Unit C1: Vectors and the Geometry of Space单元 C1：向量与空间几何

Identify the center and radius of the surface $x^2 + y^2 + z^2 - 4x + 6y - 2z + 5 = 0$.求曲面 $x^2 + y^2 + z^2 - 4x + 6y - 2z + 5 = 0$ 的球心与半径。

Group by variable and complete the square in each:按变量分组，对每个变量配方：

$$ (x^2 - 4x) + (y^2 + 6y) + (z^2 - 2z) = -5, $$ $$ (x-2)^2 - 4 + (y+3)^2 - 9 + (z-1)^2 - 1 = -5. $$

Collect the constants:合并常数项：

$$ (x-2)^2 + (y+3)^2 + (z-1)^2 = 9. $$

This is a sphere with center $(2, -3, 1)$ and radius $r = 3$.这是一个球心为 $(2, -3, 1)$、半径 $r = 3$ 的球面。

Given $A = (1, -2, 4)$ and $B = (5, 4, -2)$, find $|AB|$ and the midpoint $M$.已知 $A = (1, -2, 4)$ 与 $B = (5, 4, -2)$，求 $|AB|$ 与中点 $M$。

$$ |AB| = \sqrt{(5-1)^2 + (4-(-2))^2 + (-2-4)^2} = \sqrt{16 + 36 + 36} = \sqrt{88} = 2\sqrt{22}. $$

The midpoint averages each coordinate:中点对每个坐标取平均：

$$ M = \left( \tfrac{1+5}{2}, \tfrac{-2+4}{2}, \tfrac{4+(-2)}{2} \right) = (3, 1, 1). $$

Describe the set $x^2 + y^2 + z^2 + 2x - 4y + 6z + 20 = 0$.描述集合 $x^2 + y^2 + z^2 + 2x - 4y + 6z + 20 = 0$。

Complete the square in each variable as before:与前面一样对每个变量配方：

$$ (x+1)^2 - 1 + (y-2)^2 - 4 + (z+3)^2 - 9 + 20 = 0, $$ $$ (x+1)^2 + (y-2)^2 + (z+3)^2 = -6. $$

A sum of three squares can never be negative, so no point $(x,y,z)$ satisfies the equation. The set is empty. The same algebra that produces a sphere can produce an empty set or, when the right side is exactly $0$, a single point: the degenerate sphere of radius $0$. Always inspect the sign and size of the constant after completing the square before naming the surface.三个平方之和绝不可能为负，因此没有任何点 $(x,y,z)$ 满足该方程，该集合为空集。同样的代数运算可以给出球面，也可以给出空集；当右端恰为 $0$ 时，则给出单独一点，即半径为 $0$ 的退化球面。配方之后，务必先检查常数项的符号与大小，再判定曲面类型。

Sketch the region $1 \le x^2 + y^2 + z^2 \le 4$ together with $z \ge 0$.画出区域 $1 \le x^2 + y^2 + z^2 \le 4$ 与 $z \ge 0$ 的交集。

The inequality $x^2 + y^2 + z^2 \ge 1$ is the set of points at distance at least $1$ from the origin, the outside of the unit sphere. The inequality $x^2 + y^2 + z^2 \le 4$ is the inside of the sphere of radius $2$. Together they describe the spherical shell with inner radius $1$ and outer radius $2$. The extra condition $z \ge 0$ keeps only the upper half. The region is the top half of a thick spherical shell.不等式 $x^2 + y^2 + z^2 \ge 1$ 表示到原点距离至少为 $1$ 的点的集合，即单位球面之外的部分。不等式 $x^2 + y^2 + z^2 \le 4$ 表示半径为 $2$ 的球面之内的部分。两者合在一起描述了内半径为 $1$、外半径为 $2$ 的球壳。附加条件 $z \ge 0$ 只保留上半部分。该区域就是一层厚球壳的上半部。

Reading distances as inequalities is the bridge from algebra to solid regions, and it is exactly the language used to set up triple integrals later in the course.把距离理解为不等式，是从代数通向立体区域的桥梁，也正是课程后面建立三重积分时所用的语言。

The three-dimensional distance formula is not a new axiom; it follows from the plane Pythagorean theorem applied twice. Let $P_1 = (x_1, y_1, z_1)$ and $P_2 = (x_2, y_2, z_2)$, and introduce the corner point $Q = (x_2, y_2, z_1)$, which shares the $z$ coordinate of $P_1$ and the $x,y$ coordinates of $P_2$.三维距离公式并不是一条新的公理；它由平面勾股定理连用两次推出。设 $P_1 = (x_1, y_1, z_1)$、$P_2 = (x_2, y_2, z_2)$，并引入角点 $Q = (x_2, y_2, z_1)$，它与 $P_1$ 共享 $z$ 坐标，与 $P_2$ 共享 $x,y$ 坐标。

The segment $P_1 Q$ lies entirely in the horizontal plane $z = z_1$, so its length is the ordinary planar distance线段 $P_1 Q$ 完全位于水平面 $z = z_1$ 内，因此它的长度就是平面上的普通距离

$$ |P_1 Q| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. $$

The segment $Q P_2$ is vertical, of length $|z_2 - z_1|$, and it is perpendicular to the horizontal plane, hence perpendicular to $P_1 Q$. Triangle $P_1 Q P_2$ has a right angle at $Q$, so by the Pythagorean theorem线段 $Q P_2$ 是竖直的，长度为 $|z_2 - z_1|$，且垂直于该水平面，因而也垂直于 $P_1 Q$。三角形 $P_1 Q P_2$ 在 $Q$ 处为直角，于是由勾股定理

$$ |P_1 P_2|^2 = |P_1 Q|^2 + |Q P_2|^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2. $$

Taking the nonnegative square root gives the distance formula. The argument generalizes verbatim to $\mathbb{R}^n$, where one Pythagorean step is added per extra coordinate.取非负平方根即得距离公式。该论证可原封不动地推广到 $\mathbb{R}^n$，每多一个坐标就多用一次勾股定理。

Find the unit vector pointing in the direction of $\mathbf{a} = \langle 2, -1, 2 \rangle$.求与 $\mathbf{a} = \langle 2, -1, 2 \rangle$ 同方向的单位向量（unit vector）。

$$ \|\mathbf{a}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3, $$ $$ \hat{\mathbf{a}} = \frac{1}{3}\langle 2, -1, 2 \rangle = \left\langle \tfrac{2}{3}, -\tfrac{1}{3}, \tfrac{2}{3} \right\rangle. $$

As a check, $\|\hat{\mathbf{a}}\|^2 = \tfrac{4}{9} + \tfrac{1}{9} + \tfrac{4}{9} = 1$.作为检验，$\|\hat{\mathbf{a}}\|^2 = \tfrac{4}{9} + \tfrac{1}{9} + \tfrac{4}{9} = 1$。

Place vectors $\mathbf{a}$ and $\mathbf{b}$ tail to tail at the origin. Geometric addition $\mathbf{a} + \mathbf{b}$ is the diagonal of the parallelogram they span, obtained by translating $\mathbf{b}$ so its tail sits at the tip of $\mathbf{a}$. Translation preserves the horizontal and vertical displacements, so the displacement of the diagonal in each axis is the sum of the individual displacements. Reading off coordinates,把向量 $\mathbf{a}$ 与 $\mathbf{b}$ 的起点都放在原点。几何上的和 $\mathbf{a} + \mathbf{b}$ 就是它们张成的平行四边形的对角线，做法是把 $\mathbf{b}$ 平移，使其起点落在 $\mathbf{a}$ 的终点处。平移保持水平与竖直方向的位移不变，因此对角线在每个坐标轴上的位移等于两个向量各自位移之和。读出坐标，

$$ \mathbf{a} + \mathbf{b} = \langle a_1 + b_1, \; a_2 + b_2, \; a_3 + b_3 \rangle. $$

The same argument shows scaling by $c$ multiplies each displacement by $c$, giving $c\mathbf{a} = \langle ca_1, ca_2, ca_3 \rangle$. Thus the geometric operations agree with the algebraic ones, which is exactly the statement that $\mathbb{R}^3$ is a vector space.同样的论证表明，用 $c$ 缩放会把每个位移都乘以 $c$，得到 $c\mathbf{a} = \langle ca_1, ca_2, ca_3 \rangle$。于是几何运算与代数运算一致，这正是 $\mathbb{R}^3$ 是一个向量空间的含义。

Express $\mathbf{w} = \langle 7, 4 \rangle$ as a combination $s\,\mathbf{u} + t\,\mathbf{v}$ of $\mathbf{u} = \langle 1, 1 \rangle$ and $\mathbf{v} = \langle 3, -1 \rangle$.把 $\mathbf{w} = \langle 7, 4 \rangle$ 表示为 $\mathbf{u} = \langle 1, 1 \rangle$ 与 $\mathbf{v} = \langle 3, -1 \rangle$ 的线性组合 $s\,\mathbf{u} + t\,\mathbf{v}$。

Matching components gives a linear system:令对应分量相等，得到一个线性方程组：

$$ s + 3t = 7, \qquad s - t = 4. $$

Subtracting the second equation from the first yields $4t = 3$, so $t = \tfrac{3}{4}$, and then $s = 4 + t = \tfrac{19}{4}$. Therefore用第一式减去第二式得 $4t = 3$，故 $t = \tfrac{3}{4}$，再得 $s = 4 + t = \tfrac{19}{4}$。于是

$$ \mathbf{w} = \tfrac{19}{4}\,\mathbf{u} + \tfrac{3}{4}\,\mathbf{v}. $$

A quick check: $\tfrac{19}{4}\langle 1,1 \rangle + \tfrac{3}{4}\langle 3,-1 \rangle = \langle \tfrac{19+9}{4}, \tfrac{19-3}{4} \rangle = \langle 7, 4 \rangle$. Because $\mathbf{u}$ and $\mathbf{v}$ are not parallel they span the whole plane, so every target $\mathbf{w}$ has a unique pair of weights.快速检验：$\tfrac{19}{4}\langle 1,1 \rangle + \tfrac{3}{4}\langle 3,-1 \rangle = \langle \tfrac{19+9}{4}, \tfrac{19-3}{4} \rangle = \langle 7, 4 \rangle$。由于 $\mathbf{u}$ 与 $\mathbf{v}$ 不平行，它们张成整个平面，因此每个目标 $\mathbf{w}$ 都有唯一的一组权重。

A $50$ N weight hangs from two cables making angles of $30^\circ$ and $60^\circ$ with the horizontal ceiling. Find the magnitude of the tension in the cable at $30^\circ$.一个 $50$ N 的重物由两根缆绳悬挂，缆绳与水平天花板分别成 $30^\circ$ 与 $60^\circ$ 角。求 $30^\circ$ 那根缆绳中张力的大小。

Let $T_1$ pull up and to the left at $30^\circ$ above horizontal and $T_2$ up and to the right at $60^\circ$. In components, with right and up positive,设 $T_1$ 在水平线上方 $30^\circ$ 处向左上方拉，$T_2$ 在 $60^\circ$ 处向右上方拉。取向右、向上为正，按分量写出，

$$ \mathbf{T}_1 = T_1\langle -\cos 30^\circ, \sin 30^\circ \rangle, \qquad \mathbf{T}_2 = T_2\langle \cos 60^\circ, \sin 60^\circ \rangle. $$

Equilibrium requires $\mathbf{T}_1 + \mathbf{T}_2 + \langle 0, -50 \rangle = \mathbf{0}$. The horizontal balance gives $T_1 \cos 30^\circ = T_2 \cos 60^\circ$, so $T_2 = T_1 \dfrac{\cos 30^\circ}{\cos 60^\circ} = T_1\sqrt{3}$. The vertical balance gives $T_1 \sin 30^\circ + T_2 \sin 60^\circ = 50$. Substituting,平衡条件要求 $\mathbf{T}_1 + \mathbf{T}_2 + \langle 0, -50 \rangle = \mathbf{0}$。水平方向的平衡给出 $T_1 \cos 30^\circ = T_2 \cos 60^\circ$，故 $T_2 = T_1 \dfrac{\cos 30^\circ}{\cos 60^\circ} = T_1\sqrt{3}$。竖直方向的平衡给出 $T_1 \sin 30^\circ + T_2 \sin 60^\circ = 50$。代入得，

$$ T_1\left( \tfrac{1}{2} \right) + T_1\sqrt{3}\left( \tfrac{\sqrt{3}}{2} \right) = T_1\left( \tfrac{1}{2} + \tfrac{3}{2} \right) = 2 T_1 = 50, $$

so $T_1 = 25$ N. Vectors turn a force diagram into two scalar equations, one per axis.故 $T_1 = 25$ N。向量把一张受力图转化为两个标量方程，每个坐标轴对应一个。

Find the angle between $\mathbf{a} = \langle 1, 2, 2 \rangle$ and $\mathbf{b} = \langle 3, 0, 4 \rangle$.求 $\mathbf{a} = \langle 1, 2, 2 \rangle$ 与 $\mathbf{b} = \langle 3, 0, 4 \rangle$ 之间的夹角。

$$ \mathbf{a} \cdot \mathbf{b} = (1)(3) + (2)(0) + (2)(4) = 11, $$ $$ \|\mathbf{a}\| = 3, \quad \|\mathbf{b}\| = 5, \quad \cos\theta = \frac{11}{15}. $$ $$ \theta = \arccos\!\left(\tfrac{11}{15}\right) \approx 42.8^\circ. $$

Form the triangle with sides $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{a} - \mathbf{b}$. The law of cosines on the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ states作以 $\mathbf{a}$、$\mathbf{b}$、$\mathbf{a} - \mathbf{b}$ 为边的三角形。对 $\mathbf{a}$ 与 $\mathbf{b}$ 之间的夹角 $\theta$ 应用余弦定理，得

$$ \|\mathbf{a} - \mathbf{b}\|^2 = \|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 - 2\|\mathbf{a}\|\,\|\mathbf{b}\|\cos\theta. $$

Expand the left side using the algebraic dot product, which distributes over subtraction:用代数形式的点积展开左端，它对减法满足分配律：

$$ \|\mathbf{a} - \mathbf{b}\|^2 = (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \|\mathbf{a}\|^2 - 2\,\mathbf{a}\cdot\mathbf{b} + \|\mathbf{b}\|^2. $$

Comparing the two expressions, the $\|\mathbf{a}\|^2$ and $\|\mathbf{b}\|^2$ terms cancel and we are left with $-2\,\mathbf{a}\cdot\mathbf{b} = -2\|\mathbf{a}\|\,\|\mathbf{b}\|\cos\theta$, hence $\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\,\|\mathbf{b}\|\cos\theta$.比较两个表达式，$\|\mathbf{a}\|^2$ 与 $\|\mathbf{b}\|^2$ 项相消，剩下 $-2\,\mathbf{a}\cdot\mathbf{b} = -2\|\mathbf{a}\|\,\|\mathbf{b}\|\cos\theta$，于是 $\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\,\|\mathbf{b}\|\cos\theta$。

Decompose $\mathbf{b} = \langle 3, 2 \rangle$ into a part parallel to $\mathbf{a} = \langle 4, 0 \rangle$ and a part orthogonal to $\mathbf{a}$.把 $\mathbf{b} = \langle 3, 2 \rangle$ 分解为一个与 $\mathbf{a} = \langle 4, 0 \rangle$ 平行的部分和一个与 $\mathbf{a}$ 正交的部分。

First the parallel part, the vector projection:先求平行部分，即向量投影（projection）：

$$ \operatorname{proj}_{\mathbf{a}}\mathbf{b} = \frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|^2}\,\mathbf{a} = \frac{12}{16}\langle 4,0 \rangle = \langle 3, 0 \rangle. $$

The orthogonal part is whatever remains:正交部分就是剩下的：

$$ \mathbf{b} - \operatorname{proj}_{\mathbf{a}}\mathbf{b} = \langle 3,2 \rangle - \langle 3,0 \rangle = \langle 0, 2 \rangle. $$

Check the orthogonality: $\langle 0,2 \rangle \cdot \langle 4,0 \rangle = 0$, as required. Every vector splits uniquely into a multiple of $\mathbf{a}$ plus something perpendicular to $\mathbf{a}$, the idea behind work done by a force and behind least-squares fitting.检验正交性：$\langle 0,2 \rangle \cdot \langle 4,0 \rangle = 0$，符合要求。每个向量都能唯一地分解为 $\mathbf{a}$ 的一个倍数加上一个垂直于 $\mathbf{a}$ 的部分，这正是力做功与最小二乘拟合背后的思想。

A constant force $\mathbf{F} = \langle 4, 1, 2 \rangle$ newtons moves an object from $A = (0,0,0)$ to $B = (3, 0, 4)$ metres. Find the work done.一个恒力 $\mathbf{F} = \langle 4, 1, 2 \rangle$ 牛顿把物体从 $A = (0,0,0)$ 移动到 $B = (3, 0, 4)$ 米。求所做的功。

Work is the dot product of force with displacement, $W = \mathbf{F}\cdot\mathbf{d}$, where $\mathbf{d} = \overrightarrow{AB} = \langle 3, 0, 4 \rangle$:功是力与位移的点积，$W = \mathbf{F}\cdot\mathbf{d}$，其中 $\mathbf{d} = \overrightarrow{AB} = \langle 3, 0, 4 \rangle$：

$$ W = (4)(3) + (1)(0) + (2)(4) = 12 + 0 + 8 = 20 \text{ joules}. $$

Only the component of force along the motion contributes, which is precisely what the dot product extracts. A force perpendicular to the displacement does zero work.只有沿运动方向的力的分量才有贡献，而这正是点积所提取的量。垂直于位移的力做功为零。

The geometric form $\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\,\|\mathbf{b}\|\cos\theta$ has a consequence that does not mention angles at all. Since $|\cos\theta| \le 1$,几何形式 $\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\,\|\mathbf{b}\|\cos\theta$ 有一个完全不涉及角度的推论。由于 $|\cos\theta| \le 1$，

$$ |\mathbf{a}\cdot\mathbf{b}| \le \|\mathbf{a}\|\,\|\mathbf{b}\|, $$

which is the Cauchy-Schwarz inequality, with equality exactly when $\mathbf{a}$ and $\mathbf{b}$ are parallel. From it the triangle inequality follows. Expand the squared magnitude of a sum:这就是柯西-施瓦茨不等式（Cauchy-Schwarz inequality），当且仅当 $\mathbf{a}$ 与 $\mathbf{b}$ 平行时取等号。由它可推出三角不等式。展开和向量模长的平方：

$$ \|\mathbf{a} + \mathbf{b}\|^2 = \|\mathbf{a}\|^2 + 2\,\mathbf{a}\cdot\mathbf{b} + \|\mathbf{b}\|^2 \le \|\mathbf{a}\|^2 + 2\|\mathbf{a}\|\,\|\mathbf{b}\| + \|\mathbf{b}\|^2 = \big( \|\mathbf{a}\| + \|\mathbf{b}\| \big)^2. $$

Taking square roots gives $\|\mathbf{a} + \mathbf{b}\| \le \|\mathbf{a}\| + \|\mathbf{b}\|$: the length of a sum never exceeds the sum of the lengths, the algebraic statement that a straight path is shortest. The single inequality $|\cos\theta|\le 1$ underwrites both results.两边开方得 $\|\mathbf{a} + \mathbf{b}\| \le \|\mathbf{a}\| + \|\mathbf{b}\|$：和向量的长度绝不会超过两向量长度之和，这正是"直线最短"的代数表述。仅凭 $|\cos\theta|\le 1$ 这一条不等式，就支撑起了上述两个结论。

For $P = (1,0,0)$, $Q = (0,2,0)$, $R = (0,0,3)$, find a vector normal to triangle $PQR$ and the triangle's area.对 $P = (1,0,0)$、$Q = (0,2,0)$、$R = (0,0,3)$，求一个垂直于三角形 $PQR$ 的向量以及该三角形的面积。

$$ \overrightarrow{PQ} = \langle -1, 2, 0 \rangle, \qquad \overrightarrow{PR} = \langle -1, 0, 3 \rangle. $$ $$ \overrightarrow{PQ} \times \overrightarrow{PR} = \langle (2)(3) - (0)(0),\; (0)(-1) - (-1)(3),\; (-1)(0) - (2)(-1) \rangle = \langle 6, 3, 2 \rangle. $$

The parallelogram area is $\|\langle 6,3,2 \rangle\| = \sqrt{36 + 9 + 4} = 7$, so the triangle has area $\tfrac{7}{2}$.平行四边形的面积为 $\|\langle 6,3,2 \rangle\| = \sqrt{36 + 9 + 4} = 7$，故三角形的面积为 $\tfrac{7}{2}$。

Find the volume spanned by $\mathbf{a} = \langle 1, 0, 0 \rangle$, $\mathbf{b} = \langle 1, 2, 0 \rangle$, $\mathbf{c} = \langle 1, 2, 3 \rangle$.求由 $\mathbf{a} = \langle 1, 0, 0 \rangle$、$\mathbf{b} = \langle 1, 2, 0 \rangle$、$\mathbf{c} = \langle 1, 2, 3 \rangle$ 张成的体积。

$$ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 2 & 3 \end{vmatrix} = 1\,(2\cdot 3 - 0\cdot 2) = 6. $$

The volume is the absolute value $|6| = 6$.体积是其绝对值 $|6| = 6$。

Find the distance from $P = (1, 1, 1)$ to the line through $A = (0,0,0)$ with direction $\mathbf{v} = \langle 1, 1, 0 \rangle$.求点 $P = (1, 1, 1)$ 到过 $A = (0,0,0)$、方向为 $\mathbf{v} = \langle 1, 1, 0 \rangle$ 的直线的距离。

The distance is the height of the parallelogram with edges $\overrightarrow{AP}$ and $\mathbf{v}$, namely area over base. Here $\overrightarrow{AP} = \langle 1,1,1 \rangle$, and该距离就是以 $\overrightarrow{AP}$ 与 $\mathbf{v}$ 为边的平行四边形的高，即面积除以底。这里 $\overrightarrow{AP} = \langle 1,1,1 \rangle$，且

$$ \overrightarrow{AP} \times \mathbf{v} = \langle (1)(0) - (1)(1),\; (1)(1) - (1)(0),\; (1)(1) - (1)(1) \rangle = \langle -1, 1, 0 \rangle. $$ $$ D = \frac{\|\overrightarrow{AP} \times \mathbf{v}\|}{\|\mathbf{v}\|} = \frac{\sqrt{(-1)^2 + 1^2 + 0^2}}{\sqrt{1^2 + 1^2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1. $$

The cross product packages the perpendicular distance because its magnitude already measures area, and area divided by base length is height.叉积之所以能直接给出垂直距离，是因为它的模长本身度量面积，而面积除以底长就是高。

Decide whether $\mathbf{a} = \langle 1, 2, -1 \rangle$, $\mathbf{b} = \langle 2, -1, 1 \rangle$, $\mathbf{c} = \langle 0, 5, -3 \rangle$ are coplanar.判断 $\mathbf{a} = \langle 1, 2, -1 \rangle$、$\mathbf{b} = \langle 2, -1, 1 \rangle$、$\mathbf{c} = \langle 0, 5, -3 \rangle$ 是否共面。

Compute the scalar triple product. First $\mathbf{b}\times\mathbf{c} = \langle (-1)(-3) - (1)(5),\; (1)(0) - (2)(-3),\; (2)(5) - (-1)(0) \rangle = \langle -2, 6, 10 \rangle$. Then计算标量三重积。先求 $\mathbf{b}\times\mathbf{c} = \langle (-1)(-3) - (1)(5),\; (1)(0) - (2)(-3),\; (2)(5) - (-1)(0) \rangle = \langle -2, 6, 10 \rangle$。然后

$$ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = (1)(-2) + (2)(6) + (-1)(10) = -2 + 12 - 10 = 0. $$

The signed volume is zero, so the three vectors lie in a common plane through the origin. A nonzero result would have measured an actual solid volume and ruled out coplanarity.带符号体积为零，故这三个向量位于一个过原点的公共平面内。若结果非零，则它度量了一个实际的立体体积，从而排除共面。

Start from the algebraic definition and compute $\|\mathbf{a}\times\mathbf{b}\|^2$ by expanding the three components. A direct, if tedious, expansion gives the Lagrange identity从代数定义出发，通过展开三个分量来计算 $\|\mathbf{a}\times\mathbf{b}\|^2$。直接但略显繁琐的展开给出拉格朗日恒等式

$$ \|\mathbf{a}\times\mathbf{b}\|^2 = \|\mathbf{a}\|^2\|\mathbf{b}\|^2 - (\mathbf{a}\cdot\mathbf{b})^2. $$

Substitute the geometric form of the dot product, $\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\,\|\mathbf{b}\|\cos\theta$:代入点积的几何形式 $\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\,\|\mathbf{b}\|\cos\theta$：

$$ \|\mathbf{a}\times\mathbf{b}\|^2 = \|\mathbf{a}\|^2\|\mathbf{b}\|^2\big( 1 - \cos^2\theta \big) = \|\mathbf{a}\|^2\|\mathbf{b}\|^2 \sin^2\theta. $$

Because $0 \le \theta \le \pi$ we have $\sin\theta \ge 0$, so taking the nonnegative square root yields $\|\mathbf{a}\times\mathbf{b}\| = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\theta$. This is exactly the base times height of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$, which is its area. The cross product vanishes precisely when $\sin\theta = 0$, that is when the vectors are parallel and span no area.因为 $0 \le \theta \le \pi$，所以 $\sin\theta \ge 0$，取非负平方根即得 $\|\mathbf{a}\times\mathbf{b}\| = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\theta$。这恰好是以 $\mathbf{a}$ 与 $\mathbf{b}$ 为边的平行四边形的底乘高，也就是它的面积。叉积恰好在 $\sin\theta = 0$ 时为零，即两向量平行、张不出面积时。

Find parametric equations for the line through $A = (2, 1, 3)$ and $B = (4, -1, 0)$.求过 $A = (2, 1, 3)$ 与 $B = (4, -1, 0)$ 的直线的参数方程。

A direction vector is $\overrightarrow{AB} = \langle 2, -2, -3 \rangle$. Using $A$ as the base point,一个方向向量是 $\overrightarrow{AB} = \langle 2, -2, -3 \rangle$。取 $A$ 作为基点，

$$ x = 2 + 2t, \qquad y = 1 - 2t, \qquad z = 3 - 3t. $$

At $t = 0$ we recover $A$ and at $t = 1$ we recover $B$, confirming the line.当 $t = 0$ 时得到 $A$，当 $t = 1$ 时得到 $B$，验证了这条直线。

Do $\mathbf{r}_1(t) = \langle 1 + t, 2 - t, 3t \rangle$ and $\mathbf{r}_2(s) = \langle 2s, 3 - s, 1 + s \rangle$ intersect?$\mathbf{r}_1(t) = \langle 1 + t, 2 - t, 3t \rangle$ 与 $\mathbf{r}_2(s) = \langle 2s, 3 - s, 1 + s \rangle$ 是否相交？

Set coordinates equal: $1 + t = 2s$, $2 - t = 3 - s$, $3t = 1 + s$. From the second, $s - t = 1$, so $s = t + 1$. Substituting into the first gives $1 + t = 2(t+1) = 2t + 2$, so $t = -1$ and $s = 0$. Check the third equation: $3(-1) = -3$ while $1 + 0 = 1$, and $-3 \neq 1$.令对应坐标相等：$1 + t = 2s$、$2 - t = 3 - s$、$3t = 1 + s$。由第二式得 $s - t = 1$，故 $s = t + 1$。代入第一式得 $1 + t = 2(t+1) = 2t + 2$，于是 $t = -1$、$s = 0$。检验第三式：$3(-1) = -3$ 而 $1 + 0 = 1$，且 $-3 \neq 1$。

No parameter pair satisfies all three equations, and the directions $\langle 1,-1,3 \rangle$ and $\langle 2,-1,1 \rangle$ are not parallel, so the lines are skew.没有一对参数能同时满足这三个方程，而方向 $\langle 1,-1,3 \rangle$ 与 $\langle 2,-1,1 \rangle$ 不平行，所以这两条直线异面。

Find the distance between the skew lines $L_1: \mathbf{r}_1(t) = \langle 1 + t, 2 - t, 3t \rangle$ and $L_2: \mathbf{r}_2(s) = \langle 2s, 3 - s, 1 + s \rangle$ from the previous example.求上例中两条异面直线 $L_1: \mathbf{r}_1(t) = \langle 1 + t, 2 - t, 3t \rangle$ 与 $L_2: \mathbf{r}_2(s) = \langle 2s, 3 - s, 1 + s \rangle$ 之间的距离。

The common perpendicular direction is $\mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2$, where $\mathbf{v}_1 = \langle 1,-1,3 \rangle$ and $\mathbf{v}_2 = \langle 2,-1,1 \rangle$:公垂线方向为 $\mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2$，其中 $\mathbf{v}_1 = \langle 1,-1,3 \rangle$、$\mathbf{v}_2 = \langle 2,-1,1 \rangle$：

$$ \mathbf{n} = \langle (-1)(1) - (3)(-1),\; (3)(2) - (1)(1),\; (1)(-1) - (-1)(2) \rangle = \langle 2, 5, 1 \rangle. $$

Pick one point on each line, say $P_1 = (1,2,0)$ at $t=0$ and $P_2 = (0,3,1)$ at $s=0$, with $\overrightarrow{P_1 P_2} = \langle -1, 1, 1 \rangle$. The distance is the length of the projection of $\overrightarrow{P_1 P_2}$ onto $\mathbf{n}$:在每条直线上各取一点，比如 $t=0$ 时的 $P_1 = (1,2,0)$ 和 $s=0$ 时的 $P_2 = (0,3,1)$，得 $\overrightarrow{P_1 P_2} = \langle -1, 1, 1 \rangle$。该距离就是 $\overrightarrow{P_1 P_2}$ 在 $\mathbf{n}$ 上投影的长度：

$$ D = \frac{|\overrightarrow{P_1 P_2}\cdot \mathbf{n}|}{\|\mathbf{n}\|} = \frac{|(-1)(2) + (1)(5) + (1)(1)|}{\sqrt{4 + 25 + 1}} = \frac{4}{\sqrt{30}} = \frac{4}{\sqrt{30}}. $$

The choice of base points does not affect the answer, because shifting either point along its own line adds a multiple of $\mathbf{v}_1$ or $\mathbf{v}_2$, both orthogonal to $\mathbf{n}$, so the numerator is unchanged.基点的选取不影响结果，因为把任一点沿其所在直线移动，只会加上 $\mathbf{v}_1$ 或 $\mathbf{v}_2$ 的某个倍数，而它们都与 $\mathbf{n}$ 正交，故分子保持不变。

Find a direction vector for the line where the planes $x + y + z = 1$ and $x - 2y + 3z = 4$ meet.求平面 $x + y + z = 1$ 与 $x - 2y + 3z = 4$ 交线的一个方向向量。

The line lies in both planes, so it is perpendicular to both normals $\mathbf{n}_1 = \langle 1,1,1 \rangle$ and $\mathbf{n}_2 = \langle 1,-2,3 \rangle$. A vector perpendicular to both is their cross product:该直线同时位于两个平面内，因此它垂直于两个法向量 $\mathbf{n}_1 = \langle 1,1,1 \rangle$ 与 $\mathbf{n}_2 = \langle 1,-2,3 \rangle$。同时垂直于二者的向量就是它们的叉积：

$$ \mathbf{v} = \mathbf{n}_1 \times \mathbf{n}_2 = \langle (1)(3) - (1)(-2),\; (1)(1) - (1)(3),\; (1)(-2) - (1)(1) \rangle = \langle 5, -2, -3 \rangle. $$

So the line of intersection runs in the direction $\langle 5, -2, -3 \rangle$. The cross product turns "perpendicular to two given vectors" into a single computation.所以交线沿方向 $\langle 5, -2, -3 \rangle$ 延伸。叉积把"同时垂直于两个给定向量"化为一次计算。

Find an equation of the plane through $P = (1,0,0)$, $Q = (0,2,0)$, $R = (0,0,3)$.求过 $P = (1,0,0)$、$Q = (0,2,0)$、$R = (0,0,3)$ 的平面的方程。

Two in-plane vectors are $\overrightarrow{PQ} = \langle -1,2,0 \rangle$ and $\overrightarrow{PR} = \langle -1,0,3 \rangle$. A normal is their cross product, computed earlier as $\mathbf{n} = \langle 6,3,2 \rangle$. Using $P$,平面内的两个向量是 $\overrightarrow{PQ} = \langle -1,2,0 \rangle$ 与 $\overrightarrow{PR} = \langle -1,0,3 \rangle$。法向量就是它们的叉积，前面已算得 $\mathbf{n} = \langle 6,3,2 \rangle$。用 $P$，

$$ 6(x - 1) + 3(y - 0) + 2(z - 0) = 0 \;\Longrightarrow\; 6x + 3y + 2z = 6. $$

Let $P_0$ be any point on the plane $ax + by + cz = d$ and $P_1$ the external point. The distance from $P_1$ to the plane is the length of the projection of $\overrightarrow{P_0 P_1}$ onto the unit normal $\hat{\mathbf{n}} = \mathbf{n}/\|\mathbf{n}\|$:设 $P_0$ 为平面 $ax + by + cz = d$ 上任意一点，$P_1$ 为平面外一点。$P_1$ 到平面的距离就是 $\overrightarrow{P_0 P_1}$ 在单位法向量 $\hat{\mathbf{n}} = \mathbf{n}/\|\mathbf{n}\|$ 上投影的长度：

$$ D = \left| \overrightarrow{P_0 P_1} \cdot \hat{\mathbf{n}} \right| = \frac{|\mathbf{n} \cdot \overrightarrow{P_0 P_1}|}{\|\mathbf{n}\|}. $$

Write the numerator componentwise. Since $a x_0 + b y_0 + c z_0 = d$ (the point $P_0$ lies on the plane), the cross terms collapse:把分子按分量写开。由于 $a x_0 + b y_0 + c z_0 = d$（点 $P_0$ 在平面上），交叉项相消：

$$ \mathbf{n} \cdot \overrightarrow{P_0 P_1} = a(x_1 - x_0) + b(y_1 - y_0) + c(z_1 - z_0) = a x_1 + b y_1 + c z_1 - d. $$

Dividing by $\|\mathbf{n}\| = \sqrt{a^2 + b^2 + c^2}$ yields the stated formula.除以 $\|\mathbf{n}\| = \sqrt{a^2 + b^2 + c^2}$ 即得所述公式。

Find the acute angle between the planes $x + y + z = 1$ and $x - 2y + 3z = 4$.求平面 $x + y + z = 1$ 与 $x - 2y + 3z = 4$ 之间的锐角。

The angle between planes equals the angle between their normals $\mathbf{n}_1 = \langle 1,1,1 \rangle$ and $\mathbf{n}_2 = \langle 1,-2,3 \rangle$:两平面之间的夹角等于它们法向量 $\mathbf{n}_1 = \langle 1,1,1 \rangle$ 与 $\mathbf{n}_2 = \langle 1,-2,3 \rangle$ 之间的夹角：

$$ \cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{\|\mathbf{n}_1\|\,\|\mathbf{n}_2\|} = \frac{|1 - 2 + 3|}{\sqrt{3}\,\sqrt{14}} = \frac{2}{\sqrt{42}}. $$ $$ \theta = \arccos\!\left( \frac{2}{\sqrt{42}} \right) \approx 72.0^\circ. $$

Taking the absolute value in the numerator selects the acute angle between the planes rather than its obtuse supplement.在分子取绝对值，选取的是两平面之间的锐角，而不是它的钝角补角。

Find the point where the line $x = 1 + 2t$, $y = -1 + t$, $z = 3t$ meets the plane $2x - y + z = 7$.求直线 $x = 1 + 2t$、$y = -1 + t$、$z = 3t$ 与平面 $2x - y + z = 7$ 的交点。

Substitute the parametric coordinates into the plane equation and solve for $t$:把参数坐标代入平面方程并解出 $t$：

$$ 2(1 + 2t) - (-1 + t) + 3t = 7 \;\Longrightarrow\; 2 + 4t + 1 - t + 3t = 7 \;\Longrightarrow\; 6t + 3 = 7. $$

So $t = \tfrac{2}{3}$. Back-substituting,故 $t = \tfrac{2}{3}$。回代，

$$ \left( 1 + \tfrac{4}{3},\; -1 + \tfrac{2}{3},\; 2 \right) = \left( \tfrac{7}{3},\; -\tfrac{1}{3},\; 2 \right). $$

If substitution had produced a contradiction such as $3 = 7$, the line would be parallel to the plane and miss it; if it had produced an identity such as $7 = 7$, the line would lie inside the plane.如果代入后得到像 $3 = 7$ 这样的矛盾，则直线与平面平行而不相交；如果得到像 $7 = 7$ 这样的恒等式，则直线整个落在平面内。

Identify the surface $z = x^2 + y^2$.辨认曲面 $z = x^2 + y^2$。

The trace at $z = k > 0$ is the circle $x^2 + y^2 = k$ of radius $\sqrt{k}$, growing with height. The traces in the planes $x = 0$ and $y = 0$ are the parabolas $z = y^2$ and $z = x^2$, both opening upward.在 $z = k > 0$ 处的截痕是半径为 $\sqrt{k}$ 的圆 $x^2 + y^2 = k$，随高度增大而变大。平面 $x = 0$ 与 $y = 0$ 中的截痕是抛物线 $z = y^2$ 与 $z = x^2$，都向上开口。

Circular horizontal traces and parabolic vertical traces identify a circular paraboloid, a bowl with vertex at the origin opening up the $z$ axis.水平截痕为圆、竖直截痕为抛物线，由此辨认出这是一个圆抛物面，即顶点在原点、沿 $z$ 轴向上开口的碗形。

Classify $4x^2 + y^2 + 9z^2 = 36$.对 $4x^2 + y^2 + 9z^2 = 36$ 分类。

Divide by $36$ to reach standard form:两边除以 $36$ 化为标准形：

$$ \frac{x^2}{9} + \frac{y^2}{36} + \frac{z^2}{4} = 1. $$

All three squared terms are positive and the right side is $1$, so the surface is an ellipsoid with semi-axes $3$ along $x$, $6$ along $y$, and $2$ along $z$.三个平方项都为正且右端为 $1$，所以该曲面是椭球面，沿 $x$、$y$、$z$ 的半轴分别为 $3$、$6$、$2$。

Classify $\dfrac{x^2}{4} - \dfrac{y^2}{9} + \dfrac{z^2}{1} = 1$ and contrast it with $\dfrac{x^2}{4} - \dfrac{y^2}{9} - \dfrac{z^2}{1} = 1$.对 $\dfrac{x^2}{4} - \dfrac{y^2}{9} + \dfrac{z^2}{1} = 1$ 分类，并与 $\dfrac{x^2}{4} - \dfrac{y^2}{9} - \dfrac{z^2}{1} = 1$ 作对比。

In the first equation, exactly one squared term is negative and the right side is $+1$, so the axis is the $y$ axis, the axis of the lone minus sign. Cutting perpendicular to that axis, the traces $y = k$ are ellipses $\tfrac{x^2}{4} + z^2 = 1 + \tfrac{k^2}{9}$, which exist for every $k$, so the surface is one connected piece: a hyperboloid of one sheet opening along the $y$ axis. The vertical traces $z = k$ are instead hyperbolas $\tfrac{x^2}{4} - \tfrac{y^2}{9} = 1 - k^2$.在第一个方程中，恰有一个平方项为负且右端为 $+1$，所以轴是 $y$ 轴，即那个唯一负号所对应的轴。垂直于该轴切割，截痕 $y = k$ 是椭圆 $\tfrac{x^2}{4} + z^2 = 1 + \tfrac{k^2}{9}$，对每个 $k$ 都存在，所以曲面是连通的一整片：沿 $y$ 轴开口的单叶双曲面。而竖直截痕 $z = k$ 则是双曲线 $\tfrac{x^2}{4} - \tfrac{y^2}{9} = 1 - k^2$。

In the second equation, two squared terms are negative. The trace $y = 0$, $z = 0$ forces $x^2 = 4$, but the trace $x = 0$ gives $-\tfrac{y^2}{9} - z^2 = 1$, which has no solutions, leaving a gap. The surface is a hyperboloid of two sheets, opening along the $x$ axis (the axis of the lone plus sign). The rule: one minus sign gives one sheet, two minus signs give two sheets.在第二个方程中，有两个平方项为负。截痕 $y = 0$、$z = 0$ 迫使 $x^2 = 4$，但截痕 $x = 0$ 给出 $-\tfrac{y^2}{9} - z^2 = 1$，它无解，于是中间留出空缺。该曲面是双叶双曲面，沿 $x$ 轴（那个唯一正号所对应的轴）开口。规律是：一个负号给出单叶，两个负号给出双叶。

Classify $x^2 + 4y^2 - z^2 - 2x + 8y + 1 = 0$ and locate its center.对 $x^2 + 4y^2 - z^2 - 2x + 8y + 1 = 0$ 分类并求出其中心。

Complete the square in the variables that appear linearly:对出现一次项的变量配方：

$$ (x^2 - 2x) + 4(y^2 + 2y) - z^2 + 1 = 0, $$ $$ (x-1)^2 - 1 + 4\big[(y+1)^2 - 1\big] - z^2 + 1 = 0, $$ $$ (x-1)^2 + 4(y+1)^2 - z^2 = 4. $$

Divide by $4$ to reach standard form:两边除以 $4$ 化为标准形：

$$ \frac{(x-1)^2}{4} + (y+1)^2 - \frac{z^2}{4} = 1. $$

Two positive squared terms and one negative, equal to $+1$, mark a hyperboloid of one sheet, opening along the $z$ axis, recentered at $(1, -1, 0)$. Shifting the surface does not change its type; only the linear terms move the center.两个正平方项与一个负平方项、等于 $+1$，标志着一个单叶双曲面，沿 $z$ 轴开口，中心移到 $(1, -1, 0)$。平移曲面不改变其类型；只有一次项会移动中心。

A trace is the curve cut from a surface by a coordinate plane, found by fixing one variable. The power of traces is that each cross section is a familiar conic, and the way the conics change with the fixed value reveals the three-dimensional shape.截痕是用坐标平面从曲面上切出的曲线，通过固定一个变量得到。截痕的威力在于：每个截面都是我们熟悉的圆锥曲线，而这些圆锥曲线随固定值变化的方式揭示了三维形状。

Take the elliptic paraboloid $z = \tfrac{x^2}{a^2} + \tfrac{y^2}{b^2}$. Fixing $z = k$ gives $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = k$, an ellipse for $k > 0$ that grows with height, a single point at $k = 0$, and nothing for $k < 0$. Fixing $x = 0$ gives the parabola $z = \tfrac{y^2}{b^2}$, opening upward. Ellipses stacked along a parabolic spine assemble into the bowl.以椭圆抛物面 $z = \tfrac{x^2}{a^2} + \tfrac{y^2}{b^2}$ 为例。固定 $z = k$ 得 $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = k$：当 $k > 0$ 时是随高度增大的椭圆，$k = 0$ 时是一个点，$k < 0$ 时无图形。固定 $x = 0$ 得向上开口的抛物线 $z = \tfrac{y^2}{b^2}$。一系列椭圆沿一条抛物线脊堆叠起来，就拼成了那只碗。

For the hyperboloid of one sheet $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} - \tfrac{z^2}{c^2} = 1$, the horizontal traces $z = k$ are ellipses $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1 + \tfrac{k^2}{c^2}$, smallest at $z = 0$ and widening in both directions, while the vertical traces are hyperbolas. The waist at $z = 0$ and the absence of any forbidden height confirm the single connected sheet. Reading the three coordinate traces is therefore enough to name any quadric without plotting it.对单叶双曲面 $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} - \tfrac{z^2}{c^2} = 1$，水平截痕 $z = k$ 是椭圆 $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1 + \tfrac{k^2}{c^2}$，在 $z = 0$ 处最小，向两个方向逐渐变大，而竖直截痕是双曲线。$z = 0$ 处的"腰"以及不存在任何禁止高度，确认了这是连通的一整片。因此读出三条坐标截痕，便足以叫出任何二次曲面的名字，而无需画图。