Unit C3: Vectors单元 C3：向量
The geometry of arrows in space. Vectors carry both magnitude and direction, and the algebra that operates on them (addition, scalar multiplication, dot product, cross product) translates classical Euclidean geometry into coordinates that a calculator can handle. The HL syllabus uses vectors for lines, planes, and distances in three dimensions. Every section follows the same arc: define the operation in components, prove the geometric meaning, then apply to a canonical exam set-up.本单元研究"空间中的箭头"的几何。向量同时携带大小与方向，作用于其上的代数（加法、数乘、点积、叉积）把欧氏几何译成计算器能处理的坐标。HL 大纲用向量处理三维空间中的直线、平面与距离。每节的叙事都遵循同一弧线：先以分量定义运算，再证其几何意义，最后应用到一类典型考题。
How to use this guide本指南使用说明
C3 is one of the four super-topics in IB Math AA HL that is entirely HL only. Standard Level students do not encounter vectors in the new syllabus, so every section here is HL material. The unit is also a Paper 3 favourite: questions tend to fuse several sub-topics in one extended task (find a line, intersect with a plane, measure an angle).C3 是 IB 数学 AA HL 四个"纯 HL"超主题之一。新大纲下，标准水平的学生不学向量，因此本单元所有内容都是 HL。本单元同时是 Paper 3 的常客：考题往往把多个子主题串成一道长题（求一条直线，与一个平面相交，再求夹角）。
Memorise the three big formulas: $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$, $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$, and the line $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$. Practise one full Paper 2 question: line through two given points, intersect with a plane, find the angle between line and plane.
背熟三大公式：$|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$、$\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$、直线 $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$。完整做一道 Paper 2：求过两点的直线，与平面相交，再求线与面的夹角。
Master the three line-relationship cases: intersecting, parallel, skew. Be able to argue, with parameter algebra, which case applies. For planes, fluently switch between vector ($\mathbf{r} \cdot \mathbf{n} = d$), scalar Cartesian ($ax + by + cz = d$), and parametric ($\mathbf{r} = \mathbf{a} + \lambda \mathbf{u} + \mu \mathbf{v}$) forms.
熟练判别三种直线关系：相交、平行、异面。要能用参数方程的代数论证给出判定。对平面，要在向量式 $\mathbf{r} \cdot \mathbf{n} = d$、直角坐标式 $ax + by + cz = d$、参数式 $\mathbf{r} = \mathbf{a} + \lambda \mathbf{u} + \mu \mathbf{v}$ 之间自如切换。
Vector Arithmetic, Magnitude, Unit Vectors向量运算、模长、单位向量 AHL 3.12
Given $\mathbf{v} = (3, 4, 12)$, find $|\mathbf{v}|$ and the unit vector $\hat{\mathbf{v}}$.设 $\mathbf{v} = (3, 4, 12)$，求 $|\mathbf{v}|$ 与单位向量 $\hat{\mathbf{v}}$。
Magnitude. Apply the formula componentwise:
模长。按分量套公式：
$$ |\mathbf{v}| \;=\; \sqrt{3^{2} + 4^{2} + 12^{2}} \;=\; \sqrt{9 + 16 + 144} \;=\; \sqrt{169} \;=\; 13. $$Unit vector. Divide each component by the magnitude:
单位向量。每个分量都除以模长：
$$ \hat{\mathbf{v}} \;=\; \frac{1}{13} (3, 4, 12) \;=\; \left( \tfrac{3}{13}, \; \tfrac{4}{13}, \; \tfrac{12}{13} \right). $$Check. $\left( \tfrac{3}{13} \right)^{2} + \left( \tfrac{4}{13} \right)^{2} + \left( \tfrac{12}{13} \right)^{2} = \tfrac{9 + 16 + 144}{169} = 1$, so $|\hat{\mathbf{v}}| = 1$ as required.
验证。$\left( \tfrac{3}{13} \right)^{2} + \left( \tfrac{4}{13} \right)^{2} + \left( \tfrac{12}{13} \right)^{2} = \tfrac{9 + 16 + 144}{169} = 1$，故 $|\hat{\mathbf{v}}| = 1$，符合定义。
Going deeper: why magnitude is a Pythagorean sum深入：为何模长是勾股和
In two dimensions, the segment from the origin to $(a_1, a_2)$ is the hypotenuse of a right triangle with legs $a_1$ and $a_2$, so its length is $\sqrt{a_1^{2} + a_2^{2}}$ by Pythagoras. In three dimensions, apply Pythagoras twice: first in the $xy$-plane to get the projection length $\sqrt{a_1^{2} + a_2^{2}}$, then in the plane containing that projection and the $z$-axis to combine with $a_3$. The result is $\sqrt{(a_1^{2} + a_2^{2}) + a_3^{2}}$, which is the formula above. The pattern extends to $n$ dimensions: the magnitude of $(a_1, \dots, a_n)$ is $\sqrt{a_1^{2} + \cdots + a_n^{2}}$, which is the definition of the Euclidean norm.
二维时，从原点到 $(a_1, a_2)$ 的线段是腿长 $a_1$ 与 $a_2$ 的直角三角形的斜边，由勾股定理其长度为 $\sqrt{a_1^{2} + a_2^{2}}$。三维时连用两次勾股：先在 $xy$ 平面得投影长度 $\sqrt{a_1^{2} + a_2^{2}}$，再在含该投影与 $z$ 轴的平面中与 $a_3$ 合成，结果为 $\sqrt{(a_1^{2} + a_2^{2}) + a_3^{2}}$，即上面公式。此模式推广到 $n$ 维：$(a_1, \dots, a_n)$ 的模长为 $\sqrt{a_1^{2} + \cdots + a_n^{2}}$，即欧氏范数的定义。
Dot Product and Angle Between Vectors点积与向量夹角 AHL 3.13
The angle formula. Provided $\mathbf{u}, \mathbf{v} \ne \mathbf{0}$: $$ \cos\theta \;=\; \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}. $$ Perpendicularity test. $\mathbf{u} \perp \mathbf{v}$ if and only if $\mathbf{u} \cdot \mathbf{v} = 0$ (for nonzero vectors).
Algebraic properties. The dot product is commutative ($\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$), distributes over addition, and satisfies $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^{2}$.
夹角公式。设 $\mathbf{u}, \mathbf{v} \ne \mathbf{0}$： $$ \cos\theta \;=\; \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}. $$ 垂直判别。对非零向量：$\mathbf{u} \perp \mathbf{v}$ 当且仅当 $\mathbf{u} \cdot \mathbf{v} = 0$。
代数性质。点积满足交换律（$\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$）、对加法的分配律，以及 $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^{2}$。
Find the angle $\theta$ between $\mathbf{u} = (1, 2, 2)$ and $\mathbf{v} = (2, 1, 0)$.求 $\mathbf{u} = (1, 2, 2)$ 与 $\mathbf{v} = (2, 1, 0)$ 的夹角 $\theta$。
Dot product.
点积。
$$ \mathbf{u} \cdot \mathbf{v} \;=\; (1)(2) + (2)(1) + (2)(0) \;=\; 2 + 2 + 0 \;=\; 4. $$Magnitudes.
模长。
$$ |\mathbf{u}| \;=\; \sqrt{1 + 4 + 4} \;=\; 3, \qquad |\mathbf{v}| \;=\; \sqrt{4 + 1 + 0} \;=\; \sqrt{5}. $$Apply the angle formula.
套用夹角公式。
$$ \cos\theta \;=\; \frac{4}{3 \sqrt{5}} \;=\; \frac{4 \sqrt{5}}{15}. $$Solve for $\theta$. $\theta = \arccos \tfrac{4}{3 \sqrt{5}} \approx 53.4^{\circ}$ (or $0.9322$ rad).
解出 $\theta$。$\theta = \arccos \tfrac{4}{3 \sqrt{5}} \approx 53.4^{\circ}$（约 $0.9322$ rad）。
Going deeper: why both dot-product formulas agree深入：两种点积公式为何一致
Start with the law of cosines applied to the triangle with sides $\mathbf{u}$, $\mathbf{v}$, $\mathbf{u} - \mathbf{v}$:
对三边为 $\mathbf{u}$、$\mathbf{v}$、$\mathbf{u} - \mathbf{v}$ 的三角形使用余弦定理：
$$ |\mathbf{u} - \mathbf{v}|^{2} \;=\; |\mathbf{u}|^{2} + |\mathbf{v}|^{2} - 2 |\mathbf{u}|\,|\mathbf{v}|\,\cos\theta. $$Expand the left side in components: $|\mathbf{u} - \mathbf{v}|^{2} = (u_1 - v_1)^{2} + (u_2 - v_2)^{2} + (u_3 - v_3)^{2} = |\mathbf{u}|^{2} + |\mathbf{v}|^{2} - 2(u_1 v_1 + u_2 v_2 + u_3 v_3)$. Comparing the two expressions and cancelling gives $u_1 v_1 + u_2 v_2 + u_3 v_3 = |\mathbf{u}|\,|\mathbf{v}|\,\cos\theta$, which is the identity of the two formulas.
把左边按分量展开：$|\mathbf{u} - \mathbf{v}|^{2} = (u_1 - v_1)^{2} + (u_2 - v_2)^{2} + (u_3 - v_3)^{2} = |\mathbf{u}|^{2} + |\mathbf{v}|^{2} - 2(u_1 v_1 + u_2 v_2 + u_3 v_3)$。与上式相比并消去同类项，得 $u_1 v_1 + u_2 v_2 + u_3 v_3 = |\mathbf{u}|\,|\mathbf{v}|\,\cos\theta$，即两种公式等价。
Vector Equation of a Line直线的向量方程 AHL 3.14
Direction vector is not unique. Any nonzero scalar multiple of $\mathbf{d}$ describes the same line. Two lines are parallel exactly when one direction vector is a scalar multiple of the other.
方向向量不唯一。$\mathbf{d}$ 的任一非零数乘都描述同一条直线。两直线平行当且仅当其方向向量互为数乘。
Write the vector equation and the parametric equations of the line through $(1, 2, 3)$ with direction vector $(2, -1, 1)$. Does the point $(5, 0, 5)$ lie on this line?写出过 $(1, 2, 3)$、方向向量为 $(2, -1, 1)$ 的直线的向量方程与参数方程。判断点 $(5, 0, 5)$ 是否在该直线上。
Vector form.
向量式。
$$ \mathbf{r} \;=\; \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}. $$Parametric form.
参数式。
$$ x = 1 + 2 \lambda, \qquad y = 2 - \lambda, \qquad z = 3 + \lambda. $$Test the point $(5, 0, 5)$. If the point lies on the line, the same parameter value must satisfy all three equations. From $x = 5$: $1 + 2 \lambda = 5 \Rightarrow \lambda = 2$. Check the others: $y = 2 - 2 = 0$ ✓; $z = 3 + 2 = 5$ ✓. All three agree at $\lambda = 2$, so $(5, 0, 5)$ does lie on the line.
检验点 $(5, 0, 5)$。若点在直线上，则同一参数值需同时满足三个分量方程。由 $x = 5$：$1 + 2 \lambda = 5 \Rightarrow \lambda = 2$。验证其它：$y = 2 - 2 = 0$ ✓；$z = 3 + 2 = 5$ ✓。三式在 $\lambda = 2$ 时一致，故 $(5, 0, 5)$ 在直线上。
Intersections of Lines直线之间的位置关系 AHL 3.15
- Intersecting. The two lines share exactly one point. Direction vectors are not parallel; the system of parametric equations has a unique solution.
- Parallel. Direction vectors are scalar multiples of each other. Either the lines coincide (same line, infinitely many common points) or they share no points.
- Skew. The lines are neither parallel nor intersecting. They live in different parallel planes; no common point exists. Skew is the case that has no analogue in two dimensions.
- If all three are simultaneously satisfied: intersecting. Substitute back to find the common point.
- If no $(\lambda, \mu)$ satisfies all three: lines are skew (assuming non-parallel directions) or parallel without coincidence (parallel directions).
- 相交。两直线恰有一个公共点。方向向量不平行；参数方程组有唯一解。
- 平行。方向向量互为数乘。要么重合（同一直线、无穷多公共点），要么无公共点。
- 异面。既不平行也不相交。两直线落在两个平行平面内，无公共点。异面是二维所没有的情形。
- 若三式同时成立：相交，回代求公共点。
- 若无 $(\lambda, \mu)$ 使三式同时成立：方向不平行则异面；方向平行则平行不重合。
Determine whether the lines $L_1: \mathbf{r} = (1, 0, 2) + \lambda (1, 1, 1)$ and $L_2: \mathbf{r} = (2, 3, 1) + \mu (2, 0, -1)$ intersect. If they do, find the point of intersection.判断直线 $L_1: \mathbf{r} = (1, 0, 2) + \lambda (1, 1, 1)$ 与 $L_2: \mathbf{r} = (2, 3, 1) + \mu (2, 0, -1)$ 是否相交。若相交，求交点。
Step 1. Check whether the directions are parallel. $(1, 1, 1)$ and $(2, 0, -1)$ are not scalar multiples (the second has a zero component while the first does not), so the lines are not parallel.
第 1 步：方向是否平行。$(1, 1, 1)$ 与 $(2, 0, -1)$ 不是数乘（第二个含零分量、第一个不含），故不平行。
Step 2. Set components equal.
第 2 步：分量分别相等。
$$ 1 + \lambda \;=\; 2 + 2 \mu, \qquad \lambda \;=\; 3, \qquad 2 + \lambda \;=\; 1 - \mu. $$Step 3. Solve. From the $y$-equation, $\lambda = 3$. Substitute into the $x$-equation: $1 + 3 = 2 + 2 \mu \Rightarrow \mu = 1$. Check in the $z$-equation: $2 + 3 = 5$ on the left; $1 - 1 = 0$ on the right. The two sides disagree.
第 3 步：求解。由 $y$ 方程得 $\lambda = 3$。代入 $x$ 方程：$1 + 3 = 2 + 2 \mu \Rightarrow \mu = 1$。验证 $z$ 方程：左 $2 + 3 = 5$；右 $1 - 1 = 0$。两边不相等。
Conclusion. The system is inconsistent. Since the directions are not parallel, the lines are skew. There is no point of intersection.
结论。方程组不一致。方向不平行，故两直线异面。无公共点。
Going deeper: why "two of three" is the right verification rule深入：为何用"三式中取两式"验证
Two parametric lines in three dimensions give a system of three linear equations in two unknowns $(\lambda, \mu)$. Generically, three equations in two unknowns are overdetermined: pick any two of them and they determine $(\lambda, \mu)$ uniquely, and the third equation acts as a consistency check. If the third holds, the lines meet at that $(\lambda, \mu)$; if it fails, the system is inconsistent and there is no common point. The "two of three" procedure is the cleanest version of this argument and the one IB markers expect.
三维空间中两条参数直线给出三式两未知 $(\lambda, \mu)$ 的线性方程组。三式两未知一般属"超定"：任取两式可唯一确定 $(\lambda, \mu)$，第三式则是相容性的验证。若第三式成立，两直线在该 $(\lambda, \mu)$ 处相交；若失败则方程组不一致，无公共点。"取两式解、第三式验"是上述论证最简洁的写法，也是 IB 评卷期望的格式。
Cross Product叉积（向量积） AHL 3.16
- $\mathbf{u} \times \mathbf{v}$ is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$. Its direction is given by the right-hand rule.
- Magnitude: $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}|\,|\mathbf{v}|\,\sin\theta$, with $\theta$ the angle between $\mathbf{u}$ and $\mathbf{v}$.
- $|\mathbf{u} \times \mathbf{v}|$ equals the area of the parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$. Triangle with the same two sides has area $\tfrac{1}{2} |\mathbf{u} \times \mathbf{v}|$.
- $\mathbf{u} \times \mathbf{v}$ 同时垂直于 $\mathbf{u}$ 与 $\mathbf{v}$。方向由右手法则确定。
- 模长：$|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}|\,|\mathbf{v}|\,\sin\theta$，$\theta$ 为夹角。
- $|\mathbf{u} \times \mathbf{v}|$ 等于以 $\mathbf{u}, \mathbf{v}$ 为邻边的平行四边形面积。同邻边的三角形面积为 $\tfrac{1}{2} |\mathbf{u} \times \mathbf{v}|$。
Verify by direct computation that $(1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$.直接计算验证 $(1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$。
Apply the determinant formula. With $\mathbf{u} = \mathbf{i} = (1, 0, 0)$ and $\mathbf{v} = \mathbf{j} = (0, 1, 0)$:
套用行列式公式。$\mathbf{u} = \mathbf{i} = (1, 0, 0)$、$\mathbf{v} = \mathbf{j} = (0, 1, 0)$：
$$ \mathbf{i} \times \mathbf{j} \;=\; \bigl( (0)(0) - (0)(1), \; (0)(0) - (1)(0), \; (1)(1) - (0)(0) \bigr) \;=\; (0, 0, 1) \;=\; \mathbf{k}. $$Remark. The cyclic identities $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, $\mathbf{k} \times \mathbf{i} = \mathbf{j}$ are worth memorising. The order matters: $\mathbf{j} \times \mathbf{i} = - \mathbf{k}$ by anticommutativity.
注。循环关系 $\mathbf{i} \times \mathbf{j} = \mathbf{k}$、$\mathbf{j} \times \mathbf{k} = \mathbf{i}$、$\mathbf{k} \times \mathbf{i} = \mathbf{j}$ 值得熟记。顺序决定符号：$\mathbf{j} \times \mathbf{i} = - \mathbf{k}$（反交换律）。
Find the area of the triangle with vertices $A(1, 0, 1)$, $B(2, 1, 0)$, and $C(0, 2, 2)$.求顶点为 $A(1, 0, 1)$、$B(2, 1, 0)$、$C(0, 2, 2)$ 的三角形面积。
Form two edge vectors from $A$.
由 $A$ 出发取两条边向量。
$$ \overrightarrow{AB} \;=\; B - A \;=\; (1, 1, -1), \qquad \overrightarrow{AC} \;=\; C - A \;=\; (-1, 2, 1). $$Cross product.
叉积。
$$ \overrightarrow{AB} \times \overrightarrow{AC} \;=\; \bigl( (1)(1) - (-1)(2), \; (-1)(-1) - (1)(1), \; (1)(2) - (1)(-1) \bigr) \;=\; (3, 0, 3). $$Magnitude.
模长。
$$ \bigl| \overrightarrow{AB} \times \overrightarrow{AC} \bigr| \;=\; \sqrt{9 + 0 + 9} \;=\; \sqrt{18} \;=\; 3 \sqrt{2}. $$Triangle area. Half the parallelogram area:
三角形面积。等于平行四边形面积的一半：
$$ \text{Area} \;=\; \tfrac{1}{2} \cdot 3 \sqrt{2} \;=\; \tfrac{3 \sqrt{2}}{2}. $$Vector Equation of a Plane平面的向量方程 AHL 3.17 · 3.18
Parametric (two-direction) form. A plane through $\mathbf{a}$ containing direction vectors $\mathbf{u}$ and $\mathbf{v}$ (linearly independent, both lying in the plane) is $$ \mathbf{r} \;=\; \mathbf{a} + \lambda \mathbf{u} + \mu \mathbf{v}, \qquad \lambda, \mu \in \mathbb{R}. $$ Finding a normal from a plane through three points. Form two edge vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$; their cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ is normal to the plane. Then $d = \mathbf{a} \cdot \mathbf{n}$ using any of the three known points.
Distance from a point to a plane. For point $P$ and plane $a x + b y + c z = d$: $$ \text{dist}(P, \text{plane}) \;=\; \frac{|a x_P + b y_P + c z_P - d|}{\sqrt{a^{2} + b^{2} + c^{2}}}. $$
参数（双方向）式。过 $\mathbf{a}$、含线性无关方向向量 $\mathbf{u}$ 与 $\mathbf{v}$（均在面内）的平面： $$ \mathbf{r} \;=\; \mathbf{a} + \lambda \mathbf{u} + \mu \mathbf{v}, \qquad \lambda, \mu \in \mathbb{R}. $$ 由三点求法向量。构造两条边向量 $\overrightarrow{AB}$、$\overrightarrow{AC}$；其叉积 $\overrightarrow{AB} \times \overrightarrow{AC}$ 即为法向量。再用任一已知点求 $d = \mathbf{a} \cdot \mathbf{n}$。
点到平面的距离。点 $P$ 到平面 $a x + b y + c z = d$ 的距离： $$ \text{dist}(P, \text{plane}) \;=\; \frac{|a x_P + b y_P + c z_P - d|}{\sqrt{a^{2} + b^{2} + c^{2}}}. $$
Find the Cartesian equation of the plane through $A(1, 0, 1)$, $B(2, 1, 0)$, $C(0, 2, 2)$. Then find the (perpendicular) distance from the origin to this plane.求过 $A(1, 0, 1)$、$B(2, 1, 0)$、$C(0, 2, 2)$ 三点的平面的直角坐标方程，并求原点到该平面的距离。
Step 1. Two direction vectors in the plane.
第 1 步：取面内两个方向向量。
$$ \overrightarrow{AB} \;=\; (1, 1, -1), \qquad \overrightarrow{AC} \;=\; (-1, 2, 1). $$Step 2. Normal vector from the cross product. (Using the computation from C3.5b.)
第 2 步：用叉积求法向量。（沿用 C3.5b 的计算结果。）
$$ \mathbf{n} \;=\; \overrightarrow{AB} \times \overrightarrow{AC} \;=\; (3, 0, 3). $$For neatness, scale down by $3$: take $\mathbf{n} = (1, 0, 1)$.
为简洁，整体除以 $3$：取 $\mathbf{n} = (1, 0, 1)$。
Step 3. Find $d$. Use point $A$: $d = \mathbf{n} \cdot \overrightarrow{OA} = (1)(1) + (0)(0) + (1)(1) = 2$.
第 3 步：求 $d$。用点 $A$：$d = \mathbf{n} \cdot \overrightarrow{OA} = (1)(1) + (0)(0) + (1)(1) = 2$。
Cartesian equation.
直角坐标方程。
$$ x + z \;=\; 2. $$Step 4. Distance from origin. With $(x_P, y_P, z_P) = (0, 0, 0)$, $a = 1, b = 0, c = 1, d = 2$:
第 4 步：原点到平面的距离。$(x_P, y_P, z_P) = (0, 0, 0)$，$a = 1, b = 0, c = 1, d = 2$：
$$ \text{dist} \;=\; \frac{|0 + 0 + 0 - 2|}{\sqrt{1 + 0 + 1}} \;=\; \frac{2}{\sqrt{2}} \;=\; \sqrt{2}. $$Verification. Test that each of $A, B, C$ satisfies the equation: $A: 1 + 1 = 2$ ✓; $B: 2 + 0 = 2$ ✓; $C: 0 + 2 = 2$ ✓.
验证。把 $A, B, C$ 三点逐一代入方程：$A: 1 + 1 = 2$ ✓；$B: 2 + 0 = 2$ ✓；$C: 0 + 2 = 2$ ✓。
Going deeper: why the distance formula is what it is深入：点到平面距离公式的来由
Take a point $P$ and a plane with unit normal $\hat{\mathbf{n}} = \mathbf{n} / |\mathbf{n}|$ and known point $\mathbf{a}$. The signed distance from $P$ to the plane is the scalar projection of $\overrightarrow{aP} = P - \mathbf{a}$ onto $\hat{\mathbf{n}}$:
取点 $P$ 与平面（单位法向量 $\hat{\mathbf{n}} = \mathbf{n} / |\mathbf{n}|$、面上已知点 $\mathbf{a}$）。$P$ 到平面的带符号距离即为 $\overrightarrow{aP} = P - \mathbf{a}$ 在 $\hat{\mathbf{n}}$ 上的标量投影：
$$ \text{signed dist} \;=\; (P - \mathbf{a}) \cdot \hat{\mathbf{n}} \;=\; \frac{(P - \mathbf{a}) \cdot \mathbf{n}}{|\mathbf{n}|} \;=\; \frac{\mathbf{n} \cdot P - \mathbf{n} \cdot \mathbf{a}}{|\mathbf{n}|}. $$Writing $\mathbf{n} = (a, b, c)$ and $\mathbf{n} \cdot \mathbf{a} = d$ recovers the formula in the cram-cheat. The absolute value drops the sign, leaving the unsigned distance.
代 $\mathbf{n} = (a, b, c)$、$\mathbf{n} \cdot \mathbf{a} = d$ 即得 cheat-sheet 中的公式。绝对值去掉符号，得无向距离。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- State which point and which direction you are using. For a line, "point $\mathbf{a}$, direction $\mathbf{d}$" earns the M1. For a plane, name the point and the normal vector before writing the equation.
- 明示所用的点与方向。直线题写出"点 $\mathbf{a}$、方向 $\mathbf{d}$"可拿 M1。平面题先点名"点 + 法向量"，再写方程。
- Use different parameter letters for different lines. $L_1$ uses $\lambda$, $L_2$ uses $\mu$. Reusing the same letter is the canonical slip when checking for intersections.
- 不同直线用不同参数字母。$L_1$ 用 $\lambda$、$L_2$ 用 $\mu$。求交点时若两线复用同一字母，是最容易出现的代数错。
- Dot product is a scalar; cross product is a vector. Writing $\mathbf{u} \cdot \mathbf{v} = (3, 0, 5)$ or $\mathbf{u} \times \mathbf{v} = 7$ costs marks immediately.
- 点积是标量、叉积是向量。写出 $\mathbf{u} \cdot \mathbf{v} = (3, 0, 5)$ 或 $\mathbf{u} \times \mathbf{v} = 7$ 立刻扣分。
- For "angle between" two vectors, take an absolute value if the question asks for the acute angle. $\cos\theta$ can be negative; if the marker wants the acute angle, use $\arccos \tfrac{|\mathbf{u} \cdot \mathbf{v}|}{|\mathbf{u}||\mathbf{v}|}$.
- 题目要求"锐角"时取绝对值。$\cos\theta$ 可为负；若评卷要锐角，应用 $\arccos \tfrac{|\mathbf{u} \cdot \mathbf{v}|}{|\mathbf{u}||\mathbf{v}|}$。
- The coefficients of $x, y, z$ in the Cartesian form are the normal. Conversely, given a normal $(a, b, c)$ and a known point, the equation is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
- 直角坐标方程中 $x, y, z$ 的系数即为法向量。反之，已知法向量 $(a, b, c)$ 与一点 $(x_0, y_0, z_0)$，方程为 $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$。
- For distance from a point to a plane, divide by $|\mathbf{n}|$, not by $\mathbf{n}$ itself. The denominator is a scalar.
- 点到平面的距离要除以 $|\mathbf{n}|$，不是除以 $\mathbf{n}$。分母是标量。
- Verify by substituting the original points. If $A, B, C$ defined the plane, all three must satisfy your final equation. This is a cheap and reliable end-of-question check.
- 用原始点逐一验证。若用 $A, B, C$ 三点定义平面，三点都应满足你的最终方程。这是廉价且可靠的末段自检。
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Unit C3 Practice Quiz单元 C3 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt. The entire unit is HL only, so every item below carries the HL chip.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。本单元为纯 HL，故每一条都挂 HL 标。
- HL Translate between column-vector form and $\mathbf{i}, \mathbf{j}, \mathbf{k}$ form在列向量与 $\mathbf{i}, \mathbf{j}, \mathbf{k}$ 形式之间互换
- HL Compute magnitude $|\mathbf{v}|$ and normalise to obtain $\hat{\mathbf{v}}$算模长 $|\mathbf{v}|$ 并归一化得 $\hat{\mathbf{v}}$
- HL Compute a dot product and find the angle between two vectors算点积、求两向量夹角
- HL Test two vectors for perpendicularity using the dot product用点积判别两向量是否垂直
- HL Write the vector and parametric equations of a line through a point with a given direction (or through two given points)写出过定点指向直线（或过两点直线）的向量式与参数式
- HL Determine whether a candidate point lies on a given line判断候选点是否在给定直线上
- HL Classify two lines in 3D as intersecting, parallel, or skew, with parameter algebra用参数代数将两条三维直线分类为相交、平行或异面
- HL Find the point of intersection when two lines meet当两直线相交时求交点
- HL Compute a cross product via the $\mathbf{i}, \mathbf{j}, \mathbf{k}$ determinant用 $\mathbf{i}, \mathbf{j}, \mathbf{k}$ 行列式计算叉积
- HL Use the cross product to find the area of a parallelogram or triangle用叉积求平行四边形或三角形的面积
- HL Write the Cartesian equation of a plane given a point and a normal vector已知一点与法向量，写出平面的直角坐标方程
- HL Find a normal to a plane defined by three points using the cross product of two edge vectors对由三点定义的平面，用两边向量的叉积求法向量
- HL Compute the perpendicular distance from a point to a plane in Cartesian form由直角坐标方程算点到平面的垂直距离
IB Paper-Style PracticeIB 试卷风格练习
C3 Practice and Solutions are on the roadmap. They will ship under Practice Questions/Unit_C3_*.html with the bilingual built-in pattern.
C3 配套的 Practice 与 Solutions 已在排期，上线后位于 Practice Questions/Unit_C3_*.html，采用双语内嵌格式。