Vectors and the Geometry of Space · Solutions向量与空间解析几何 · 解答

(a) Cross product via the determinant(a) 用行列式计算叉积 M1·M1·A1·A1

Expand along the first row of the $3\times 3$ determinant: (M1)沿 $3\times 3$ 行列式第一行展开：(M1)

$$ \mathbf{p}\times\mathbf{q} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & -2 \\ -2 & 1 & 4 \end{vmatrix}. $$

$\mathbf{i}$-component: $3\cdot 4 - (-2)\cdot 1 = 12 + 2 = 14$. (M1)$\mathbf{i}$ 分量：$3\cdot 4 - (-2)\cdot 1 = 12 + 2 = 14$。(M1)

$\mathbf{j}$-component: $-(1\cdot 4 - (-2)\cdot(-2)) = -(4-4) = 0$.$\mathbf{j}$ 分量：$-(1\cdot 4 - (-2)\cdot(-2)) = -(4-4) = 0$。

$\mathbf{k}$-component: $1\cdot 1 - 3\cdot(-2) = 1+6 = 7$. (A1 for each pair; A1 for assembled answer)$\mathbf{k}$ 分量：$1\cdot 1 - 3\cdot(-2) = 1+6 = 7$。（每对 A1；组合结果 A1）

$$ \mathbf{p}\times\mathbf{q} = \langle 14,\,0,\,7\rangle. $$

(b) Orthogonality checks(b) 正交性验证 A1·A1

Compute $\mathbf{p}\cdot(\mathbf{p}\times\mathbf{q}) = \langle 1,3,-2\rangle\cdot\langle 14,0,7\rangle = 14 + 0 - 14 = 0$. (A1)计算 $\mathbf{p}\cdot(\mathbf{p}\times\mathbf{q}) = \langle 1,3,-2\rangle\cdot\langle 14,0,7\rangle = 14 + 0 - 14 = 0$。(A1)

Compute $\mathbf{q}\cdot(\mathbf{p}\times\mathbf{q}) = \langle -2,1,4\rangle\cdot\langle 14,0,7\rangle = -28 + 0 + 28 = 0$. (A1)计算 $\mathbf{q}\cdot(\mathbf{p}\times\mathbf{q}) = \langle -2,1,4\rangle\cdot\langle 14,0,7\rangle = -28 + 0 + 28 = 0$。(A1)

Both are zero, confirming orthogonality.两者均为零，证实了正交性。

(c) Areas of parallelogram and triangle(c) 平行四边形与三角形面积 M1·A1

The area of the parallelogram is $|\mathbf{p}\times\mathbf{q}|$: (M1)平行四边形的面积为 $|\mathbf{p}\times\mathbf{q}|$：(M1)

$$ |\langle 14,0,7\rangle| = \sqrt{196+0+49} = \sqrt{245} = 7\sqrt{5}. $$

The triangle has half that area: $\dfrac{7\sqrt{5}}{2}$. (A1)三角形面积为其一半：$\dfrac{7\sqrt{5}}{2}$。(A1)

(d) Scalar triple product and parallelepiped volume(d) 混合积与平行六面体体积 M1·A1

First compute $\mathbf{q}\times\mathbf{r} = \langle -2,1,4\rangle\times\langle 0,2,1\rangle$: (M1)先计算 $\mathbf{q}\times\mathbf{r} = \langle -2,1,4\rangle\times\langle 0,2,1\rangle$：(M1)

$$ \mathbf{q}\times\mathbf{r} = \langle 1\cdot 1 - 4\cdot 2,\;4\cdot 0-(-2)\cdot 1,\;(-2)\cdot 2 - 1\cdot 0\rangle = \langle 1-8,\;0+2,\;-4-0\rangle = \langle -7,2,-4\rangle. $$

Then $\mathbf{p}\cdot(\mathbf{q}\times\mathbf{r}) = \langle 1,3,-2\rangle\cdot\langle -7,2,-4\rangle = -7+6+8 = 7$.再计算 $\mathbf{p}\cdot(\mathbf{q}\times\mathbf{r}) = \langle 1,3,-2\rangle\cdot\langle -7,2,-4\rangle = -7+6+8 = 7$。

Volume $= |\,7\,| = 7$. (A1)体积 $= |\,7\,| = 7$。(A1)

(a) The determinant definition(a) 行列式定义 B1·A1

The cross product is defined by expanding the formal $3\times 3$ determinant: (B1)叉积由展开形式 $3\times 3$ 行列式来定义：(B1)

$$ \mathbf{u}\times\mathbf{v} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix} = \langle u_2v_3-u_3v_2,\;u_3v_1-u_1v_3,\;u_1v_2-u_2v_1\rangle. $$

So $w_1=u_2v_3-u_3v_2$, $w_2=u_3v_1-u_1v_3$, $w_3=u_1v_2-u_2v_1$. (A1)故 $w_1=u_2v_3-u_3v_2$，$w_2=u_3v_1-u_1v_3$，$w_3=u_1v_2-u_2v_1$。(A1)

(b) Proof that $\mathbf{u}\cdot\mathbf{w}=0$(b) 证明 $\mathbf{u}\cdot\mathbf{w}=0$ M1·A1·R1

Compute $\mathbf{u}\cdot\mathbf{w} = u_1w_1+u_2w_2+u_3w_3$. Substituting the components from (a): (M1)计算 $\mathbf{u}\cdot\mathbf{w} = u_1w_1+u_2w_2+u_3w_3$。代入 (a) 的分量：(M1)

$$ \mathbf{u}\cdot\mathbf{w} = u_1(u_2v_3-u_3v_2) + u_2(u_3v_1-u_1v_3) + u_3(u_1v_2-u_2v_1). $$

Expand: (A1)展开：(A1)

$$ = u_1u_2v_3 - u_1u_3v_2 + u_2u_3v_1 - u_1u_2v_3 + u_1u_3v_2 - u_2u_3v_1. $$

Each term $u_iu_jv_k$ appears exactly twice with opposite signs and cancels, leaving $0$. (R1)每个形如 $u_iu_jv_k$ 的项都以相反符号出现恰好两次，相消后剩余 $0$。(R1)

(c) Analogous argument for $\mathbf{v}$ and geometric conclusion(c) 对 $\mathbf{v}$ 的类似论证及几何结论 M1·A1·R1

Compute $\mathbf{v}\cdot\mathbf{w} = v_1w_1+v_2w_2+v_3w_3$. Substituting: (M1)计算 $\mathbf{v}\cdot\mathbf{w} = v_1w_1+v_2w_2+v_3w_3$。代入：(M1)

$$ \mathbf{v}\cdot\mathbf{w} = v_1(u_2v_3-u_3v_2)+v_2(u_3v_1-u_1v_3)+v_3(u_1v_2-u_2v_1). $$

Expanding gives terms of the form $v_iu_jv_k$ and $-v_iu_jv_k$ (with $i=k$ in each pair), so every term cancels and $\mathbf{v}\cdot\mathbf{w}=0$. (A1)展开后得到形如 $v_iu_jv_k$ 与 $-v_iu_jv_k$（每对中 $i=k$）的项，所有项相消，故 $\mathbf{v}\cdot\mathbf{w}=0$。(A1)

Since $\mathbf{w}$ is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$, it is perpendicular to every vector in the plane spanned by $\mathbf{u}$ and $\mathbf{v}$. Therefore $\mathbf{u}\times\mathbf{v}$ is a normal vector to that plane. (R1)由于 $\mathbf{w}$ 同时垂直于 $\mathbf{u}$ 和 $\mathbf{v}$，它垂直于 $\mathbf{u}$ 和 $\mathbf{v}$ 张成平面中的每一个向量。因此 $\mathbf{u}\times\mathbf{v}$ 是该平面的法向量。(R1)

(a) Proof of the distance formula(a) 距离公式的证明 M1·M1·A1·M1·A1

Decompose $\mathbf{v}$ into the part along $\mathbf{d}$ and the part perpendicular to $\mathbf{d}$. The foot of the perpendicular is $F = A + \operatorname{proj}_{\mathbf{d}}\mathbf{v}$, so: (M1)将 $\mathbf{v}$ 分解为沿 $\mathbf{d}$ 方向的部分与垂直于 $\mathbf{d}$ 的部分。垂足为 $F = A + \operatorname{proj}_{\mathbf{d}}\mathbf{v}$，故：(M1)

$$ \operatorname{proj}_{\mathbf{d}}\mathbf{v} = \frac{\mathbf{v}\cdot\mathbf{d}}{|\mathbf{d}|^2}\,\mathbf{d}, \qquad \overrightarrow{FP} = \mathbf{v} - \frac{\mathbf{v}\cdot\mathbf{d}}{|\mathbf{d}|^2}\,\mathbf{d}. $$

The distance from $P$ to $\ell$ is $|\overrightarrow{FP}|$. (M1) Now use the cross product: since $\mathbf{v}\times\mathbf{d} = |\mathbf{v}||\mathbf{d}|\sin\theta\,\hat{\mathbf{n}}$ where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{d}$, we have $|\mathbf{v}\times\mathbf{d}|=|\mathbf{v}||\mathbf{d}|\sin\theta$. (A1) The perpendicular distance from $P$ to the line is$P$ 到 $\ell$ 的距离为 $|\overrightarrow{FP}|$。(M1) 利用叉积：由于 $\mathbf{v}\times\mathbf{d} = |\mathbf{v}||\mathbf{d}|\sin\theta\,\hat{\mathbf{n}}$，其中 $\theta$ 为 $\mathbf{v}$ 与 $\mathbf{d}$ 的夹角，故 $|\mathbf{v}\times\mathbf{d}|=|\mathbf{v}||\mathbf{d}|\sin\theta$。(A1) $P$ 到直线的垂直距离为

$$ d(P,\ell) = |\mathbf{v}|\sin\theta = \frac{|\mathbf{v}||\mathbf{d}|\sin\theta}{|\mathbf{d}|} = \frac{|\mathbf{v}\times\mathbf{d}|}{|\mathbf{d}|}. $$

(M1 for identifying $|\mathbf{v}|\sin\theta$ as the perpendicular component; A1 for the final assembly)（M1：将 $|\mathbf{v}|\sin\theta$ 识别为垂直分量；A1：最终整合）

(b) Applying the formula(b) 应用公式 M1·A1·A1

Here $\mathbf{v}=\overrightarrow{AP}=\langle 2,1,-1\rangle$ and $\mathbf{d}=\langle 1,2,2\rangle$. Compute $\mathbf{v}\times\mathbf{d}$: (M1)此处 $\mathbf{v}=\overrightarrow{AP}=\langle 2,1,-1\rangle$，$\mathbf{d}=\langle 1,2,2\rangle$。计算 $\mathbf{v}\times\mathbf{d}$：(M1)

$$ \mathbf{v}\times\mathbf{d} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&1&-1\\1&2&2\end{vmatrix} = \langle 1\cdot 2-(-1)\cdot 2,\;(-1)\cdot 1-2\cdot 2,\;2\cdot 2-1\cdot 1\rangle = \langle 4,-5,3\rangle. $$

(A1) $|\mathbf{v}\times\mathbf{d}|=\sqrt{16+25+9}=\sqrt{50}=5\sqrt{2}$. Then $|\mathbf{d}|=\sqrt{1+4+4}=3$, so (A1)(A1) $|\mathbf{v}\times\mathbf{d}|=\sqrt{16+25+9}=\sqrt{50}=5\sqrt{2}$。又 $|\mathbf{d}|=\sqrt{1+4+4}=3$，故 (A1)

$$ d(P,\ell) = \frac{5\sqrt{2}}{3}. $$

(c) Confirmation via the perpendicular vector(c) 通过垂直向量验证 M1·A1

The scalar projection of $\mathbf{v}$ onto $\mathbf{d}$ is $\dfrac{\mathbf{v}\cdot\mathbf{d}}{|\mathbf{d}|^2} = \dfrac{2+2-2}{9} = \dfrac{2}{9}$. (M1) So:$\mathbf{v}$ 在 $\mathbf{d}$ 上的标量系数为 $\dfrac{\mathbf{v}\cdot\mathbf{d}}{|\mathbf{d}|^2} = \dfrac{2+2-2}{9} = \dfrac{2}{9}$。(M1) 故：

$$ \operatorname{proj}_{\mathbf{d}}\mathbf{v} = \frac{2}{9}\langle 1,2,2\rangle = \left\langle\frac{2}{9},\frac{4}{9},\frac{4}{9}\right\rangle. $$ $$ \overrightarrow{FP} = \left\langle 2-\frac{2}{9},\;1-\frac{4}{9},\;-1-\frac{4}{9}\right\rangle = \left\langle\frac{16}{9},\,\frac{5}{9},\,-\frac{13}{9}\right\rangle. $$ $$ |\overrightarrow{FP}| = \frac{1}{9}\sqrt{256+25+169} = \frac{\sqrt{450}}{9} = \frac{15\sqrt{2}}{9} = \frac{5\sqrt{2}}{3}. \quad \checkmark $$

(A1)

(a) Volume formula(a) 体积公式 M1·A1·M1·A1

The base of the parallelepiped is the parallelogram spanned by $\mathbf{b}$ and $\mathbf{c}$. Its area is $|\mathbf{b}\times\mathbf{c}|$. (M1) Let $\hat{\mathbf{n}} = \dfrac{\mathbf{b}\times\mathbf{c}}{|\mathbf{b}\times\mathbf{c}|}$ be the unit normal to the base.平行六面体的底面是由 $\mathbf{b}$ 和 $\mathbf{c}$ 张成的平行四边形，面积为 $|\mathbf{b}\times\mathbf{c}|$。(M1) 设 $\hat{\mathbf{n}} = \dfrac{\mathbf{b}\times\mathbf{c}}{|\mathbf{b}\times\mathbf{c}|}$ 为底面的单位法向量。

The height of the parallelepiped is the length of the component of $\mathbf{a}$ in the direction of $\hat{\mathbf{n}}$, that is, $h = |\mathbf{a}\cdot\hat{\mathbf{n}}|$. (A1) Therefore: (M1)平行六面体的高为 $\mathbf{a}$ 在 $\hat{\mathbf{n}}$ 方向上分量的长度，即 $h = |\mathbf{a}\cdot\hat{\mathbf{n}}|$。(A1) 因此：(M1)

$$ V = \text{base area}\times h = |\mathbf{b}\times\mathbf{c}|\,\left|\mathbf{a}\cdot\frac{\mathbf{b}\times\mathbf{c}}{|\mathbf{b}\times\mathbf{c}|}\right| = \left|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\right|. $$

(A1)

(b) Coplanarity criterion(b) 共面性判定准则 R1·A1

Statement: $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are coplanar if and only if $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=0$.命题：$\mathbf{a}$、$\mathbf{b}$、$\mathbf{c}$ 共面，当且仅当 $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=0$。

Proof: by part (a), $V=|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|$. The parallelepiped has volume zero if and only if its three edge vectors are coplanar (i.e. all lie in a common plane through the origin, which happens exactly when $\mathbf{a}$ is a linear combination of $\mathbf{b}$ and $\mathbf{c}$, so the height $h=0$). Therefore $V=0 \Leftrightarrow \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=0 \Leftrightarrow$ the vectors are coplanar. (R1·A1)证明：由 (a)，$V=|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|$。平行六面体体积为零当且仅当三条棱向量共面（即均位于过原点的某一平面内，这恰好发生在 $\mathbf{a}$ 是 $\mathbf{b}$ 与 $\mathbf{c}$ 的线性组合时，高 $h=0$）。因此 $V=0 \Leftrightarrow \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=0 \Leftrightarrow$ 向量共面。(R1·A1)

(c) Numerical test(c) 数值验证 M1·A1

Compute $\mathbf{b}\times\mathbf{c} = \langle 0,1,-1\rangle\times\langle 2,3,7\rangle$: (M1)计算 $\mathbf{b}\times\mathbf{c} = \langle 0,1,-1\rangle\times\langle 2,3,7\rangle$：(M1)

$$ \mathbf{b}\times\mathbf{c} = \langle 1\cdot 7-(-1)\cdot 3,\;(-1)\cdot 2-0\cdot 7,\;0\cdot 3-1\cdot 2\rangle = \langle 10,-2,-2\rangle. $$

Then $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \langle 1,2,3\rangle\cdot\langle 10,-2,-2\rangle = 10-4-6 = 0$. (A1)再计算 $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \langle 1,2,3\rangle\cdot\langle 10,-2,-2\rangle = 10-4-6 = 0$。(A1)

Since the triple product is zero, the three vectors are coplanar by the criterion in (b).由于混合积为零，由 (b) 的判定准则可知三个向量共面。

(a) Vector and parametric equations(a) 向量方程与参数方程 B1·A1

The vector equation is $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{d}$: (B1)向量方程为 $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{d}$：(B1)

$$ \mathbf{r}(t) = \langle 1,-2,3\rangle + t\langle 2,1,-1\rangle = \langle 1+2t,\;-2+t,\;3-t\rangle. $$

Parametric form: $x=1+2t$, $y=-2+t$, $z=3-t$. (A1)参数形式：$x=1+2t$，$y=-2+t$，$z=3-t$。(A1)

(b) Symmetric equations(b) 对称式方程 A1

Solve each parametric equation for $t$ and equate: (A1)从每个参数方程解出 $t$ 并令其相等：(A1)

$$ \frac{x-1}{2} = \frac{y+2}{1} = \frac{z-3}{-1}. $$

(c) Does $Q=(5,0,1)$ lie on $\ell$?(c) 点 $Q=(5,0,1)$ 是否在 $\ell$ 上？ M1·A1·A1

Substitute into the parametric equations and test consistency. From $x=5$: $1+2t=5 \Rightarrow t=2$. (M1) Check $y$: $-2+2=0$ ✓. Check $z$: $3-2=1$ ✓. (A1) All three equations are satisfied at $t=2$, so $Q$ lies on $\ell$. (A1)代入参数方程并验证一致性。由 $x=5$：$1+2t=5 \Rightarrow t=2$。(M1) 验证 $y$：$-2+2=0$ ✓。验证 $z$：$3-2=1$ ✓。(A1) 三个方程在 $t=2$ 时均满足，故 $Q$ 在 $\ell$ 上。(A1)

(d) The parallel line $m$ through $R=(0,1,-1)$(d) 过 $R=(0,1,-1)$ 且与 $\ell$ 平行的直线 $m$ A1·A1

A line parallel to $\ell$ has the same direction vector $\mathbf{d}=\langle 2,1,-1\rangle$. Using $R$ as the base point: (A1)与 $\ell$ 平行的直线具有相同方向向量 $\mathbf{d}=\langle 2,1,-1\rangle$。以 $R$ 为基点：(A1)

$$ m:\quad x=2t,\quad y=1+t,\quad z=-1-t. $$

To confirm $m\ne\ell$, check whether $R=(0,1,-1)\in\ell$. From $x=0$: $t=-\tfrac{1}{2}$; then $y=-2-\tfrac{1}{2}=-\tfrac{5}{2}\ne 1$. So $R\notin\ell$, confirming $m$ and $\ell$ are distinct parallel lines. (A1)为确认 $m\ne\ell$，验证 $R=(0,1,-1)$ 是否在 $\ell$ 上。由 $x=0$：$t=-\tfrac{1}{2}$；则 $y=-2-\tfrac{1}{2}=-\tfrac{5}{2}\ne 1$。故 $R\notin\ell$，确认 $m$ 与 $\ell$ 是不同的平行直线。(A1)

(a) Normal vector from the cross product(a) 用叉积求法向量 M1·M1·A1

Two vectors lying in $\Pi$ are $\overrightarrow{AB}=\langle -1,2,0\rangle$ and $\overrightarrow{AC}=\langle -1,0,3\rangle$. (M1) Compute $\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}$: (M1)平面 $\Pi$ 内的两个向量为 $\overrightarrow{AB}=\langle -1,2,0\rangle$ 和 $\overrightarrow{AC}=\langle -1,0,3\rangle$。(M1) 计算 $\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}$：(M1)

$$ \mathbf{n} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&2&0\\-1&0&3\end{vmatrix} = \langle 2\cdot 3-0\cdot 0,\;0\cdot(-1)-(-1)\cdot 3,\;(-1)\cdot 0-2\cdot(-1)\rangle = \langle 6,3,2\rangle. $$

(A1)

(b) Scalar plane equation(b) 平面的标量方程 M1·A1

Using point $A=(1,0,0)$ and normal $\langle 6,3,2\rangle$: $6(x-1)+3(y-0)+2(z-0)=0$. (M1) Simplify: $6x+3y+2z=6$. (A1)用点 $A=(1,0,0)$ 和法向量 $\langle 6,3,2\rangle$：$6(x-1)+3(y-0)+2(z-0)=0$。(M1) 化简：$6x+3y+2z=6$。(A1)

(c) Distance from the origin to $\Pi$(c) 原点到 $\Pi$ 的距离 M1·A1

The point-to-plane distance formula gives, for the origin $(0,0,0)$ and plane $6x+3y+2z=6$: (M1)对原点 $(0,0,0)$ 和平面 $6x+3y+2z=6$ 应用点到平面距离公式：(M1)

$$ d = \frac{|6(0)+3(0)+2(0)-6|}{\sqrt{6^2+3^2+2^2}} = \frac{6}{\sqrt{36+9+4}} = \frac{6}{\sqrt{49}} = \frac{6}{7}. $$

(A1)

(d) Relationship between $\Pi$ and $\Pi'$(d) $\Pi$ 与 $\Pi'$ 的位置关系 A1

The normal to $\Pi'$ is $\mathbf{n}'=\langle 6,3,2\rangle$, which is identical to $\mathbf{n}$, the normal to $\Pi$. Since the normals are parallel, the planes are parallel. (Note that $\Pi'$ passes through the origin, since $6(0)+3(0)+2(0)=0$, while $\Pi$ does not. So the planes are distinct and parallel.) (A1)$\Pi'$ 的法向量 $\mathbf{n}'=\langle 6,3,2\rangle$ 与 $\Pi$ 的法向量 $\mathbf{n}$ 完全相同。由于法向量平行，两平面平行。（注意 $\Pi'$ 过原点，因为 $6(0)+3(0)+2(0)=0$，而 $\Pi$ 不过原点。故两平面不同且平行。）(A1)

(a) Skew lines and the distance formula(a) 异面直线与距离公式 M1·A1·M1·A1

The direction vectors are $\mathbf{d}_1=\langle 1,1,0\rangle$ and $\mathbf{d}_2=\langle 0,1,1\rangle$. Since $\mathbf{d}_1$ is not a scalar multiple of $\mathbf{d}_2$, the lines are not parallel. (M1)方向向量为 $\mathbf{d}_1=\langle 1,1,0\rangle$ 和 $\mathbf{d}_2=\langle 0,1,1\rangle$。由于 $\mathbf{d}_1$ 不是 $\mathbf{d}_2$ 的数量倍，两直线不平行。(M1)

Compute the common perpendicular direction: $\mathbf{d}_1\times\mathbf{d}_2 = \langle 1,-1,1\rangle$ (since $\langle 1\cdot 1-0\cdot 1,\;0\cdot 0-1\cdot 1,\;1\cdot 1-1\cdot 0\rangle=\langle 1,-1,1\rangle$). Now check whether the lines actually intersect: the vector connecting a point on $L_1$ to a point on $L_2$ is $\overrightarrow{P_1P_2}=\langle 0,1,0\rangle-\langle 1,0,0\rangle=\langle -1,1,0\rangle$. For intersection we would need $\overrightarrow{P_1P_2}=t\mathbf{d}_1-s\mathbf{d}_2$ to be solvable; computing $\overrightarrow{P_1P_2}\cdot(\mathbf{d}_1\times\mathbf{d}_2)=\langle -1,1,0\rangle\cdot\langle 1,-1,1\rangle=-1-1+0=-2\ne 0$, which proves the lines are skew. (A1)计算公垂线方向：$\mathbf{d}_1\times\mathbf{d}_2 = \langle 1,-1,1\rangle$（因为 $\langle 1\cdot 1-0\cdot 1,\;0\cdot 0-1\cdot 1,\;1\cdot 1-1\cdot 0\rangle=\langle 1,-1,1\rangle$）。检验两直线是否相交：连接 $L_1$ 上一点与 $L_2$ 上一点的向量为 $\overrightarrow{P_1P_2}=\langle 0,1,0\rangle-\langle 1,0,0\rangle=\langle -1,1,0\rangle$。若相交，方程 $\overrightarrow{P_1P_2}=t\mathbf{d}_1-s\mathbf{d}_2$ 须有解；计算 $\overrightarrow{P_1P_2}\cdot(\mathbf{d}_1\times\mathbf{d}_2)=\langle -1,1,0\rangle\cdot\langle 1,-1,1\rangle=-1-1+0=-2\ne 0$，证明两直线为异面直线。(A1)

The distance between skew lines is $d = \dfrac{|\overrightarrow{P_1P_2}\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}$. (M1)异面直线间的距离为 $d = \dfrac{|\overrightarrow{P_1P_2}\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}$。(M1)

$$ |\mathbf{d}_1\times\mathbf{d}_2| = |\langle 1,-1,1\rangle| = \sqrt{3}. \qquad d = \frac{|-2|}{\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}. $$

(A1)

(b) Line-plane intersection(b) 直线与平面的交点 M1·A1

Substitute the parametric equations $x=2+t$, $y=-t$, $z=-1+3t$ into the plane $x+2y+z=7$: (M1)将参数方程 $x=2+t$，$y=-t$，$z=-1+3t$ 代入平面 $x+2y+z=7$：(M1)

$$ (2+t)+2(-t)+(-1+3t)=7 \implies 2+t-2t-1+3t=7 \implies 1+2t=7 \implies t=3. $$

(A1) The intersection point is $\langle 2+3,\;0-3,\;-1+9\rangle = (5,-3,8)$. Check: $5+2(-3)+8=5-6+8=7$ ✓.(A1) 交点为 $\langle 2+3,\;0-3,\;-1+9\rangle = (5,-3,8)$。验证：$5+2(-3)+8=5-6+8=7$ ✓。

(a) Torque as a cross product(a) 用叉积计算力矩 M1·A1·A1

Compute $\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}$ using the $3\times 3$ determinant: (M1)用 $3\times 3$ 行列式计算 $\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}$：(M1)

$$ \boldsymbol{\tau} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-1&3\\0&4&-2\end{vmatrix} = \langle (-1)(-2)-3\cdot 4,\;3\cdot 0-2\cdot(-2),\;2\cdot 4-(-1)\cdot 0\rangle = \langle 2-12,\;0+4,\;8-0\rangle = \langle -10,4,8\rangle. $$

(A1) The magnitude is $|\boldsymbol{\tau}|=\sqrt{100+16+64}=\sqrt{180}=6\sqrt{5}$ N m. (A1)(A1) 模长为 $|\boldsymbol{\tau}|=\sqrt{100+16+64}=\sqrt{180}=6\sqrt{5}$ N m。(A1)

The torque vector $\langle -10,4,8\rangle$ is perpendicular to both $\mathbf{r}$ and $\mathbf{F}$; its direction gives the axis of rotation by the right-hand rule.力矩向量 $\langle -10,4,8\rangle$ 同时垂直于 $\mathbf{r}$ 和 $\mathbf{F}$；其方向由右手定则给出转轴方向。

(b) Intersection line of two planes(b) 两平面的交线 M1·A1

Direction vector: $\mathbf{n}_1\times\mathbf{n}_2 = \langle 1,1,1\rangle\times\langle 2,-1,1\rangle$. (M1)方向向量：$\mathbf{n}_1\times\mathbf{n}_2 = \langle 1,1,1\rangle\times\langle 2,-1,1\rangle$。(M1)

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

Find a point: set $z=0$. Then $x+y=6$ and $2x-y=3$. Adding: $3x=9$, $x=3$, $y=3$. Point $(3,3,0)$ lies on both planes (verify: $3+3+0=6$ ✓; $6-3+0=3$ ✓). (A1) Parametric equations of the intersection line:求一个点：令 $z=0$。则 $x+y=6$，$2x-y=3$。两式相加：$3x=9$，$x=3$，$y=3$。点 $(3,3,0)$ 在两平面上（验证：$3+3+0=6$ ✓；$6-3+0=3$ ✓）。(A1) 交线的参数方程：

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

(c) Quadric surface identification(c) 二次曲面识别 A1·A1

(i) $\dfrac{x^2}{4}+\dfrac{y^2}{9}+z^2=1$: this matches $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$ with $a=2$, $b=3$, $c=1$. It is an ellipsoid.(i) $\dfrac{x^2}{4}+\dfrac{y^2}{9}+z^2=1$：符合 $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$，其中 $a=2$，$b=3$，$c=1$。这是椭球面。

(ii) $z = x^2+\dfrac{y^2}{4}$: this matches $z = \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}$ (opening in the positive $z$-direction). It is an elliptic paraboloid with axis along the $z$-axis.(ii) $z = x^2+\dfrac{y^2}{4}$：符合 $z = \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}$（沿正 $z$ 方向开口）。这是轴沿 $z$ 轴的椭圆抛物面。

(iii) $x^2+y^2-z^2=1$: this matches $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=1$ with positive sum minus one squared term. It is a hyperboloid of one sheet with axis along the $z$-axis. (A1 for two correct; A1 for all three correct)(iii) $x^2+y^2-z^2=1$：符合 $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=1$，正项之和减去一个平方项。这是轴沿 $z$ 轴的单叶双曲面。（答对两个 A1；答对全部三个 A1）