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Unit C3 · Geometry & Trigonometry (HL only)Unit C3 · 几何与三角（HL 专属）

Vectors向量

IB-Style Practice Questions · Paper 1A · Paper 1B · Paper 2 · Paper 3IB 风格练习题 · 第一卷 A 节 · 第一卷 B 节 · 第二卷 · 第三卷

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2 Paper 3

Syllabus AHL 3.12 to 3.18考纲 AHL 3.12 至 3.18AA HL

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PART I  ·  PAPER 1 SECTION A第一部分  ·  第一卷 A 节No calculator · short response · 20 marks不可使用计算器 · 简答题 · 20 分

Section A · Short ResponseA 节 · 简答题

Show every component computation. State direction vectors explicitly when writing a vector equation of a line. For dot and cross products, write the formula in symbols before substituting. Angles between vectors must be quoted as $\arccos(\cdot)$ with the cosine in exact form, then converted to degrees only at the final step.分量运算每一步都要写。写直线向量方程时须显式标出方向向量。点积与叉积须先以符号形式写出公式，再代入。两向量的夹角应先写成 $\arccos(\cdot)$（cos 取精确形式），最后一步再化为度数。

Q1EASY Paper 1A AHL 3.12 Magnitude & Unit Vector [4 marks]

Consider the vector $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$.考虑向量 $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$。

(a) Find $|\mathbf{a}|$.求 $|\mathbf{a}|$。 [2]
(b) Hence find the unit vector $\hat{\mathbf{a}}$ in the direction of $\mathbf{a}$.由此求与 $\mathbf{a}$ 同向的单位向量 $\hat{\mathbf{a}}$。 [1]
(c) Write down the vector of magnitude $26$ in the direction opposite to $\mathbf{a}$.写出与 $\mathbf{a}$ 反向、模长为 $26$ 的向量。 [1]
Q2MEDIUM Paper 1A AHL 3.13 Dot Product & Angle [5 marks]

Let $\mathbf{u} = (1, 2, 2)$ and $\mathbf{v} = (2, 1, 0)$.设 $\mathbf{u} = (1, 2, 2)$，$\mathbf{v} = (2, 1, 0)$。

(a) Find $\mathbf{u} \cdot \mathbf{v}$.求 $\mathbf{u} \cdot \mathbf{v}$。 [1]
(b) Find $|\mathbf{u}|$ and $|\mathbf{v}|$ in exact form.求 $|\mathbf{u}|$ 与 $|\mathbf{v}|$（精确形式）。 [2]
(c) Hence find the angle $\theta$ between $\mathbf{u}$ and $\mathbf{v}$. Give $\cos\theta$ in exact form and $\theta$ to one decimal place in degrees.由此求 $\mathbf{u}$ 与 $\mathbf{v}$ 的夹角 $\theta$。$\cos\theta$ 给精确形式，$\theta$ 给度数（保留一位小数）。 [2]
Q3MEDIUM Paper 1A AHL 3.14 Vector Equation of a Line [6 marks]

The points $A$ and $B$ have position vectors $\vec{OA} = (1, 2, 3)$ and $\vec{OB} = (4, 0, 5)$.点 $A$、$B$ 的位置向量分别为 $\vec{OA} = (1, 2, 3)$、$\vec{OB} = (4, 0, 5)$。

(a) Find a direction vector for the line $\ell$ through $A$ and $B$.求过 $A$、$B$ 的直线 $\ell$ 的一个方向向量。 [1]
(b) Write a vector equation of $\ell$ in the form $\mathbf{r} = \mathbf{a} + t\mathbf{d}$, taking $\mathbf{a} = \vec{OA}$.以 $\mathbf{a} = \vec{OA}$ 为定点，写出 $\ell$ 的向量方程 $\mathbf{r} = \mathbf{a} + t\mathbf{d}$。 [2]
(c) Determine whether the point $C(7, -2, 7)$ lies on $\ell$. Justify with the value of $t$ (or show that no consistent value exists).判定点 $C(7, -2, 7)$ 是否在 $\ell$ 上。给出对应的 $t$ 值（或说明不存在一致的 $t$）。 [3]
Q4HARD Paper 1A AHL 3.16 Cross Product (HL) [5 marks]

Let $\mathbf{p} = (1, 0, 2)$ and $\mathbf{q} = (3, 1, -1)$.设 $\mathbf{p} = (1, 0, 2)$，$\mathbf{q} = (3, 1, -1)$。

(a) Compute $\mathbf{p} \times \mathbf{q}$.求 $\mathbf{p} \times \mathbf{q}$。 [2]
(b) Verify by dot product that $\mathbf{p} \times \mathbf{q}$ is perpendicular to both $\mathbf{p}$ and $\mathbf{q}$.用点积验证 $\mathbf{p} \times \mathbf{q}$ 同时垂直于 $\mathbf{p}$ 与 $\mathbf{q}$。 [2]
(c) Find the area of the parallelogram with adjacent sides $\mathbf{p}$ and $\mathbf{q}$. Give the answer in exact form.求以 $\mathbf{p}$、$\mathbf{q}$ 为相邻边的平行四边形面积（精确形式）。 [1]
PART II  ·  PAPER 1 SECTION B第二部分  ·  第一卷 B 节No calculator · extended response · 11 marks不可使用计算器 · 长答题 · 11 分

Section B · Extended ResponseB 节 · 长答题

When classifying two lines, work in three steps. Step 1: compare direction vectors; if proportional, the lines are parallel (then test a point on one against the other). Step 2: if not parallel, equate parametric forms and solve the first two scalar equations for the two parameters; Step 3: substitute back into the third equation. Consistent gives intersection; inconsistent gives skew. State your final conclusion in one sentence.判定两条直线分三步走。第一步：比较方向向量；若成比例则平行（再用一条线上的点测试是否在另一条上）。第二步：若不平行，令两条直线的参数方程相等，用前两个标量方程解出两个参数；第三步：把参数代回第三个方程。一致即相交；不一致即异面。最后用一句话给出结论。

Q5HARD Paper 1B AHL 3.15 Classify Two Lines (HL) [11 marks]

Consider the two lines考虑两条直线

$$ L_{1}: \mathbf{r} \;=\; \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}, \qquad L_{2}: \mathbf{r} \;=\; \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} + s \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}. $$
(a) Show that the direction vectors of $L_{1}$ and $L_{2}$ are not parallel.证明 $L_{1}$、$L_{2}$ 的方向向量不平行。 [2]
(b) Set up three scalar equations by equating components of the position vectors. Solve the first two for $s$ and $t$, then verify (or refute) consistency with the third equation. Conclude whether the lines intersect or are skew.令位置向量分量相等，建立三个标量方程。用前两个解出 $s$、$t$，再代入第三个方程验证一致性。判断两直线是相交还是异面。 [5]
(c) If the lines intersect, find the point of intersection. If they are skew, write down the values of $s$ and $t$ from (b) and state the discrepancy.若相交，求交点；若异面，写出 (b) 中的 $s$、$t$，并说明矛盾出现在哪一分量。 [2]
(d) Find the acute angle between $L_{1}$ and $L_{2}$. Give $\cos\theta$ in exact form.求 $L_{1}$ 与 $L_{2}$ 之间的锐角。$\cos\theta$ 给精确形式。 [2]
PART III  ·  PAPER 2第三部分  ·  第二卷Calculator · mixed response · 17 marks可使用计算器 · 混合题型 · 17 分

Paper 2 · Calculator Permitted第二卷 · 允许使用计算器

A graphing calculator is required. For "plane through three points" problems, always take two edge vectors out of one common point and feed their cross product as the normal. For line meets plane, substitute the parametric line into the scalar plane equation and solve for the parameter. Exact fractions beat decimals: keep fractions until the very last numerical step.需要图形计算器（GDC）。求"过三点的平面"时，固定一点向其余两点引出两条边向量，叉积即为法向量。求"直线与平面交点"时，把参数化直线代入标量平面方程，解出参数。能保留分数就别用小数：直到最后一步才取数值。

Q6MEDIUM Paper 2 AHL 3.17 Plane through Three Points (HL) [7 marks]

The points $A(1, 0, 0)$, $B(0, 2, 0)$, $C(0, 0, 3)$ lie in a plane $\Pi$.点 $A(1, 0, 0)$、$B(0, 2, 0)$、$C(0, 0, 3)$ 共面于 $\Pi$。

(a) Find the edge vectors $\vec{AB}$ and $\vec{AC}$.求边向量 $\vec{AB}$ 与 $\vec{AC}$。 [2]
(b) Compute $\mathbf{n} = \vec{AB} \times \vec{AC}$ and explain why $\mathbf{n}$ is normal to $\Pi$.求 $\mathbf{n} = \vec{AB} \times \vec{AC}$，并说明 $\mathbf{n}$ 为何是 $\Pi$ 的法向量。 [3]
(c) Hence write the Cartesian equation of $\Pi$ in the form $ax + by + cz = d$.由此写出 $\Pi$ 的直角坐标方程，形式为 $ax + by + cz = d$。 [2]
Q7HARD Paper 2 AHL 3.18 Line Meets Plane (HL) [10 marks]

The line $\ell$ has vector equation $\mathbf{r} = (1, 2, 3) + t(3, -2, 2)$. The plane $\Pi$ has equation $6x + 3y + 2z = 6$.直线 $\ell$ 的向量方程为 $\mathbf{r} = (1, 2, 3) + t(3, -2, 2)$。平面 $\Pi$ 的方程为 $6x + 3y + 2z = 6$。

(a) Substitute the parametric form of $\ell$ into the equation of $\Pi$ and solve for $t$.把 $\ell$ 的参数形式代入 $\Pi$ 的方程，解出 $t$。 [3]
(b) Hence find the coordinates of the point of intersection $P$. Give exact fractions.由此求交点 $P$ 的坐标（精确分数形式）。 [2]
(c) Find the acute angle $\alpha$ between $\ell$ and $\Pi$. (Recall $\sin\alpha = \dfrac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$, where $\mathbf{d}$ is a direction vector of $\ell$ and $\mathbf{n}$ is a normal to $\Pi$.) Give $\sin\alpha$ in exact form and $\alpha$ to one decimal place in degrees.求 $\ell$ 与 $\Pi$ 的锐角 $\alpha$。（提示：$\sin\alpha = \dfrac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$，其中 $\mathbf{d}$ 为 $\ell$ 的方向向量，$\mathbf{n}$ 为 $\Pi$ 的法向量。）$\sin\alpha$ 给精确形式，$\alpha$ 给度数（一位小数）。 [3]
(d) State the geometric configuration that would arise if $\mathbf{d} \cdot \mathbf{n} = 0$ instead.若 $\mathbf{d} \cdot \mathbf{n} = 0$，请说明此时的几何情形。 [2]
PART IV  ·  PAPER 3第四部分  ·  第三卷Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分

Paper 3 · HL Extended Problem第三卷 · HL 长题探究

A graphing calculator is required. Method marks dominate. For "distance from a point to a line" there are two equivalent routes: the cross-product formula $d = |\vec{AP} \times \mathbf{d}| / |\mathbf{d}|$ and the foot-of-perpendicular method via parameter projection. Both are accepted; for full marks, demonstrate that you understand they give the same number.需要图形计算器（GDC）。方法分占主导。"点到直线距离"有两条等价路径：叉积公式 $d = |\vec{AP} \times \mathbf{d}| / |\mathbf{d}|$ 与"垂足参数投影法"。两者都被接受；为拿满分，需说明二者给出同一数值。

Q8HARD Paper 3 AHL 3.15 / 3.18 Distances in Space (HL) [15 marks]

Let $P = (2, 1, 0)$. Consider the line $L: \mathbf{r} = (1, 0, 2) + t(2, 1, -1)$ with point $A = (1, 0, 2)$ on $L$ and direction $\mathbf{d} = (2, 1, -1)$. Consider also the plane $\Pi: 6x + 3y + 2z = 6$ with normal $\mathbf{n} = (6, 3, 2)$.设 $P = (2, 1, 0)$。考虑直线 $L: \mathbf{r} = (1, 0, 2) + t(2, 1, -1)$，其中 $A = (1, 0, 2) \in L$，方向 $\mathbf{d} = (2, 1, -1)$。再考虑平面 $\Pi: 6x + 3y + 2z = 6$，法向量 $\mathbf{n} = (6, 3, 2)$。

(a) Compute $\vec{AP}$ and $\vec{AP} \times \mathbf{d}$. Show the component working.求 $\vec{AP}$ 与 $\vec{AP} \times \mathbf{d}$，写出分量运算。 [3]
(b) Hence show that the distance from $P$ to $L$ equals $\sqrt{\tfrac{11}{6}}$ (equivalently, $\tfrac{\sqrt{66}}{6}$).由此证明 $P$ 到 $L$ 的距离为 $\sqrt{\tfrac{11}{6}}$（即 $\tfrac{\sqrt{66}}{6}$）。 [2]
(c) Verify the distance found in (b) by the foot-of-perpendicular method: parameterise points $Q(t)$ on $L$, minimise $|PQ(t)|^{2}$ in $t$ (equivalently, impose $\vec{PQ}(t) \cdot \mathbf{d} = 0$), and compute $|PQ|$ at the optimal $t$.用"垂足法"复核 (b) 的距离：将 $L$ 上的点参数化为 $Q(t)$，对 $t$ 极小化 $|PQ(t)|^{2}$（等价于令 $\vec{PQ}(t) \cdot \mathbf{d} = 0$），并在最优 $t$ 处计算 $|PQ|$。 [4]
(d) Compute the perpendicular distance from $P$ to the plane $\Pi$ using the formula $d_{\Pi} = \dfrac{|\mathbf{n} \cdot \vec{OP} - 6|}{|\mathbf{n}|}$. Give an exact value.用公式 $d_{\Pi} = \dfrac{|\mathbf{n} \cdot \vec{OP} - 6|}{|\mathbf{n}|}$ 计算 $P$ 到平面 $\Pi$ 的垂直距离（精确值）。 [3]
(e) Compare the two distances numerically (to three decimal places) and identify which of $L$ and $\Pi$ is closer to $P$.将两个距离化为数值（保留三位小数）并比较，指出 $L$ 与 $\Pi$ 哪一个离 $P$ 更近。 [3]