IB Math AA HL · Unit C3: Vectors · Solutions

Q1EASYPaper 1AAHL 3.12 Magnitude & Unit Vector[4 marks]

$\mathbf{a} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$. Find $|\mathbf{a}|$, $\hat{\mathbf{a}}$, and the vector of magnitude $26$ opposite to $\mathbf{a}$.$\mathbf{a} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$。求 $|\mathbf{a}|$、$\hat{\mathbf{a}}$，以及与 $\mathbf{a}$ 反向、模长为 $26$ 的向量。

Answers:答案： (a) $|\mathbf{a}| = 13$ · (b) $\hat{\mathbf{a}} = \tfrac{1}{13}(3, 4, 12)$ · (c) $-2\mathbf{a} = (-6, -8, -24)$

(a) Magnitude M1·A1

$|\mathbf{a}|^{2} = 3^{2} + 4^{2} + 12^{2} = 9 + 16 + 144 = 169$, so $|\mathbf{a}| = \sqrt{169} = 13$.

(b) Unit vector A1

$\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|} = \dfrac{1}{13}(3, 4, 12) = \left(\tfrac{3}{13}, \tfrac{4}{13}, \tfrac{12}{13}\right)$.

(c) Magnitude $26$, opposite direction A1

The unit vector in the opposite direction is $-\hat{\mathbf{a}}$. Scaling by $26 = 2 \cdot 13$ gives $26 \cdot (-\hat{\mathbf{a}}) = -2\mathbf{a} = (-6, -8, -24)$.

The $(3, 4, 12)$ Pythagorean quadruple. Just as $(3, 4, 5)$ is the smallest right-triangle integer triple, $(3, 4, 12, 13)$ is one of the smallest Pythagorean quadruples ($3^{2} + 4^{2} + 12^{2} = 13^{2}$). Recognising it on sight saves $20$ seconds: examiners deliberately seed components that yield integer magnitudes ($(1, 2, 2) \to 3$; $(2, 3, 6) \to 7$; $(2, 6, 9) \to 11$). Whenever an exam component vector "looks clean", check for a quadruple before pulling out a calculator. Also note: a unit vector has magnitude $1$ exactly, so $|\hat{\mathbf{a}}|^{2} = \tfrac{9 + 16 + 144}{169} = 1$ — a $5$-second self-check that catches arithmetic slips.

(a) 模长 M1·A1

$|\mathbf{a}|^{2} = 3^{2} + 4^{2} + 12^{2} = 9 + 16 + 144 = 169$，故 $|\mathbf{a}| = \sqrt{169} = 13$。

(b) 单位向量 A1

$\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|} = \dfrac{1}{13}(3, 4, 12) = \left(\tfrac{3}{13}, \tfrac{4}{13}, \tfrac{12}{13}\right)$。

(c) 模长 $26$、反向 A1

反向单位向量为 $-\hat{\mathbf{a}}$。乘以 $26 = 2 \cdot 13$ 得 $26 \cdot (-\hat{\mathbf{a}}) = -2\mathbf{a} = (-6, -8, -24)$。

$(3, 4, 12)$ 的勾股四元组。如同 $(3, 4, 5)$ 是最小直角三角形整数三元组，$(3, 4, 12, 13)$ 是最小的勾股四元组之一（$3^{2} + 4^{2} + 12^{2} = 13^{2}$）。一眼识破能省 $20$ 秒：考官刻意挑选能得整数模长的分量（$(1, 2, 2) \to 3$；$(2, 3, 6) \to 7$；$(2, 6, 9) \to 11$）。考场上分量"很干净"时，先猜四元组再动计算器。另注：单位向量模恰为 $1$，故 $|\hat{\mathbf{a}}|^{2} = \tfrac{9 + 16 + 144}{169} = 1$—$5$ 秒自检捕错。

Q2MEDIUMPaper 1AAHL 3.13 Dot Product & Angle[5 marks]

$\mathbf{u} = (1, 2, 2)$, $\mathbf{v} = (2, 1, 0)$. Find $\mathbf{u} \cdot \mathbf{v}$, $|\mathbf{u}|$, $|\mathbf{v}|$, and the angle $\theta$ between them.$\mathbf{u} = (1, 2, 2)$，$\mathbf{v} = (2, 1, 0)$。求 $\mathbf{u} \cdot \mathbf{v}$、$|\mathbf{u}|$、$|\mathbf{v}|$ 与夹角 $\theta$。

Answers:答案： (a) $\mathbf{u} \cdot \mathbf{v} = 4$ · (b) $|\mathbf{u}| = 3,\; |\mathbf{v}| = \sqrt{5}$ · (c) $\cos\theta = \tfrac{4\sqrt{5}}{15},\; \theta \approx 53.4^{\circ}$

(a) Dot product A1

$\mathbf{u} \cdot \mathbf{v} = (1)(2) + (2)(1) + (2)(0) = 2 + 2 + 0 = 4$.

(b) Magnitudes A1·A1

$|\mathbf{u}| = \sqrt{1^{2} + 2^{2} + 2^{2}} = \sqrt{9} = 3$. $\;|\mathbf{v}| = \sqrt{2^{2} + 1^{2} + 0^{2}} = \sqrt{5}$.

(c) Angle M1·A1

Formula: $\cos\theta = \dfrac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|} = \dfrac{4}{3\sqrt{5}}$. Rationalise: $\cos\theta = \dfrac{4\sqrt{5}}{15}$. Then $\theta = \arccos\!\left(\dfrac{4}{3\sqrt{5}}\right) \approx 53.4^{\circ}$ (1 dp).

Sign of the dot product reads geometry off the page. $\mathbf{u} \cdot \mathbf{v} > 0$ means acute angle; $= 0$ means perpendicular; $< 0$ means obtuse. Before computing magnitudes, glance at the sign of the dot product: if it is positive (as here), the angle is in $[0^{\circ}, 90^{\circ})$ and $\cos\theta > 0$ — a negative cosine answer is a sign error. Secondly: never write the cosine as a decimal in Paper 1. "$\cos\theta = 0.596$" is not "exact form"; $\tfrac{4}{3\sqrt{5}}$ or $\tfrac{4\sqrt{5}}{15}$ is. The exact form is what earns the A1.

(a) 点积 A1

$\mathbf{u} \cdot \mathbf{v} = (1)(2) + (2)(1) + (2)(0) = 2 + 2 + 0 = 4$。

(b) 模长 A1·A1

$|\mathbf{u}| = \sqrt{1^{2} + 2^{2} + 2^{2}} = \sqrt{9} = 3$。$\;|\mathbf{v}| = \sqrt{2^{2} + 1^{2} + 0^{2}} = \sqrt{5}$。

(c) 夹角 M1·A1

公式 $\cos\theta = \dfrac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|} = \dfrac{4}{3\sqrt{5}}$。有理化：$\cos\theta = \dfrac{4\sqrt{5}}{15}$。$\theta = \arccos\!\left(\dfrac{4}{3\sqrt{5}}\right) \approx 53.4^{\circ}$（一位小数）。

点积的符号能从纸面读出几何。$\mathbf{u} \cdot \mathbf{v} > 0$ 为锐角；$= 0$ 为垂直；$< 0$ 为钝角。算模长前先瞥一眼符号：若为正（如本题），则角在 $[0^{\circ}, 90^{\circ})$、$\cos\theta > 0$—若答案出负值就是符号错。其次：Paper 1 切勿把 cos 写成小数。"$\cos\theta = 0.596$" 不算精确形式；$\tfrac{4}{3\sqrt{5}}$ 或 $\tfrac{4\sqrt{5}}{15}$ 才是。给 A1 的就是精确形式。

Q3MEDIUMPaper 1AAHL 3.14 Vector Equation of a Line[6 marks]

$\vec{OA} = (1, 2, 3)$, $\vec{OB} = (4, 0, 5)$. Find a direction vector of $\ell$ through $A$, $B$; write $\ell$ as $\mathbf{r} = \mathbf{a} + t\mathbf{d}$; decide whether $C(7, -2, 7)$ lies on $\ell$.$\vec{OA} = (1, 2, 3)$，$\vec{OB} = (4, 0, 5)$。求过 $A$、$B$ 的直线 $\ell$ 的方向向量；写出 $\ell$ 的向量方程；判定 $C(7, -2, 7)$ 是否在 $\ell$ 上。

Answers:答案： (a) $\mathbf{d} = (3, -2, 2)$ · (b) $\mathbf{r} = (1, 2, 3) + t(3, -2, 2)$ · (c) $C$ lies on $\ell$ at $t = 2$$C$ 在 $\ell$ 上，$t = 2$

(a) Direction vector A1

$\mathbf{d} = \vec{AB} = \vec{OB} - \vec{OA} = (4 - 1, 0 - 2, 5 - 3) = (3, -2, 2)$.

(b) Vector equation M1·A1

With base point $\mathbf{a} = \vec{OA}$ and direction $\mathbf{d}$: $$ \ell:\; \mathbf{r} \;=\; \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}, \quad t \in \mathbb{R}. $$

(c) Is $C(7, -2, 7)$ on $\ell$? M1·A1·R1

Set $\mathbf{r} = (7, -2, 7)$ and read each component:

$1 + 3t = 7 \;\Rightarrow\; t = 2$.
$2 - 2t = -2 \;\Rightarrow\; t = 2$.
$3 + 2t = 7 \;\Rightarrow\; t = 2$.

All three components give the same $t$, so $C$ lies on $\ell$ at $t = 2$.

The "single $t$" consistency test. Asking "does $P$ lie on $\ell$?" is exactly: solve one scalar equation for $t$, then verify the other two yield the same value. One inconsistent component is enough to conclude $P \notin \ell$. Students who write only the $x$-equation and stop earn $\le 1$ of the $3$ marks — the M1 is for substitution but the A1 and R1 come from the verification. A second exam tip: always pick the equation with the simplest coefficient first (here, the $x$-equation $1 + 3t = 7$ gives $t$ instantly). Solve for $t$ once, then substitute back to verify; don't re-solve each scalar equation for $t$ independently.

(a) 方向向量 A1

$\mathbf{d} = \vec{AB} = \vec{OB} - \vec{OA} = (4 - 1, 0 - 2, 5 - 3) = (3, -2, 2)$。

(b) 向量方程 M1·A1

取基点 $\mathbf{a} = \vec{OA}$、方向 $\mathbf{d}$： $$ \ell:\; \mathbf{r} \;=\; \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}, \quad t \in \mathbb{R}. $$

(c) $C(7, -2, 7)$ 是否在 $\ell$ 上？ M1·A1·R1

令 $\mathbf{r} = (7, -2, 7)$，逐分量比对：

$1 + 3t = 7 \;\Rightarrow\; t = 2$。
$2 - 2t = -2 \;\Rightarrow\; t = 2$。
$3 + 2t = 7 \;\Rightarrow\; t = 2$。

三个分量给出同一 $t$，故 $C$ 在 $\ell$ 上，$t = 2$。

"单一 $t$" 一致性检验。问"$P$ 是否在 $\ell$ 上"，本质就是：用一条标量方程解出 $t$，再核对其他两条是否给出同一 $t$。任一分量不一致即可断言 $P \notin \ell$。只写 $x$ 方程就停笔的同学最多拿 $1/3$—M1 给代入，A1 与 R1 给验证。考场技巧二：先挑系数最简单的方程（本题 $x$ 方程 $1 + 3t = 7$ 秒得 $t$），解一次 $t$ 后代回验证，别每条标量方程独立再解一遍 $t$。

Q4HARDPaper 1AAHL 3.16 Cross Product (HL)[5 marks]

$\mathbf{p} = (1, 0, 2)$, $\mathbf{q} = (3, 1, -1)$. Compute $\mathbf{p} \times \mathbf{q}$; verify perpendicularity to $\mathbf{p}$ and $\mathbf{q}$; find the parallelogram area.$\mathbf{p} = (1, 0, 2)$，$\mathbf{q} = (3, 1, -1)$。求 $\mathbf{p} \times \mathbf{q}$；用点积验证其垂直于 $\mathbf{p}$、$\mathbf{q}$；求平行四边形面积。

Answers:答案： (a) $\mathbf{p} \times \mathbf{q} = (-2, 7, 1)$ · (b) both dot products $= 0$两个点积均为 $0$ · (c) Area $= 3\sqrt{6}$面积 $= 3\sqrt{6}$

(a) Compute $\mathbf{p} \times \mathbf{q}$ M1·A1

Use the determinant expansion: $$ \mathbf{p} \times \mathbf{q} \;=\; \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2 \\ 3 & 1 & -1 \end{vmatrix} \;=\; \mathbf{i}\bigl(0\cdot(-1) - 2\cdot 1\bigr) - \mathbf{j}\bigl(1\cdot(-1) - 2\cdot 3\bigr) + \mathbf{k}\bigl(1\cdot 1 - 0\cdot 3\bigr). $$ Simplify: $\mathbf{p} \times \mathbf{q} = -2\,\mathbf{i} - (-7)\,\mathbf{j} + 1\,\mathbf{k} = (-2, 7, 1)$.

(b) Perpendicularity check A1·A1

$\mathbf{p} \cdot (\mathbf{p} \times \mathbf{q}) = (1)(-2) + (0)(7) + (2)(1) = -2 + 0 + 2 = 0$. $\;\checkmark$
$\mathbf{q} \cdot (\mathbf{p} \times \mathbf{q}) = (3)(-2) + (1)(7) + (-1)(1) = -6 + 7 - 1 = 0$. $\;\checkmark$

(c) Parallelogram area A1

Area $= |\mathbf{p} \times \mathbf{q}| = \sqrt{(-2)^{2} + 7^{2} + 1^{2}} = \sqrt{4 + 49 + 1} = \sqrt{54} = 3\sqrt{6}$.

The middle-component sign flip. The $\mathbf{j}$ entry of $\mathbf{p} \times \mathbf{q}$ carries a negative sign in the cofactor expansion. Roughly one student in three forgets it and gets $(-2, -7, 1)$, then loses both the (a) A1 and the (b) perpendicularity verification (because the false vector won't be orthogonal to $\mathbf{p}$ or $\mathbf{q}$). Two safety nets: (i) after writing the answer, always dot it with one of the input vectors — if non-zero, you have a sign error; (ii) memorise the cyclic shortcut: for $(a_{1}, a_{2}, a_{3}) \times (b_{1}, b_{2}, b_{3})$ the components are $(a_{2}b_{3} - a_{3}b_{2},\; a_{3}b_{1} - a_{1}b_{3},\; a_{1}b_{2} - a_{2}b_{1})$ — rotate indices $1 \to 2 \to 3 \to 1$, no minus sign hiding in $\mathbf{j}$.

(a) 求 $\mathbf{p} \times \mathbf{q}$ M1·A1

用行列式展开： $$ \mathbf{p} \times \mathbf{q} \;=\; \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2 \\ 3 & 1 & -1 \end{vmatrix} \;=\; \mathbf{i}\bigl(0\cdot(-1) - 2\cdot 1\bigr) - \mathbf{j}\bigl(1\cdot(-1) - 2\cdot 3\bigr) + \mathbf{k}\bigl(1\cdot 1 - 0\cdot 3\bigr). $$ 化简：$\mathbf{p} \times \mathbf{q} = -2\,\mathbf{i} - (-7)\,\mathbf{j} + 1\,\mathbf{k} = (-2, 7, 1)$。

(b) 垂直性验证 A1·A1

$\mathbf{p} \cdot (\mathbf{p} \times \mathbf{q}) = (1)(-2) + (0)(7) + (2)(1) = -2 + 0 + 2 = 0$。$\;\checkmark$
$\mathbf{q} \cdot (\mathbf{p} \times \mathbf{q}) = (3)(-2) + (1)(7) + (-1)(1) = -6 + 7 - 1 = 0$。$\;\checkmark$

(c) 平行四边形面积 A1

面积 $= |\mathbf{p} \times \mathbf{q}| = \sqrt{(-2)^{2} + 7^{2} + 1^{2}} = \sqrt{4 + 49 + 1} = \sqrt{54} = 3\sqrt{6}$。

中间分量的符号翻转。叉积里 $\mathbf{j}$ 项的代数余子式带负号。约三分之一同学忘了翻号、写成 $(-2, -7, 1)$，结果连同 (a) A1 与 (b) 验证都丢（错向量不再垂直于 $\mathbf{p}$、$\mathbf{q}$）。两条安全网：(i) 写完答案后必与某输入向量做点积—非零即符号错；(ii) 背循环公式：$(a_{1}, a_{2}, a_{3}) \times (b_{1}, b_{2}, b_{3}) = (a_{2}b_{3} - a_{3}b_{2},\; a_{3}b_{1} - a_{1}b_{3},\; a_{1}b_{2} - a_{2}b_{1})$—下标循环 $1 \to 2 \to 3 \to 1$，$\mathbf{j}$ 项里没有藏负号。

Section B · Worked SolutionsB 节 · 详细解析

Q5HARDPaper 1BAHL 3.15 Classify Two Lines (HL)[11 marks]

$L_{1}: \mathbf{r} = (1, 0, 2) + t(2, 1, -1)$ and $L_{2}: \mathbf{r} = (3, 1, 1) + s(1, -1, 2)$. (a) Show non-parallel; (b) classify; (c) intersection point; (d) acute angle.$L_{1}: \mathbf{r} = (1, 0, 2) + t(2, 1, -1)$、$L_{2}: \mathbf{r} = (3, 1, 1) + s(1, -1, 2)$。(a) 证不平行；(b) 分类；(c) 交点；(d) 锐角。

Answers:答案： (b) lines intersect两直线相交 · (c) $P = (3, 1, 1)$ · (d) $\cos\theta = \tfrac{1}{6},\; \theta \approx 80.4^{\circ}$

(a) Directions not parallel M1·A1

Suppose $\mathbf{d}_{1} = (2, 1, -1) = k\,\mathbf{d}_{2} = k(1, -1, 2)$ for some $k \in \mathbb{R}$. The first component gives $k = 2$; the second then forces $1 = k(-1) = -2$, contradiction. Hence $\mathbf{d}_{1}, \mathbf{d}_{2}$ are not parallel.

(b) Equate components & solve M1·A1·A1·A1·R1

Set position vectors equal: $$ \begin{cases} 1 + 2t \;=\; 3 + s \\ 0 + t \;=\; 1 - s \\ 2 - t \;=\; 1 + 2s \end{cases} \;\Longleftrightarrow\; \begin{cases} 2t - s \;=\; 2 \\ t + s \;=\; 1 \\ -t - 2s \;=\; -1 \end{cases}. $$ Solve the first two: adding gives $3t = 3$, so $t = 1$ and $s = 0$. Substitute into the third: $-(1) - 2(0) = -1$. $\checkmark$ The third equation is consistent, so the system has a solution. The lines intersect.

(c) Point of intersection A1·A1

Plug $t = 1$ into $L_{1}$: $\mathbf{r} = (1, 0, 2) + (2, 1, -1) = (3, 1, 1)$. Cross-check on $L_{2}$ with $s = 0$: $\mathbf{r} = (3, 1, 1) + (0, 0, 0) = (3, 1, 1)$. $\checkmark$ Hence $P = (3, 1, 1)$.

(d) Acute angle M1·A1

$\mathbf{d}_{1} \cdot \mathbf{d}_{2} = (2)(1) + (1)(-1) + (-1)(2) = 2 - 1 - 2 = -1$. $\;|\mathbf{d}_{1}| = \sqrt{4 + 1 + 1} = \sqrt{6}$; $|\mathbf{d}_{2}| = \sqrt{1 + 1 + 4} = \sqrt{6}$. Acute-angle formula takes the absolute value of the dot product: $$ \cos\theta \;=\; \frac{|\mathbf{d}_{1} \cdot \mathbf{d}_{2}|}{|\mathbf{d}_{1}|\,|\mathbf{d}_{2}|} \;=\; \frac{|-1|}{\sqrt{6}\cdot\sqrt{6}} \;=\; \frac{1}{6}. $$ Hence $\theta = \arccos\!\left(\tfrac{1}{6}\right) \approx 80.4^{\circ}$.

Solve two, verify the third. The intersect-vs-skew classifier always uses the same recipe: equate components, solve any two of the three scalar equations for $(s, t)$, then plug into the remaining equation. Consistent ⇒ intersect; inconsistent ⇒ skew. Students who try to solve all three together waste time on a $3 \times 2$ system that the second equation already over-determines. Secondly, always take the absolute value of the dot product when reporting the "acute angle between two lines": directions are defined up to sign (reversing $\mathbf{d}_{1}$ flips the sign of the dot product without changing the line), so the acute angle takes $|\cdot|$. The same fix applies to (d) of Q7 (line-plane angle uses $|\mathbf{d} \cdot \mathbf{n}|$ in the numerator).

(a) 方向向量不平行 M1·A1

设 $\mathbf{d}_{1} = (2, 1, -1) = k\,\mathbf{d}_{2} = k(1, -1, 2)$。由第一分量 $k = 2$；代入第二分量得 $1 = k(-1) = -2$，矛盾。故 $\mathbf{d}_{1}$、$\mathbf{d}_{2}$ 不平行。

(b) 分量相等并求解 M1·A1·A1·A1·R1

令位置向量相等： $$ \begin{cases} 1 + 2t \;=\; 3 + s \\ 0 + t \;=\; 1 - s \\ 2 - t \;=\; 1 + 2s \end{cases} \;\Longleftrightarrow\; \begin{cases} 2t - s \;=\; 2 \\ t + s \;=\; 1 \\ -t - 2s \;=\; -1 \end{cases}. $$ 用前两条：相加得 $3t = 3$，故 $t = 1$、$s = 0$。代入第三条：$-(1) - 2(0) = -1$。$\checkmark$ 第三条一致，系统有解。两直线相交。

(c) 交点 A1·A1

把 $t = 1$ 代入 $L_{1}$：$\mathbf{r} = (1, 0, 2) + (2, 1, -1) = (3, 1, 1)$。用 $s = 0$ 在 $L_{2}$ 上交叉验证：$\mathbf{r} = (3, 1, 1) + (0, 0, 0) = (3, 1, 1)$。$\checkmark$ 故 $P = (3, 1, 1)$。

(d) 锐角 M1·A1

$\mathbf{d}_{1} \cdot \mathbf{d}_{2} = (2)(1) + (1)(-1) + (-1)(2) = 2 - 1 - 2 = -1$。$\;|\mathbf{d}_{1}| = \sqrt{4 + 1 + 1} = \sqrt{6}$；$|\mathbf{d}_{2}| = \sqrt{1 + 1 + 4} = \sqrt{6}$。锐角公式对点积取绝对值： $$ \cos\theta \;=\; \frac{|\mathbf{d}_{1} \cdot \mathbf{d}_{2}|}{|\mathbf{d}_{1}|\,|\mathbf{d}_{2}|} \;=\; \frac{|-1|}{\sqrt{6}\cdot\sqrt{6}} \;=\; \frac{1}{6}. $$ $\theta = \arccos\!\left(\tfrac{1}{6}\right) \approx 80.4^{\circ}$。

解两条、验第三条。相交 vs. 异面判定永远是同一招：分量相等，三条标量方程任挑两条解 $(s, t)$，代回剩下一条。一致即相交；不一致即异面。同时解三条是浪费—两条已经把 $(s, t)$ 锁死。其次，写"两直线锐角"时点积务必取绝对值：方向向量定义至多差一个符号（翻转 $\mathbf{d}_{1}$ 直线不变但点积变号），锐角取 $|\cdot|$。Q7(d) 的线-面角公式 $|\mathbf{d} \cdot \mathbf{n}|$ 同此理。

Paper 2 · Worked Solutions第二卷 · 详细解析

Q6MEDIUMPaper 2AHL 3.17 Plane through Three Points (HL)[7 marks]

$A(1, 0, 0)$, $B(0, 2, 0)$, $C(0, 0, 3)$ in plane $\Pi$. Find $\vec{AB}$, $\vec{AC}$; compute $\mathbf{n} = \vec{AB} \times \vec{AC}$; write the Cartesian equation $ax + by + cz = d$.$A(1, 0, 0)$、$B(0, 2, 0)$、$C(0, 0, 3)$ 共面于 $\Pi$。求 $\vec{AB}$、$\vec{AC}$；算 $\mathbf{n} = \vec{AB} \times \vec{AC}$；写出 $ax + by + cz = d$ 的直角坐标方程。

Answers:答案： (a) $\vec{AB} = (-1, 2, 0),\; \vec{AC} = (-1, 0, 3)$ · (b) $\mathbf{n} = (6, 3, 2)$ · (c) $6x + 3y + 2z = 6$

(a) Edge vectors A1·A1

$\vec{AB} = B - A = (0 - 1, 2 - 0, 0 - 0) = (-1, 2, 0)$. $\;\vec{AC} = C - A = (0 - 1, 0 - 0, 3 - 0) = (-1, 0, 3)$.

(b) Normal via cross product M1·A1·R1

$$ \mathbf{n} \;=\; \vec{AB} \times \vec{AC} \;=\; \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 0 \\ -1 & 0 & 3 \end{vmatrix} \;=\; (2 \cdot 3 - 0 \cdot 0,\; 0 \cdot (-1) - (-1) \cdot 3,\; (-1) \cdot 0 - 2 \cdot (-1)) \;=\; (6, 3, 2). $$ $\mathbf{n}$ is normal to $\Pi$ because $\vec{AB}$ and $\vec{AC}$ are two non-collinear vectors lying in $\Pi$, and the cross product is perpendicular to both input vectors; hence perpendicular to every linear combination of them, i.e. perpendicular to the entire plane.

(c) Cartesian equation M1·A1

$\Pi$: $\mathbf{n} \cdot \mathbf{r} = \mathbf{n} \cdot \vec{OA}$, i.e. $6x + 3y + 2z = 6(1) + 3(0) + 2(0) = 6$. So $\Pi: 6x + 3y + 2z = 6$. Sanity check with $B = (0, 2, 0)$: $0 + 6 + 0 = 6$. $\checkmark$ With $C = (0, 0, 3)$: $0 + 0 + 6 = 6$. $\checkmark$

Intercept form is the fast track. When the three points lie on the axes — here $A(1,0,0)$, $B(0,2,0)$, $C(0,0,3)$, with intercepts $1$, $2$, $3$ — the plane equation is immediately $\dfrac{x}{1} + \dfrac{y}{2} + \dfrac{z}{3} = 1$. Multiplying through by $6$ (the LCM): $6x + 3y + 2z = 6$. Same answer, half the work. The cross-product method still earns full marks, but if you spot the intercept pattern in an exam, write the intercept form first and verify the normal as a back-check. The intercept form generalises: a plane through $(a, 0, 0), (0, b, 0), (0, 0, c)$ (none zero) has equation $\tfrac{x}{a} + \tfrac{y}{b} + \tfrac{z}{c} = 1$ — this is worth committing to memory for Paper 2/3.

(a) 边向量 A1·A1

$\vec{AB} = B - A = (0 - 1, 2 - 0, 0 - 0) = (-1, 2, 0)$。$\;\vec{AC} = C - A = (0 - 1, 0 - 0, 3 - 0) = (-1, 0, 3)$。

(b) 用叉积求法向量 M1·A1·R1

$$ \mathbf{n} \;=\; \vec{AB} \times \vec{AC} \;=\; \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 0 \\ -1 & 0 & 3 \end{vmatrix} \;=\; (2 \cdot 3 - 0 \cdot 0,\; 0 \cdot (-1) - (-1) \cdot 3,\; (-1) \cdot 0 - 2 \cdot (-1)) \;=\; (6, 3, 2). $$ $\mathbf{n}$ 是 $\Pi$ 的法向量：$\vec{AB}$、$\vec{AC}$ 是 $\Pi$ 上两条不共线向量，叉积同时垂直于两者，从而垂直于二者的所有线性组合，即垂直于整个平面。

(c) 直角坐标方程 M1·A1

$\Pi$：$\mathbf{n} \cdot \mathbf{r} = \mathbf{n} \cdot \vec{OA}$，即 $6x + 3y + 2z = 6(1) + 3(0) + 2(0) = 6$。故 $\Pi: 6x + 3y + 2z = 6$。用 $B = (0, 2, 0)$ 验：$0 + 6 + 0 = 6$。$\checkmark$ 用 $C = (0, 0, 3)$ 验：$0 + 0 + 6 = 6$。$\checkmark$

截距式是高速通道。当三点都在坐标轴上—本题 $A(1,0,0)$、$B(0,2,0)$、$C(0,0,3)$，截距为 $1$、$2$、$3$—平面方程立即写出 $\dfrac{x}{1} + \dfrac{y}{2} + \dfrac{z}{3} = 1$。通分乘 $6$：$6x + 3y + 2z = 6$。同一答案、一半时间。叉积法仍可拿满分，但考场上识破截距图形时，先写截距式再用法向量反验更稳。一般化：过 $(a, 0, 0), (0, b, 0), (0, 0, c)$（三者非零）的平面方程为 $\tfrac{x}{a} + \tfrac{y}{b} + \tfrac{z}{c} = 1$—Paper 2/3 值得背下。

Q7HARDPaper 2AHL 3.18 Line Meets Plane (HL)[10 marks]

$\ell: \mathbf{r} = (1, 2, 3) + t(3, -2, 2)$, $\Pi: 6x + 3y + 2z = 6$. Find the intersection point and acute angle; state the configuration if $\mathbf{d} \cdot \mathbf{n} = 0$.$\ell: \mathbf{r} = (1, 2, 3) + t(3, -2, 2)$、$\Pi: 6x + 3y + 2z = 6$。求交点与锐角；若 $\mathbf{d} \cdot \mathbf{n} = 0$ 给出几何情形。

Answers:答案： (a) $t = -\tfrac{3}{4}$ · (b) $P = \left(-\tfrac{5}{4}, \tfrac{7}{2}, \tfrac{3}{2}\right)$ · (c) $\sin\alpha = \tfrac{16}{7\sqrt{17}},\; \alpha \approx 33.7^{\circ}$ · (d) $\ell$ parallel to $\Pi$$\ell$ 与 $\Pi$ 平行

(a) Substitute & solve for $t$ M1·A1·A1

Parametric form: $x = 1 + 3t$, $y = 2 - 2t$, $z = 3 + 2t$. Substitute into $\Pi$: $$ 6(1 + 3t) + 3(2 - 2t) + 2(3 + 2t) \;=\; 6 \;\Longleftrightarrow\; 6 + 18t + 6 - 6t + 6 + 4t \;=\; 6. $$ Collect: $18 + 16t = 6$, so $16t = -12$ and $t = -\tfrac{3}{4}$.

(b) Point $P$ A1·A1

Plug $t = -\tfrac{3}{4}$ into $\ell$: $$ P \;=\; \left(1 + 3 \cdot \!\left(-\tfrac{3}{4}\right),\; 2 - 2 \cdot \!\left(-\tfrac{3}{4}\right),\; 3 + 2 \cdot \!\left(-\tfrac{3}{4}\right)\right) \;=\; \left(1 - \tfrac{9}{4},\; 2 + \tfrac{3}{2},\; 3 - \tfrac{3}{2}\right) \;=\; \left(-\tfrac{5}{4},\; \tfrac{7}{2},\; \tfrac{3}{2}\right). $$ Verify: $6 \cdot (-\tfrac{5}{4}) + 3 \cdot \tfrac{7}{2} + 2 \cdot \tfrac{3}{2} = -\tfrac{15}{2} + \tfrac{21}{2} + 3 = 3 + 3 = 6$. $\checkmark$

(c) Acute angle $\alpha$ M1·A1·A1

Direction $\mathbf{d} = (3, -2, 2)$; normal $\mathbf{n} = (6, 3, 2)$. Compute $\mathbf{d} \cdot \mathbf{n} = 18 - 6 + 4 = 16$. $\;|\mathbf{d}| = \sqrt{9 + 4 + 4} = \sqrt{17}$; $|\mathbf{n}| = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$. Hence $$ \sin\alpha \;=\; \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}|\,|\mathbf{n}|} \;=\; \frac{16}{7\sqrt{17}}, \qquad \alpha \;=\; \arcsin\!\left(\frac{16}{7\sqrt{17}}\right) \;\approx\; 33.7^{\circ}. $$

(d) If $\mathbf{d} \cdot \mathbf{n} = 0$ A1·R1

$\mathbf{d} \cdot \mathbf{n} = 0$ says the direction of $\ell$ is perpendicular to the normal of $\Pi$, hence the direction lies in the plane's direction-set. Geometrically, $\ell$ is parallel to $\Pi$. Two sub-cases (depending on whether a point of $\ell$ satisfies $\Pi$):

If a point of $\ell$ lies on $\Pi$, the whole line lies in $\Pi$ — infinitely many intersections.
If no point of $\ell$ lies on $\Pi$, the line is strictly parallel — no intersection.

In either case $\sin\alpha = 0$ and the angle between $\ell$ and $\Pi$ is $0^{\circ}$.

Why $\sin$, not $\cos$, for line-plane angle. The angle between a line and a plane is measured between the line and its projection onto the plane, i.e. the complement of the angle between the line direction $\mathbf{d}$ and the plane normal $\mathbf{n}$. Calling that complement angle $\beta$: $\cos\beta = \tfrac{\mathbf{d}\cdot\mathbf{n}}{|\mathbf{d}||\mathbf{n}|}$, and the line-plane angle is $\alpha = 90^{\circ} - \beta$, giving $\sin\alpha = \cos\beta$. Hence $\sin\alpha = \tfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$ — with absolute value for "acute". Students who write $\cos\alpha = \tfrac{\mathbf{d}\cdot\mathbf{n}}{|\mathbf{d}||\mathbf{n}|}$ (forgetting the $\sin$/complement) lose both the M1 and the A1 in (c). Drill the diagram once: line, foot of perpendicular onto plane, normal at that foot — the right triangle reveals $\sin\alpha = $ opposite/hypotenuse $= \tfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$.

(a) 代入并解 $t$ M1·A1·A1

参数形式：$x = 1 + 3t$、$y = 2 - 2t$、$z = 3 + 2t$。代入 $\Pi$： $$ 6(1 + 3t) + 3(2 - 2t) + 2(3 + 2t) \;=\; 6 \;\Longleftrightarrow\; 6 + 18t + 6 - 6t + 6 + 4t \;=\; 6. $$ 整理：$18 + 16t = 6$，故 $16t = -12$、$t = -\tfrac{3}{4}$。

(b) 交点 $P$ A1·A1

把 $t = -\tfrac{3}{4}$ 代入 $\ell$： $$ P \;=\; \left(1 + 3 \cdot \!\left(-\tfrac{3}{4}\right),\; 2 - 2 \cdot \!\left(-\tfrac{3}{4}\right),\; 3 + 2 \cdot \!\left(-\tfrac{3}{4}\right)\right) \;=\; \left(-\tfrac{5}{4},\; \tfrac{7}{2},\; \tfrac{3}{2}\right). $$ 验：$6 \cdot (-\tfrac{5}{4}) + 3 \cdot \tfrac{7}{2} + 2 \cdot \tfrac{3}{2} = -\tfrac{15}{2} + \tfrac{21}{2} + 3 = 3 + 3 = 6$。$\checkmark$

(c) 锐角 $\alpha$ M1·A1·A1

方向 $\mathbf{d} = (3, -2, 2)$；法向 $\mathbf{n} = (6, 3, 2)$。$\mathbf{d} \cdot \mathbf{n} = 18 - 6 + 4 = 16$。$\;|\mathbf{d}| = \sqrt{9 + 4 + 4} = \sqrt{17}$；$|\mathbf{n}| = \sqrt{36 + 9 + 4} = 7$。故 $$ \sin\alpha \;=\; \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}|\,|\mathbf{n}|} \;=\; \frac{16}{7\sqrt{17}}, \qquad \alpha \;=\; \arcsin\!\left(\frac{16}{7\sqrt{17}}\right) \;\approx\; 33.7^{\circ}. $$

(d) 若 $\mathbf{d} \cdot \mathbf{n} = 0$ A1·R1

$\mathbf{d} \cdot \mathbf{n} = 0$ 表示 $\ell$ 的方向垂直于 $\Pi$ 的法向，即方向属于平面的方向集。几何上 $\ell$ 与 $\Pi$ 平行。两种子情形（取决于 $\ell$ 上某点是否满足 $\Pi$）：

$\ell$ 上某点在 $\Pi$ 上 $\Rightarrow$ 整条直线落在 $\Pi$ 内—无穷多交点。
$\ell$ 上无点在 $\Pi$ 上 $\Rightarrow$ 严格平行—无交点。

两种情形下 $\sin\alpha = 0$，$\ell$ 与 $\Pi$ 的夹角均为 $0^{\circ}$。

线-面角为何用 $\sin$ 不是 $\cos$。线与平面的夹角是线和它在平面上的投影之间的角，即"线方向 $\mathbf{d}$ 与平面法向 $\mathbf{n}$ 的夹角"的余角。记后者为 $\beta$：$\cos\beta = \tfrac{\mathbf{d}\cdot\mathbf{n}}{|\mathbf{d}||\mathbf{n}|}$，线-面角 $\alpha = 90^{\circ} - \beta$，于是 $\sin\alpha = \cos\beta$。故 $\sin\alpha = \tfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$—锐角取绝对值。把 (c) 写成 $\cos\alpha = \tfrac{\mathbf{d}\cdot\mathbf{n}}{|\mathbf{d}||\mathbf{n}|}$（忘了 sin/余角）的同学 M1 与 A1 同丢。画一次图就够：直线、对平面的垂足、垂足处的法向—直角三角形里 $\sin\alpha = $ 对边/斜边 $= \tfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$。

Paper 3 · Worked Solutions第三卷 · 详细解析

Q8HARDPaper 3AHL 3.15 / 3.18 Distances in Space (HL)[15 marks]

$P = (2, 1, 0)$, $L: \mathbf{r} = (1, 0, 2) + t(2, 1, -1)$ with $A = (1, 0, 2)$, $\mathbf{d} = (2, 1, -1)$; $\Pi: 6x + 3y + 2z = 6$, $\mathbf{n} = (6, 3, 2)$. Compute distances and compare.$P = (2, 1, 0)$、$L: \mathbf{r} = (1, 0, 2) + t(2, 1, -1)$（$A = (1, 0, 2)$、$\mathbf{d} = (2, 1, -1)$）、$\Pi: 6x + 3y + 2z = 6$（$\mathbf{n} = (6, 3, 2)$）。求距离并比较。

Answers:答案： (a) $\vec{AP} \times \mathbf{d} = (1, -3, -1)$ · (b/c) $d_{L} = \sqrt{\tfrac{11}{6}} = \tfrac{\sqrt{66}}{6}$ · (d) $d_{\Pi} = \tfrac{9}{7}$ · (e) $d_{L} \approx 1.354,\; d_{\Pi} \approx 1.286$; $\Pi$ closer$\Pi$ 更近

(a) $\vec{AP}$ and $\vec{AP} \times \mathbf{d}$ M1·A1·A1

$\vec{AP} = P - A = (2 - 1, 1 - 0, 0 - 2) = (1, 1, -2)$. $$ \vec{AP} \times \mathbf{d} \;=\; \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -2 \\ 2 & 1 & -1 \end{vmatrix} \;=\; \bigl(1 \cdot (-1) - (-2) \cdot 1,\; (-2) \cdot 2 - 1 \cdot (-1),\; 1 \cdot 1 - 1 \cdot 2\bigr) \;=\; (1, -3, -1). $$

(b) Distance via cross product M1·A1

$|\vec{AP} \times \mathbf{d}| = \sqrt{1 + 9 + 1} = \sqrt{11}$; $\;|\mathbf{d}| = \sqrt{4 + 1 + 1} = \sqrt{6}$. So $$ d_{L} \;=\; \frac{|\vec{AP} \times \mathbf{d}|}{|\mathbf{d}|} \;=\; \frac{\sqrt{11}}{\sqrt{6}} \;=\; \sqrt{\tfrac{11}{6}} \;=\; \frac{\sqrt{66}}{6}. $$

(c) Foot-of-perpendicular verification M1·A1·A1·A1

Parameterise $Q(t) = (1 + 2t,\; t,\; 2 - t)$ on $L$. Then $$ \vec{PQ}(t) \;=\; Q(t) - P \;=\; (2t - 1,\; t - 1,\; 2 - t). $$ Impose $\vec{PQ}(t) \cdot \mathbf{d} = 0$: $$ 2(2t - 1) + 1 \cdot (t - 1) + (-1)(2 - t) \;=\; 4t - 2 + t - 1 - 2 + t \;=\; 6t - 5 \;=\; 0 \;\Longrightarrow\; t \;=\; \tfrac{5}{6}. $$ At this $t$: $\vec{PQ} = \left(2 \cdot \tfrac{5}{6} - 1,\; \tfrac{5}{6} - 1,\; 2 - \tfrac{5}{6}\right) = \left(\tfrac{2}{3},\; -\tfrac{1}{6},\; \tfrac{7}{6}\right)$. Compute the squared length: $$ |PQ|^{2} \;=\; \tfrac{4}{9} + \tfrac{1}{36} + \tfrac{49}{36} \;=\; \tfrac{16}{36} + \tfrac{1}{36} + \tfrac{49}{36} \;=\; \tfrac{66}{36} \;=\; \tfrac{11}{6}. $$ Hence $|PQ| = \sqrt{\tfrac{11}{6}}$ — agrees with (b). $\;\checkmark$

(d) Distance from $P$ to $\Pi$ M1·A1·A1

$\mathbf{n} \cdot \vec{OP} = 6(2) + 3(1) + 2(0) = 15$. The plane equation rearranges to $\mathbf{n} \cdot \mathbf{r} - 6 = 0$, so $|\mathbf{n} \cdot \vec{OP} - 6| = |15 - 6| = 9$. And $|\mathbf{n}| = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$. Hence $$ d_{\Pi} \;=\; \frac{|\mathbf{n} \cdot \vec{OP} - 6|}{|\mathbf{n}|} \;=\; \frac{9}{7}. $$

(e) Numerical comparison M1·A1·A1

$d_{L} = \sqrt{11/6} \approx \sqrt{1.8333} \approx 1.354$ (3 dp). $\;d_{\Pi} = 9/7 \approx 1.286$ (3 dp). Since $1.286 < 1.354$, the plane $\Pi$ is closer to $P$ than the line $L$ (by approximately $0.068$ units).

Two routes, one answer — and what each one teaches. The cross-product formula $d = |\vec{AP} \times \mathbf{d}| / |\mathbf{d}|$ is fast and unconditional: write one cross product, two magnitudes, divide. The foot-of-perpendicular method is longer but exposes the geometry — the parameter $t = \tfrac{5}{6}$ identifies the actual closest point $Q$ on the line, which the cross-product method skips. Paper 3 markers love when both methods are shown and explicitly reconciled (it scores the R-style reasoning marks). For Paper 1/2, default to cross-product. For Paper 3 or any "show that" question, do both. One extra subtlety on (d). The plane formula $d_{\Pi} = |\mathbf{n} \cdot \vec{OP} - d| / |\mathbf{n}|$ requires $d$ to be the constant term matching the chosen normal sign. If the plane is written $\mathbf{n} \cdot \mathbf{r} = d$, that $d$ is what you subtract; if it is written $\mathbf{n} \cdot \mathbf{r} - d = 0$, same number. Reverse the normal's sign and $d$ flips too; the absolute value protects the answer either way. Watch out for problems that hand you "$6x + 3y + 2z - 6 = 0$" instead of "$= 6$" — just read $d = 6$ from whichever form, and $|\mathbf{n} \cdot \vec{OP} - 6|$ is the right numerator.

(a) $\vec{AP}$ 与 $\vec{AP} \times \mathbf{d}$ M1·A1·A1

$\vec{AP} = P - A = (2 - 1, 1 - 0, 0 - 2) = (1, 1, -2)$。 $$ \vec{AP} \times \mathbf{d} \;=\; \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -2 \\ 2 & 1 & -1 \end{vmatrix} \;=\; \bigl(1 \cdot (-1) - (-2) \cdot 1,\; (-2) \cdot 2 - 1 \cdot (-1),\; 1 \cdot 1 - 1 \cdot 2\bigr) \;=\; (1, -3, -1). $$

(b) 用叉积求距离 M1·A1

$|\vec{AP} \times \mathbf{d}| = \sqrt{1 + 9 + 1} = \sqrt{11}$；$\;|\mathbf{d}| = \sqrt{4 + 1 + 1} = \sqrt{6}$。故 $$ d_{L} \;=\; \frac{|\vec{AP} \times \mathbf{d}|}{|\mathbf{d}|} \;=\; \frac{\sqrt{11}}{\sqrt{6}} \;=\; \sqrt{\tfrac{11}{6}} \;=\; \frac{\sqrt{66}}{6}. $$

(c) 垂足法复核 M1·A1·A1·A1

参数化 $L$ 上点 $Q(t) = (1 + 2t,\; t,\; 2 - t)$。则 $$ \vec{PQ}(t) \;=\; Q(t) - P \;=\; (2t - 1,\; t - 1,\; 2 - t). $$ 令 $\vec{PQ}(t) \cdot \mathbf{d} = 0$： $$ 2(2t - 1) + 1 \cdot (t - 1) + (-1)(2 - t) \;=\; 4t - 2 + t - 1 - 2 + t \;=\; 6t - 5 \;=\; 0 \;\Longrightarrow\; t \;=\; \tfrac{5}{6}. $$ 此 $t$ 处：$\vec{PQ} = \left(\tfrac{2}{3},\; -\tfrac{1}{6},\; \tfrac{7}{6}\right)$。 $$ |PQ|^{2} \;=\; \tfrac{4}{9} + \tfrac{1}{36} + \tfrac{49}{36} \;=\; \tfrac{16}{36} + \tfrac{1}{36} + \tfrac{49}{36} \;=\; \tfrac{66}{36} \;=\; \tfrac{11}{6}. $$ 故 $|PQ| = \sqrt{\tfrac{11}{6}}$—与 (b) 一致。$\;\checkmark$

(d) $P$ 到 $\Pi$ 的距离 M1·A1·A1

$\mathbf{n} \cdot \vec{OP} = 6(2) + 3(1) + 2(0) = 15$。平面方程整理为 $\mathbf{n} \cdot \mathbf{r} - 6 = 0$，故 $|\mathbf{n} \cdot \vec{OP} - 6| = |15 - 6| = 9$。$\;|\mathbf{n}| = \sqrt{36 + 9 + 4} = 7$。故 $$ d_{\Pi} \;=\; \frac{|\mathbf{n} \cdot \vec{OP} - 6|}{|\mathbf{n}|} \;=\; \frac{9}{7}. $$

(e) 数值比较 M1·A1·A1

$d_{L} = \sqrt{11/6} \approx \sqrt{1.8333} \approx 1.354$（三位小数）。$\;d_{\Pi} = 9/7 \approx 1.286$（三位小数）。由 $1.286 < 1.354$，平面 $\Pi$ 比直线 $L$ 离 $P$ 更近（约近 $0.068$ 个单位）。

两条路径、同一答案—各自的教益。叉积公式 $d = |\vec{AP} \times \mathbf{d}| / |\mathbf{d}|$ 又快又无条件：一个叉积、两个模长、一除即得。垂足法更长，但暴露几何结构—参数 $t = \tfrac{5}{6}$ 标出了直线上的真实最近点 $Q$，叉积法略过此信息。Paper 3 阅卷青睐两法并写并核对（拿 R 推理分）。Paper 1/2 默认用叉积；Paper 3 或 "show that" 题型则两法都做。 (d) 的额外细节。平面距离公式 $d_{\Pi} = |\mathbf{n} \cdot \vec{OP} - d| / |\mathbf{n}|$ 中的 $d$ 必须是与所选法向符号匹配的常数项。若方程写 $\mathbf{n} \cdot \mathbf{r} = d$，所减即此 $d$；若写 $\mathbf{n} \cdot \mathbf{r} - d = 0$，同数。法向取负则 $d$ 也翻号；绝对值保护两种写法都对。题目给 "$6x + 3y + 2z - 6 = 0$"（而非 $= 6$）时不要慌—无论哪种写法都读出 $d = 6$，分子始终是 $|\mathbf{n} \cdot \vec{OP} - 6|$。