice Set · Mark-by-mark walkthroughs · ON MCV4U + US HSN-VM (+) Honors配套练习解析 · 逐分讲解 · ON MCV4U + US HSN-VM (+) 荣誉

EASY MEDIUM HARD 🇨🇦 ON 🇺🇸 US ON Provincial-style AP-feeder FRQ Honors
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  简答题 · 解析ON MCV4U + US HSN-VM (+) · 18 marksON MCV4U + US HSN-VM (+) · 18 分

Section A · Worked SolutionsA 节 · 详细解答

Q1EASY 🇨🇦 ON 🇺🇸 US ON Provincial-style §1 Directed Segments · HSN-VM.A.2 (+) [3 marks]

$A(-1,4)$, $B(5,-4)$. Component form and magnitude of $\vec{AB}$.$A(-1,4)$，$B(5,-4)$。求 $\vec{AB}$ 的分量形式与模长。

Answer:答案：  (A)  $\vec{AB} = \langle 6, -8\rangle$, $|\vec{AB}| = 10$

(a) Component form via terminal minus initial用"终点减起点"求分量形式 M1·A1

The HSN-VM.A.2 rule: $\vec{AB} = B - A$ (terminal coordinates minus initial coordinates):HSN-VM.A.2 规则：$\vec{AB} = B - A$（终点坐标减起点坐标）： $$ \vec{AB} \;=\; \langle 5 - (-1),\; -4 - 4\rangle \;=\; \langle 6,\; -8\rangle. $$

(b) Magnitude via Pythagoras用勾股定理求模长 A1

$$ |\vec{AB}| \;=\; \sqrt{6^{2} + (-8)^{2}} \;=\; \sqrt{36 + 64} \;=\; \sqrt{100} \;=\; 10. $$
Why the wrong choices fail.错误选项分析。
  • (B) $\langle -6, 8\rangle$, $10$, signs flipped: this is $\vec{BA} = A - B$, the reverse vector. Order matters in vector notation.，符号颠倒：这是 $\vec{BA} = A - B$，是反向向量。向量记号中顺序很重要。
  • (C) $\langle 6, -8\rangle$, $14$, correct components but $|6| + |-8| = 14$ is the "taxicab" (Manhattan) length, not the Euclidean magnitude.，分量正确，但 $|6| + |-8| = 14$ 是"出租车"（曼哈顿）距离，不是欧几里得模长。
  • (D) $\langle 4, 0\rangle$, $4$, midpoint thinking: $(-1+5)/2 = 2$ is the midpoint $x$-coordinate, not the displacement. Vectors are not midpoints.，中点思路：$(-1+5)/2 = 2$ 是中点的 $x$ 坐标，不是位移。向量不是中点。
"Terminal minus initial" is the only rule you need for $\vec{AB}$.求 $\vec{AB}$ 只要"终点减起点"一条规则。 The mnemonic is the arrow direction: write the head ($B$) coordinates first, then subtract the tail ($A$) coordinates. The reverse, $\vec{BA} = A - B$, gives the same arrow length but the opposite direction (a sign flip in every component). Markers split a 2-mark "component form" item into M1 (set up the subtraction in the correct order) and A1 (arithmetic). Reverse the order and you forfeit both.记法跟随箭头方向：先写箭头头部 ($B$) 的坐标，再减去箭头尾部 ($A$) 的坐标。反向的 $\vec{BA} = A - B$ 长度相同但方向相反（每个分量翻号）。阅卷把 2 分的"分量形式"题拆成 M1（顺序设置正确）与 A1（算术正确），顺序写反两分都丢。
Q2EASY 🇨🇦 ON 🇺🇸 US AP-feeder FRQ §2 Magnitude & Unit Vector · HSN-VM.A.1 (+) [3 marks]

$\vec{v} = \langle -3, 4\rangle$. Magnitude and unit vector.$\vec{v} = \langle -3, 4\rangle$。求模长与单位向量。

Answer:答案：  (B)  $|\vec{v}| = 5$, $\hat{v} = \left\langle -\tfrac{3}{5}, \tfrac{4}{5}\right\rangle$

(a) Magnitude模长 M1·A1

$$ |\vec{v}| \;=\; \sqrt{(-3)^{2} + 4^{2}} \;=\; \sqrt{9 + 16} \;=\; \sqrt{25} \;=\; 5. $$

(b) Unit vector by scalar division通过数除得到单位向量 A1

The unit vector in the direction of $\vec{v}$ is $\hat{v} = \vec{v}/|\vec{v}|$:$\vec{v}$ 方向上的单位向量为 $\hat{v} = \vec{v}/|\vec{v}|$： $$ \hat{v} \;=\; \frac{1}{5}\langle -3, 4\rangle \;=\; \left\langle -\tfrac{3}{5},\; \tfrac{4}{5}\right\rangle. $$ Sanity check: $|\hat{v}|^{2} = \tfrac{9}{25} + \tfrac{16}{25} = 1$. $\checkmark$合理性检验：$|\hat{v}|^{2} = \tfrac{9}{25} + \tfrac{16}{25} = 1$。$\checkmark$
Why the wrong choices fail.错误选项分析。
  • (A) $|\vec{v}| = 1$, $\hat{v} = \langle -3, 4\rangle$, confuses "unit vector" with the input vector; only a vector that already has magnitude $1$ is its own unit vector.，把"单位向量"与原向量混为一谈；只有本身模长已等于 $1$ 的向量才是自己的单位向量。
  • (C) $|\vec{v}| = 5$, $\hat{v} = \langle 3/5, -4/5\rangle$, correct magnitude but signs flipped; the student divided $-\vec{v}$ by $|\vec{v}|$, producing the unit vector in the opposite direction.，模长正确但符号颠倒；学生将 $-\vec{v}$ 除以 $|\vec{v}|$，得到反方向的单位向量。
  • (D) $|\vec{v}| = 7$, $\hat{v} = \langle -3/7, 4/7\rangle$, computed $|\vec{v}| = |-3| + |4| = 7$ (taxicab length again); the unit vector is consistent with the wrong magnitude.，把 $|\vec{v}| = |-3| + |4| = 7$（又是出租车距离）；单位向量与错误的模长一致。
$\hat{v}$ keeps the direction of $\vec{v}$, scaled down to length $1$.$\hat{v}$ 保持 $\vec{v}$ 的方向，长度缩为 $1$。 Two ways to verify: (i) the components of $\hat{v}$ should have the same signs as $\vec{v}$'s components (a sign flip is direction reversal, not normalization); (ii) the sum of squares of $\hat{v}$'s components must equal $1$ exactly. AP-feeder markers reward students who write the verification step "$|\hat{v}|^{2} = 1$" explicitly, it earns the A1 even when the arithmetic of the magnitude itself was sloppy.两种验证方式：(i) $\hat{v}$ 的分量应与 $\vec{v}$ 的分量同号（翻号是方向反转，不是单位化）；(ii) $\hat{v}$ 各分量平方和必须恰好等于 $1$。AP 衔接阅卷会奖励显式写出"$|\hat{v}|^{2} = 1$"验证步骤的学生，即便模长本身的算术粗糙也能拿到 A1。
Q3MEDIUM 🇨🇦 ON 🇺🇸 US AP-feeder FRQ §3 Addition + Scalar Multiplication · HSN-VM.B.4 & B.5 (+) [3 marks]

$\vec{u} = \langle 2, -1\rangle$, $\vec{w} = \langle -4, 3\rangle$. Compute $2\vec{u} - \vec{w}$.$\vec{u} = \langle 2, -1\rangle$，$\vec{w} = \langle -4, 3\rangle$。求 $2\vec{u} - \vec{w}$。

Answer:答案：  (A)  $\langle 8, -5\rangle$

(a) Scalar multiplication of $\vec{u}$$\vec{u}$ 的数乘 M1

$2\vec{u} = 2 \langle 2, -1\rangle = \langle 4, -2\rangle$ (HSN-VM.B.5: scalar multiplies every component).$2\vec{u} = 2 \langle 2, -1\rangle = \langle 4, -2\rangle$（HSN-VM.B.5：标量乘以每一个分量）。

(b) Component subtraction按分量相减 A1·A1

$$ 2\vec{u} - \vec{w} \;=\; \langle 4, -2\rangle - \langle -4, 3\rangle \;=\; \langle 4 - (-4),\; -2 - 3\rangle \;=\; \langle 8, -5\rangle. $$ Component-by-component (HSN-VM.B.4); the second-component subtraction $-2 - 3 = -5$ is the trap point.按分量逐一处理（HSN-VM.B.4）；第二分量的减法 $-2 - 3 = -5$ 是陷阱。
Why the wrong choices fail.错误选项分析。
  • (B) $\langle 0, -5\rangle$, scaled $\vec{u}$ by $2$ but then added $\vec{w}$ instead of subtracting in the first component: $4 + (-4) = 0$. Half the sign-error.，把 $\vec{u}$ 乘以 $2$ 后在第一分量上加上了 $\vec{w}$ 而非相减：$4 + (-4) = 0$。半个符号错误。
  • (C) $\langle 8, 1\rangle$, correct first component, but second-component sign slip: $-2 - 3 = 1$ is wrong, the student computed $-2 - (-3) = 1$ as if subtracting the negative of $\vec{w}$'s second component.，第一分量正确，但第二分量符号弄反：$-2 - 3 = 1$ 是错的，学生算成了 $-2 - (-3) = 1$，仿佛减去的是 $\vec{w}$ 第二分量的相反数。
  • (D) $\langle -8, 5\rangle$, this is $\vec{w} - 2\vec{u}$ instead of $2\vec{u} - \vec{w}$; the entire result is negated.，这是 $\vec{w} - 2\vec{u}$ 而非 $2\vec{u} - \vec{w}$；整个结果取了负号。
Scale before subtracting, and write the substitution before doing arithmetic.先数乘再相减，先写代入再算术。 The single most reliable scaffold for $a\vec{u} + b\vec{w}$ items is to (1) write $a\vec{u}$ in full, (2) write $b\vec{w}$ in full, (3) then stack and subtract componentwise on paper. Trying to combine the scaling and subtraction in one step is where the sign flips creep in. The double-negative trap in the second component ($-2 - 3$ vs $-2 - (-3)$) appears on essentially every SAT-style vector item with mixed signs.解 $a\vec{u} + b\vec{w}$ 类题目最可靠的脚手架是：(1) 完整写出 $a\vec{u}$；(2) 完整写出 $b\vec{w}$；(3) 再在纸上按分量上下对齐并相减。把数乘与减法挤在一步里，正是符号翻号悄悄发生的地方。第二分量的双负号陷阱（$-2 - 3$ 与 $-2 - (-3)$）几乎出现在每一道含混合符号的 SAT 风格向量题里。
Q4MEDIUM 🇨🇦 ON ON Provincial-style §1 & §2 Magnitude / Direction · MCV4U Strand C1 [4 marks]

Wind $|\vec{w}| = 20$ km/h, $35^{\circ}$ N of E. (a) Component form. (b) Verify $|\vec{w}|$.风 $|\vec{w}| = 20$ km/h，方向北偏东 $35^{\circ}$（东向上方 $35^{\circ}$）。(a) 分量形式。(b) 验算 $|\vec{w}|$。

Answer:答案：  (a) $\vec{w} \approx \langle 16.38, 11.47\rangle$ km/h  ·  (b) $|\vec{w}| \approx \sqrt{399.87} \approx 20.00$ km/h $\checkmark$

(a) Decompose into east and north components分解为东向与北向分量 M1·A1

With $|\vec{w}| = 20$ and the angle $\theta = 35^{\circ}$ measured from the positive $x$-axis (east), the components are已知 $|\vec{w}| = 20$，自正 $x$ 轴（东）逆时针测得 $\theta = 35^{\circ}$，分量为 $$ w_{x} \;=\; |\vec{w}| \cos\theta \;=\; 20 \cos 35^{\circ} \;\approx\; 20 (0.81915) \;\approx\; 16.38, $$ $$ w_{y} \;=\; |\vec{w}| \sin\theta \;=\; 20 \sin 35^{\circ} \;\approx\; 20 (0.57358) \;\approx\; 11.47. $$ So $\vec{w} \approx \langle 16.38,\; 11.47\rangle$ km/h.因此 $\vec{w} \approx \langle 16.38,\; 11.47\rangle$ km/h。

(b) Pythagorean check on the components用分量做勾股验算 M1·A1

$$ |\vec{w}| \;=\; \sqrt{16.38^{2} + 11.47^{2}} \;=\; \sqrt{268.30 + 131.56} \;=\; \sqrt{399.86} \;\approx\; 19.997 \;\approx\; 20.00. $$ Matches the given magnitude to two decimals. $\checkmark$ (Any small drift is rounding in the trig.)与给定模长在两位小数内一致。$\checkmark$（微小偏差来自三角函数的舍入。）
"North of east" places the angle in the standard position, $35^{\circ}$ from $+x$ toward $+y$."北偏东"把角放在标准位置：自 $+x$ 朝 $+y$ 方向 $35^{\circ}$。 Compass conventions (N of E, E of N, bearings) are the most common source of dropped marks on Ontario MCV4U C1 short-answer items. Always restate the convention you are using in one short sentence ("take east as $+x$, measure $\theta$ counterclockwise from $+x$"); markers will give the M1 for the correct setup even if the arithmetic is off. Watch for "N$35^{\circ}$E", which is a bearing notation and means $35^{\circ}$ measured east of north, i.e. an angle of $90 - 35 = 55^{\circ}$ from the $+x$-axis, not $35^{\circ}$.罗盘约定（北偏东、东偏北、方位角）是 Ontario MCV4U C1 简答题中最常见的失分原因。务必用一句短话重申所用约定（"取东为 $+x$，自 $+x$ 逆时针测 $\theta$"）；只要前置设置正确，即便算术出错也能拿到 M1。注意"N$35^{\circ}$E"是方位角记法，表示自正北顺时针测 $35^{\circ}$，相当于自 $+x$ 轴量 $90 - 35 = 55^{\circ}$，而非 $35^{\circ}$。
Q5MEDIUMHonors 🇨🇦 ON 🇺🇸 US AP-feeder FRQ §4 Dot Product / Perpendicularity · MCV4U C3 / HSN-VM (+) [5 marks]

$\vec{a} = \langle 3, -2\rangle$, $\vec{b} = \langle 4, k\rangle$. (a) $\vec{a} \cdot \vec{b}$ in terms of $k$. (b) $k$ for perpendicularity.$\vec{a} = \langle 3, -2\rangle$，$\vec{b} = \langle 4, k\rangle$。(a) 用 $k$ 表示 $\vec{a} \cdot \vec{b}$。(b) 求使二者垂直的 $k$。

Answer:答案：  (a) $\vec{a} \cdot \vec{b} = 12 - 2k$  ·  (b) $k = 6$ (perpendicularity test: $\vec{a} \cdot \vec{b} = 0$)（垂直判别式：$\vec{a} \cdot \vec{b} = 0$）

(a) Componentwise dot product按分量计算点积 M1·A1

$$ \vec{a} \cdot \vec{b} \;=\; (3)(4) + (-2)(k) \;=\; 12 - 2k. $$

(b) Apply the perpendicularity test应用垂直判别式 R1·M1·A1

Two non-zero vectors are perpendicular iff their dot product is zero (this is the algebraic form of the perpendicularity test, equivalent to $\cos\theta = 0 \Leftrightarrow \theta = 90^{\circ}$). Setting $\vec{a} \cdot \vec{b} = 0$:两个非零向量垂直当且仅当其点积为零（这是垂直判别式的代数形式，等价于 $\cos\theta = 0 \Leftrightarrow \theta = 90^{\circ}$）。令 $\vec{a} \cdot \vec{b} = 0$： $$ 12 - 2k \;=\; 0 \;\Longrightarrow\; k \;=\; 6. $$ Verify: with $k = 6$, $\vec{b} = \langle 4, 6\rangle$ and $\vec{a} \cdot \vec{b} = 12 - 12 = 0$. $\checkmark$验算：当 $k = 6$ 时，$\vec{b} = \langle 4, 6\rangle$，$\vec{a} \cdot \vec{b} = 12 - 12 = 0$。$\checkmark$
The dot-product perpendicularity test is the fastest way to find a missing component.点积垂直判别式是求未知分量最快的途径。 Setting up the equation $\vec{a} \cdot \vec{b} = 0$ converts a geometric question ("perpendicular?") into a one-line linear equation in the unknown. Compare with the alternative, using slopes: you would compute $m_{a} = -2/3$, then require $m_{b} \cdot m_{a} = -1$, then solve for $k/4$. Both work, but the dot-product route is dimension-agnostic (works in 3-D where slopes do not) and never requires dividing by zero (vertical-vector edge case). MCV4U C3 markers split this item: R1 for stating the test, M1 for setting up the equation, A1 for solving.列出 $\vec{a} \cdot \vec{b} = 0$ 即可把"是否垂直？"这一几何问题转成关于未知数的一行线性方程。对比另一思路——用斜率：先算 $m_{a} = -2/3$，再要求 $m_{b} \cdot m_{a} = -1$，最后解 $k/4$。两种方法都可行，但点积路径与维度无关（三维下斜率失效），且从不除以零（避免竖直向量的边界情形）。MCV4U C3 阅卷拆分本题：R1 写判别式，M1 列方程，A1 解出。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  长答题 · 解析AP-feeder FRQ + ON MCV4U C3 · 35 marksAP 衔接 FRQ + ON MCV4U C3 · 35 分

Section B · Worked SolutionsB 节 · 详细解答

Q6MEDIUMHonors 🇨🇦 ON 🇺🇸 US AP-feeder FRQ §4 Dot Product / Angle · MCV4U C3 / HSN-VM (+) [8 marks]

$\vec{u} = \langle 2, 3\rangle$, $\vec{v} = \langle -1, 5\rangle$. (a) $\vec{u} \cdot \vec{v}$. (b) $|\vec{u}|, |\vec{v}|$ exact. (c) Angle $\theta$. (d) Classify $\theta$.$\vec{u} = \langle 2, 3\rangle$，$\vec{v} = \langle -1, 5\rangle$。(a) 求 $\vec{u} \cdot \vec{v}$。(b) 精确求 $|\vec{u}|, |\vec{v}|$。(c) 夹角 $\theta$。(d) 将 $\theta$ 分类。

Answer:答案：  (a) $\vec{u} \cdot \vec{v} = 13$  ·  (b) $|\vec{u}| = \sqrt{13}$, $|\vec{v}| = \sqrt{26}$  ·  (c) $\theta = 45.0^{\circ}$  ·  (d) acute锐角 ($\vec{u} \cdot \vec{v} > 0$)

(a) Algebraic dot product代数法求点积 M1·A1

$$ \vec{u} \cdot \vec{v} \;=\; (2)(-1) + (3)(5) \;=\; -2 + 15 \;=\; 13. $$

(b) Magnitudes in exact-radical form精确根式求模长 A1·A1

$|\vec{u}| = \sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13}$.
$|\vec{v}| = \sqrt{(-1)^{2} + 5^{2}} = \sqrt{1 + 25} = \sqrt{26}$.

(c) Angle from the geometric form用几何形式求夹角 M1·A1·A1

$$ \cos\theta \;=\; \frac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|} \;=\; \frac{13}{\sqrt{13}\,\sqrt{26}} \;=\; \frac{13}{\sqrt{338}}. $$ Simplify the denominator: $\sqrt{338} = \sqrt{169 \cdot 2} = 13\sqrt{2}$. Hence化简分母：$\sqrt{338} = \sqrt{169 \cdot 2} = 13\sqrt{2}$。因此 $$ \cos\theta \;=\; \frac{13}{13\sqrt{2}} \;=\; \frac{1}{\sqrt{2}} \;=\; \frac{\sqrt{2}}{2}. $$ So $\theta = \arccos\left(\tfrac{\sqrt{2}}{2}\right) = 45.0^{\circ}$ (exact).所以 $\theta = \arccos\left(\tfrac{\sqrt{2}}{2}\right) = 45.0^{\circ}$（精确值）。

(d) Classification分类 A1

$\theta = 45.0^{\circ}$ is between $0^{\circ}$ and $90^{\circ}$, so $\theta$ is acute. This matches the sign of the dot product: $\vec{u} \cdot \vec{v} = 13 > 0 \Leftrightarrow \cos\theta > 0 \Leftrightarrow \theta$ acute.$\theta = 45.0^{\circ}$ 介于 $0^{\circ}$ 与 $90^{\circ}$ 之间，所以 $\theta$ 是锐角。这与点积符号一致：$\vec{u} \cdot \vec{v} = 13 > 0 \Leftrightarrow \cos\theta > 0 \Leftrightarrow \theta$ 为锐角。
The sign of $\vec{u} \cdot \vec{v}$ classifies the angle, no inverse-cosine needed.$\vec{u} \cdot \vec{v}$ 的符号即可给夹角分类，无需反余弦。 A four-way diagnostic: $\vec{u} \cdot \vec{v} > 0 \Leftrightarrow$ acute; $= 0 \Leftrightarrow$ right; $< 0 \Leftrightarrow$ obtuse; $\pm |\vec{u}||\vec{v}| \Leftrightarrow$ parallel ($0^{\circ}$ or $180^{\circ}$). Part (d)'s mark is free if you state this rule explicitly. Cross-check on the algebra: simplify $\sqrt{13 \cdot 26}$ before reaching for the calculator, $\sqrt{a^{2} \cdot b} = a\sqrt{b}$ takes a messy decimal and turns it into the exact $13\sqrt{2}$, which gives an exact $\theta = 45^{\circ}$. Markers reward the exact-form route over decimal approximation.四档判别：$\vec{u} \cdot \vec{v} > 0 \Leftrightarrow$ 锐角；$= 0 \Leftrightarrow$ 直角；$< 0 \Leftrightarrow$ 钝角；$\pm |\vec{u}||\vec{v}| \Leftrightarrow$ 平行（$0^{\circ}$ 或 $180^{\circ}$）。明示此规则即可白拿 (d) 那一分。代数自查：动计算器前先化简 $\sqrt{13 \cdot 26}$，$\sqrt{a^{2} \cdot b} = a\sqrt{b}$ 把一串杂乱小数变成精确的 $13\sqrt{2}$，从而得到精确的 $\theta = 45^{\circ}$。阅卷会偏爱精确根式而非小数近似。
Q7HARDHonors 🇨🇦 ON ON Provincial-style §5 Vectors in 3-D · MCV4U C1-C3 [9 marks]

$\vec{p} = \langle 1, -2, 2\rangle$, $\vec{q} = \langle 3, 0, -4\rangle$. (a) Magnitudes. (b) $\hat{p}$. (c) $\vec{p} \cdot \vec{q}$. (d) Angle.

Answer:  (a) $|\vec{p}| = 3$, $|\vec{q}| = 5$  ·  (b) $\hat{p} = \langle \tfrac{1}{3}, -\tfrac{2}{3}, \tfrac{2}{3}\rangle$  ·  (c) $\vec{p} \cdot \vec{q} = -5$  ·  (d) $\theta \approx 109.5^{\circ}$ (obtuse)

(a) 3-D magnitudes A1·A1

$|\vec{p}| = \sqrt{1^{2} + (-2)^{2} + 2^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{q}| = \sqrt{3^{2} + 0^{2} + (-4)^{2}} = \sqrt{9 + 0 + 16} = \sqrt{25} = 5$.

(b) Unit vector in the direction of $\vec{p}$ M1·A1

$$ \hat{p} \;=\; \frac{\vec{p}}{|\vec{p}|} \;=\; \frac{1}{3} \langle 1, -2, 2\rangle \;=\; \left\langle \tfrac{1}{3},\; -\tfrac{2}{3},\; \tfrac{2}{3}\right\rangle. $$ Verification: $|\hat{p}|^{2} = \tfrac{1}{9} + \tfrac{4}{9} + \tfrac{4}{9} = \tfrac{9}{9} = 1$. $\checkmark$

(c) 3-D dot product M1·A1

$$ \vec{p} \cdot \vec{q} \;=\; (1)(3) + (-2)(0) + (2)(-4) \;=\; 3 + 0 - 8 \;=\; -5. $$ Note the sign is negative, signalling an obtuse angle.

(d) Angle via the geometric form M1·A1·A1

$$ \cos\theta \;=\; \frac{\vec{p} \cdot \vec{q}}{|\vec{p}|\,|\vec{q}|} \;=\; \frac{-5}{3 \cdot 5} \;=\; -\tfrac{1}{3}. $$ $$ \theta \;=\; \arccos\!\left(-\tfrac{1}{3}\right) \;\approx\; 109.471^{\circ} \;\approx\; 109.5^{\circ}. $$ Confirmed obtuse: $\cos\theta < 0$.
3-D dot product is just 2-D with one extra term, the magnitude jump is the only conceptual step. Students who handle $\vec{u} \cdot \vec{v} = u_{1}v_{1} + u_{2}v_{2}$ in 2-D have no trouble extending to $u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}$; the trap is in the magnitude: $\sqrt{a^{2} + b^{2} + c^{2}}$, not $\sqrt{a^{2} + b^{2}}$ with $c$ forgotten. The $\arccos\!\left(-\tfrac{1}{3}\right) \approx 109.5^{\circ}$ angle is the famous tetrahedral bond angle from chemistry, methane molecules have $\vec{p}$ and $\vec{q}$-style C−H bonds at exactly this angle. Worth remembering as a numerical sanity check.
Q8HARDHonors 🇨🇦 ON 🇺🇸 US AP-feeder FRQ §6 Cross Product · MCV4U C3 / HSN-VM (+) [8 marks]

$\vec{a} = \langle 2, -1, 3\rangle$, $\vec{b} = \langle 1, 4, -2\rangle$. (a) $\vec{a} \times \vec{b}$ via determinant. (b) Perpendicularity check. (c) $|\vec{a} \times \vec{b}|$ and its geometric meaning.

Answer:  (a) $\vec{a} \times \vec{b} = \langle -10,\; 7,\; 9\rangle$  ·  (b) both dot products $= 0$ $\checkmark$  ·  (c) $|\vec{a} \times \vec{b}| = \sqrt{230} \approx 15.17$, area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$

(a) Cofactor expansion of the determinant M1·A1·A1·A1

$$ \vec{a} \times \vec{b} \;=\; \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 1 & 4 & -2 \end{vmatrix}. $$ Expand along the top row: $$ \vec{a} \times \vec{b} \;=\; \hat{i} \begin{vmatrix} -1 & 3 \\ 4 & -2 \end{vmatrix} \;-\; \hat{j} \begin{vmatrix} 2 & 3 \\ 1 & -2 \end{vmatrix} \;+\; \hat{k} \begin{vmatrix} 2 & -1 \\ 1 & 4 \end{vmatrix}. $$ Compute each $2 \times 2$:
  • $\hat{i}$-component: $(-1)(-2) - (3)(4) = 2 - 12 = -10$.
  • $\hat{j}$-component: the extra minus sign on $\hat{j}$ flips the bracket, $-\bigl[(2)(-2) - (3)(1)\bigr] = -\bigl[-4 - 3\bigr] = -(-7) = +7$.
  • $\hat{k}$-component: $(2)(4) - (-1)(1) = 8 + 1 = 9$.
So $\vec{a} \times \vec{b} = \langle -10,\; 7,\; 9\rangle$.

(b) Perpendicularity check A1·A1

$(\vec{a} \times \vec{b}) \cdot \vec{a} = (-10)(2) + (7)(-1) + (9)(3) = -20 - 7 + 27 = 0$. $\checkmark$
$(\vec{a} \times \vec{b}) \cdot \vec{b} = (-10)(1) + (7)(4) + (9)(-2) = -10 + 28 - 18 = 0$. $\checkmark$
Both dot products vanish, confirming $\vec{a} \times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.

(c) Magnitude and geometric meaning A1·A1

$$ |\vec{a} \times \vec{b}| \;=\; \sqrt{(-10)^{2} + 7^{2} + 9^{2}} \;=\; \sqrt{100 + 49 + 81} \;=\; \sqrt{230} \;\approx\; 15.17. $$ Geometric meaning: $|\vec{a} \times \vec{b}|$ is the area of the parallelogram with sides $\vec{a}$ and $\vec{b}$. (The triangle with the same two sides has half that area, $\tfrac{1}{2}\sqrt{230} \approx 7.58$.)
The $\hat{j}$ minus sign is the #1 cross-product trap. Cofactor expansion of a $3 \times 3$ determinant alternates signs $+, -, +$ across the top row. That middle $-$ is in front of the $\hat{j}$ bracket, and forgetting it gives the negative of the correct middle component, so the entire cross product comes out wrong by a single sign flip. Self-audit: after you compute $\vec{a} \times \vec{b}$, always immediately check $(\vec{a} \times \vec{b}) \cdot \vec{a} = 0$. If it's not zero, the middle-component sign is the first place to look. AP-feeder graders deduct one A1 for the sign-flip even if the magnitudes of all three components are right.
Q9HARDHonors 🇨🇦 ON 🇺🇸 US AP-feeder FRQ §4-§6 Mixed Vector Algebra · MCV4U C3 / HSN-VM (+) [10 marks]

$\vec{u} = \langle 1, 2, -1\rangle$, $\vec{v} = \langle 2, 0, 1\rangle$. (a) $\vec{u} \cdot \vec{v}$ + angle. (b) $\vec{u} \times \vec{v}$. (c) Lagrange's identity check. (d) $\vec{v} \times \vec{u}$.

Answer:  (a) $\vec{u} \cdot \vec{v} = 1$, $\theta \approx 79.5^{\circ}$  ·  (b) $\vec{u} \times \vec{v} = \langle 2, -3, -4\rangle$  ·  (c) both sides $= 30$ $\checkmark$  ·  (d) $\vec{v} \times \vec{u} = \langle -2, 3, 4\rangle$

(a) Dot product and angle M1·A1·A1

$\vec{u} \cdot \vec{v} = (1)(2) + (2)(0) + (-1)(1) = 2 + 0 - 1 = 1$.
$|\vec{u}| = \sqrt{1 + 4 + 1} = \sqrt{6}$, $|\vec{v}| = \sqrt{4 + 0 + 1} = \sqrt{5}$. $$ \cos\theta \;=\; \frac{1}{\sqrt{6}\sqrt{5}} \;=\; \frac{1}{\sqrt{30}} \;\approx\; 0.1826 \;\Longrightarrow\; \theta \;\approx\; 79.48^{\circ} \;\approx\; 79.5^{\circ}. $$

(b) Cross product via determinant M1·A1·A1

$$ \vec{u} \times \vec{v} \;=\; \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 0 & 1 \end{vmatrix}. $$
  • $\hat{i}$: $(2)(1) - (-1)(0) = 2 - 0 = 2$.
  • $\hat{j}$: $-\bigl[(1)(1) - (-1)(2)\bigr] = -\bigl[1 + 2\bigr] = -3$. (Note the minus sign trap again.)
  • $\hat{k}$: $(1)(0) - (2)(2) = 0 - 4 = -4$.
So $\vec{u} \times \vec{v} = \langle 2,\; -3,\; -4\rangle$. (Perpendicularity audit: $\langle 2, -3, -4\rangle \cdot \langle 1, 2, -1\rangle = 2 - 6 + 4 = 0$ $\checkmark$; $\langle 2, -3, -4\rangle \cdot \langle 2, 0, 1\rangle = 4 + 0 - 4 = 0$ $\checkmark$.)

(c) Lagrange's identity M1·A1·A1

Left-hand side: $|\vec{u}|^{2} |\vec{v}|^{2} = 6 \cdot 5 = 30$.
Right-hand side: $(\vec{u} \cdot \vec{v})^{2} + |\vec{u} \times \vec{v}|^{2} = 1^{2} + (2^{2} + (-3)^{2} + (-4)^{2}) = 1 + (4 + 9 + 16) = 1 + 29 = 30$.
Both sides equal $30$. $\checkmark$ Lagrange's identity holds, this is the algebraic statement of $\sin^{2}\theta + \cos^{2}\theta = 1$ in vector form.

(d) Antisymmetry A1

By antisymmetry $\vec{v} \times \vec{u} = -(\vec{u} \times \vec{v}) = -\langle 2, -3, -4\rangle = \langle -2,\; 3,\; 4\rangle$. No fresh computation needed.
Lagrange ties dot and cross together, $|\vec{u}|^{2} |\vec{v}|^{2}$ splits into "the cos part squared plus the sin part squared". The identity $$ |\vec{u}|^{2} |\vec{v}|^{2} \;=\; (\vec{u} \cdot \vec{v})^{2} \;+\; |\vec{u} \times \vec{v}|^{2} $$ is exactly $|\vec{u}|^{2}|\vec{v}|^{2}(\cos^{2}\theta + \sin^{2}\theta) = |\vec{u}|^{2}|\vec{v}|^{2}$. Numerically it's a free audit on both your dot and your cross product: if part (c) doesn't balance, one of (a) or (b) is wrong. The antisymmetry $\vec{v} \times \vec{u} = -(\vec{u} \times \vec{v})$ from the right-hand rule is the most-tested cross-product property on MCV4U C3 exams, swap inputs, negate every component.
PART III  ·  MODELING / APPLIED · SOLUTIONSAP Physics-feeder + ON MCV4U C2-C3 · 28 marks

Section C · Modeling and Applications

Q10MEDIUMHonors 🇨🇦 ON 🇺🇸 US AP-feeder FRQ §7 Force Resolution / Inclined Plane · HSN-VM.A.3 (+) [9 marks]

10 kg crate, $25^{\circ}$ frictionless ramp, $g = 9.8$. (a) $W$. (b) $W_{\parallel}, W_{\perp}$. (c) Holding force. (d) Pythagoras check.

Answer:  (a) $W = 98$ N  ·  (b) $W_{\parallel} = W \sin 25^{\circ} \approx 41.4$ N, $W_{\perp} = W \cos 25^{\circ} \approx 88.8$ N  ·  (c) $\approx 41.4$ N up the slope  ·  (d) $41.4^{2} + 88.8^{2} \approx 9604 = 98^{2}$ $\checkmark$

(a) Weight magnitude A1

$W = mg = (10)(9.8) = 98$ N (straight down).

(b) Resolve along and perpendicular to the slope M1·A1·A1·A1

Set up axes along the ramp. The angle between $\vec{W}$ (which points straight down) and the inward normal to the ramp is the same as the ramp incline $\alpha = 25^{\circ}$ (basic geometry: tilt the ground, the gravity vector tilts the same amount relative to the new normal).
  • Parallel-to-ramp component (down the slope): $W_{\parallel} = W \sin\alpha = 98 \sin 25^{\circ} \approx 98 (0.4226) \approx 41.4$ N.
  • Perpendicular-to-ramp component (into the surface): $W_{\perp} = W \cos\alpha = 98 \cos 25^{\circ} \approx 98 (0.9063) \approx 88.8$ N.

(c) Holding force A1·A1

For the crate to be stationary on a frictionless slope, the applied force up the ramp must exactly cancel $W_{\parallel}$ (the perpendicular component is cancelled by the normal force from the ramp). So the required force is $\approx 41.4$ N directed up the slope.

(d) Pythagoras sanity check M1·A1

$W_{\parallel}^{2} + W_{\perp}^{2} = 41.4^{2} + 88.8^{2} \approx 1714 + 7886 = 9600 \approx 98^{2} = 9604$. $\checkmark$ (Small drift is rounding; algebraically $W^{2}\sin^{2}\alpha + W^{2}\cos^{2}\alpha = W^{2}$ exactly.)
$\sin$ goes with parallel, $\cos$ goes with perpendicular, because the parallel component is opposite the incline angle. Students who swap $\sin$ and $\cos$ on an inclined-plane problem are the #1 source of dropped AP Physics marks. The clean geometric story: draw the weight vector and the slope; the angle the weight makes with the normal to the slope equals the incline angle $\alpha$; therefore $W_{\perp} = W\cos\alpha$ (adjacent) and $W_{\parallel} = W\sin\alpha$ (opposite). Sanity check: as $\alpha \to 0$ (flat ground), $W_{\parallel} \to 0$ and $W_{\perp} \to W$, all weight presses into the floor, nothing pulls along. As $\alpha \to 90^{\circ}$ (vertical wall), $W_{\parallel} \to W$, all weight pulls "down the slope" since the slope is straight down. The limits decide the formula.
Q11MEDIUMHonors 🇨🇦 ON 🇺🇸 US AP-feeder FRQ §7 Work as Dot Product · HSN-VM.A.3 (+) [9 marks]

Rope tension $80$ N at $30^{\circ}$ above horizontal; $\vec{d} = \langle 12, 0\rangle$ m east. (a) $\vec{F}$ components. (b) $W = \vec{F} \cdot \vec{d}$. (c) Geometric verification. (d) Why no vertical work?

Answer:  (a) $\vec{F} \approx \langle 69.3,\; 40.0\rangle$ N  ·  (b) $W \approx 831.4$ J  ·  (c) $|\vec{F}||\vec{d}|\cos 30^{\circ} \approx 831.4$ J $\checkmark$  ·  (d) $\vec{d}$ has zero $y$-component, so $F_{y} d_{y} = 0$.

(a) Resolve $\vec{F}$ into components M1·A1·A1

With $|\vec{F}| = 80$ N at $\theta = 30^{\circ}$ above the $+x$-axis: $$ F_{x} \;=\; 80 \cos 30^{\circ} \;\approx\; 80 \cdot \tfrac{\sqrt{3}}{2} \;\approx\; 69.282 \;\approx\; 69.3, $$ $$ F_{y} \;=\; 80 \sin 30^{\circ} \;=\; 80 \cdot \tfrac{1}{2} \;=\; 40.0. $$ So $\vec{F} \approx \langle 69.3,\; 40.0\rangle$ N.

(b) Work via algebraic dot product M1·A1·A1

$$ W \;=\; \vec{F} \cdot \vec{d} \;=\; (69.282)(12) + (40.0)(0) \;=\; 831.38 + 0 \;\approx\; 831.4 \;\text{J}. $$ Units: newton-metres are joules ($1 \,\text{N} \cdot \text{m} = 1\,\text{J}$).

(c) Geometric verification M1·A1

$$ W \;=\; |\vec{F}|\,|\vec{d}|\cos\theta \;=\; (80)(12) \cos 30^{\circ} \;=\; 960 \cdot \tfrac{\sqrt{3}}{2} \;\approx\; 960 (0.8660) \;\approx\; 831.4 \;\text{J}. $$ Both forms agree. $\checkmark$

(d) Why the vertical $\vec{F}$-component does no work A1

Work is $\vec{F} \cdot \vec{d} = F_{x} d_{x} + F_{y} d_{y}$. Here $d_{y} = 0$ (the sled moves horizontally), so the second term vanishes regardless of $F_{y}$, the vertical pull does no work because there is no vertical displacement to be done over.
Work is a dot product, and dot products kill perpendicular components. The two equivalent statements of work, algebraic ($\vec{F}\cdot\vec{d} = \sum F_i d_i$) and geometric ($|\vec{F}||\vec{d}|\cos\theta$), are the cleanest "two-form sanity check" in vector physics. If part (c) doesn't match part (b), the most likely error is a degree-mode / radian-mode calculator slip, or a missing minus sign on a force directed against motion (negative work). The general lesson from (d): only the component of force along the displacement does work; perpendicular components produce zero work, exactly the AP Physics "centripetal force does no work on a particle in uniform circular motion" result, packaged at the high-school vector level.
Q12HARDHonors 🇨🇦 ON 🇺🇸 US AP-feeder FRQ §6 + §7 Torque (Moment) as Cross Product · HSN-VM.A.3 (+) [10 marks]

Wrench: $\vec{F} = \langle 0, 10, 0\rangle$ N, $\vec{r} = \langle 0.30, 0, 0\rangle$ m. (a) $\vec{\tau}$. (b) $|\vec{\tau}|$ + direction. (c) New position $\vec{r}' = \langle 0.30, 0.10, 0\rangle$ m, recompute $\vec{\tau}'$. (d) Why no torque gain?

Answer:  (a) $\vec{\tau} = \langle 0, 0, 3\rangle$ N·m  ·  (b) $|\vec{\tau}| = 3$ N·m, $+\hat{k}$ direction (out of the page)  ·  (c) $\vec{\tau}' = \langle 0, 0, 3\rangle$ N·m  ·  (d) $\vec{r}'$ added a $\hat{j}$-component parallel to $\vec{F}$, which contributes $0$ to the cross product ($\sin 0^{\circ} = 0$).

(a) Compute $\vec{\tau} = \vec{r} \times \vec{F}$ M1·A1·A1

$$ \vec{\tau} \;=\; \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.30 & 0 & 0 \\ 0 & 10 & 0 \end{vmatrix}. $$
  • $\hat{i}$: $(0)(0) - (0)(10) = 0$.
  • $\hat{j}$: $-\bigl[(0.30)(0) - (0)(0)\bigr] = 0$.
  • $\hat{k}$: $(0.30)(10) - (0)(0) = 3$.
$\vec{\tau} = \langle 0,\; 0,\; 3\rangle$ N·m.

(b) Magnitude and direction A1·A1

$|\vec{\tau}| = 3$ N·m. The torque points in the $+\hat{k}$ direction (positive $z$, "out of the page" in the right-hand convention with $+x$ east and $+y$ north). This matches the right-hand rule: fingers curl from $\vec{r}$ ($+\hat{i}$) toward $\vec{F}$ ($+\hat{j}$), thumb points to $+\hat{k}$.

(c) Recompute with $\vec{r}' = \langle 0.30, 0.10, 0\rangle$ M1·A1·A1

$$ \vec{\tau}' \;=\; \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.30 & 0.10 & 0 \\ 0 & 10 & 0 \end{vmatrix}. $$
  • $\hat{i}$: $(0.10)(0) - (0)(10) = 0$.
  • $\hat{j}$: $-\bigl[(0.30)(0) - (0)(0)\bigr] = 0$.
  • $\hat{k}$: $(0.30)(10) - (0.10)(0) = 3 - 0 = 3$.
$\vec{\tau}' = \langle 0,\; 0,\; 3\rangle$ N·m. Identical to $\vec{\tau}$.

(d) Why no torque gain M1·A1

$|\vec{\tau}| = |\vec{r}|\,|\vec{F}|\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. Only the component of $\vec{r}$ perpendicular to $\vec{F}$ contributes to the cross product; the parallel component contributes nothing ($\sin 0^{\circ} = 0$). The added $\langle 0, 0.10, 0\rangle$ piece of $\vec{r}'$ points along $+\hat{j}$, which is the same direction as $\vec{F} = \langle 0, 10, 0\rangle$, so it is the parallel component and adds zero torque. Only the original $\langle 0.30, 0, 0\rangle$ part of $\vec{r}'$ (perpendicular to $\vec{F}$) generates torque, hence $|\vec{\tau}'| = |\vec{\tau}| = 3$ N·m.
Cross product picks out the perpendicular part, parallel components vanish. This is the geometric soul of the cross product, and it is exactly the dual of the dot-product lesson from Q11(d): dot product picks the parallel component (and kills perpendiculars), cross product picks the perpendicular component (and kills parallels). For a torque problem, the practical move is the "moment arm" simplification: drop a perpendicular from the line of action of $\vec{F}$ to the pivot; the length of that perpendicular is the moment arm, and $|\vec{\tau}| = (\text{moment arm}) \cdot |\vec{F}|$. Adding length to $\vec{r}$ in the direction of $\vec{F}$ slides the point of application along the line of action and changes nothing, exactly what part (c) showed numerically. AP Physics graders explicitly probe this geometric understanding on rotational-dynamics free-response.