9 月 2022 - 第3页共5页

两个平行直线间的性质（B009）

问题

若直线 $L_{1}$ 的表达式为 $\frac{x-x_{1}}{l_{1}}$ $=$ $\frac{y-y_{1}}{m_{1}}$ $=$ $\frac{z-z_{1}}{n_{1}}$, 直线 $L_{2}$ 的表达式为 $\frac{x-x_{2}}{l_{2}}$ $=$ $\frac{y-y_{2}}{m_{2}}$ $=$ $\frac{z-z_{2}}{n_{2}}$. 此外，直线 $L_{1}$ 和 $L_{2}$ 的方向向量分别为 $\vec{s_{1}}$ $=$ $(l_{1}, m_{1}, n_{1})$ 和 $\vec{s_{2}}$ $=$ $(l_{2}, m_{2}, n_{2})$.

那么，若 $L_{1}$ $//$ $L_{2}$, 则可以引申出来哪些性质？

选项

[A]. $L_{1}$ $//$ $L_{2}$ $\Leftrightarrow$ $\vec{s_{1}}$ $//$ $\vec{s_{2}}$ $\Leftrightarrow$ $l_{1}l_{2}$ $=$ $m_{1}m_{2}$ $=$ $n_{1}n_{2}$ $\Leftrightarrow$ $\vec{s_{1}}$ $\times$ $\vec{s_{2}}$ $=$ $0$

[B]. $L_{1}$ $//$ $L_{2}$ $\Leftrightarrow$ $\vec{s_{1}}$ $\perp$ $\vec{s_{2}}$ $\Leftrightarrow$ $\frac{l_{1}}{l_{2}}$ $=$ $\frac{m_{1}}{m_{2}}$ $=$ $\frac{n_{1}}{n_{2}}$ $\Leftrightarrow$ $\vec{s_{1}}$ $\times$ $\vec{s_{2}}$ $=$ $1$

[C]. $L_{1}$ $//$ $L_{2}$ $\Leftrightarrow$ $\vec{s_{1}}$ $//$ $\vec{s_{2}}$ $\Leftrightarrow$ $\frac{l_{1}}{l_{2}}$ $=$ $\frac{m_{1}}{m_{2}}$ $=$ $\frac{n_{1}}{n_{2}}$ $\Leftrightarrow$ $\vec{s_{1}}$ $\cdot$ $\vec{s_{2}}$ $=$ $1$

[D]. $L_{1}$ $//$ $L_{2}$ $\Leftrightarrow$ $\vec{s_{1}}$ $//$ $\vec{s_{2}}$ $\Leftrightarrow$ $\frac{l_{1}}{l_{2}}$ $=$ $\frac{m_{1}}{m_{2}}$ $=$ $\frac{n_{1}}{n_{2}}$ $\Leftrightarrow$ $\vec{s_{1}}$ $\times$ $\vec{s_{2}}$ $=$ $0$

答案

$L_{\textcolor{orange}{1}}$ $//$ $L_{\textcolor{cyan}{2}}$ $\textcolor{yellow}{\Leftrightarrow}$ $\vec{s_{\textcolor{orange}{1}}}$ $//$ $\vec{s_{\textcolor{cyan}{2}}}$ $\textcolor{yellow}{\Leftrightarrow}$ $\frac{l_{\textcolor{orange}{1}}}{l_{\textcolor{cyan}{2}}}$ $=$ $\frac{m_{\textcolor{orange}{1}}}{m_{\textcolor{cyan}{2}}}$ $=$ $\frac{n_{\textcolor{orange}{1}}}{n_{\textcolor{cyan}{2}}}$ $\textcolor{yellow}{\Leftrightarrow}$ $\vec{s_{\textcolor{orange}{1}}}$ $\times$ $\vec{s_{\textcolor{cyan}{2}}}$ $=$ $\textcolor{red}{0}$

两个呈夹角 $\theta$ 的平面间的性质（B009）

问题

若平面 $\pi_{1}$ 的表达式为 $A_{1} x$ $+$ $B_{1} y$ $+$ $C_{1} z$ $+$ $D_{1}$ $=$ $0$, 平面 $\pi_{2}$ 的表达式为 $A_{2} x$ $+$ $B_{2} y$ $+$ $C_{2} z$ $+$ $D_{2}$ $=$ $0$. 此外，平面 $\pi_{1}$ 和 $\pi_{2}$ 的法向量分别为 $\vec{n_{1}}$ $=$ $(A_{1}, B_{1}, C_{1})$ 和 $\vec{n_{2}}$ $=$ $(A_{1}, B_{2}, C_{2})$.

那么，若 $\pi_{1}$ 与 $\pi_{2}$ 之间的夹角为 $\theta$ $(0 \leqslant \theta \leqslant \frac{\pi}{2})$, 则 $\cos \theta$ $=$ $?$

选项

[A]. $\cos \theta$ $=$ $\frac{|A_{1}A_{2} + B_{1}B_{2} + C_{1} C_{2}|}{(A_{1}^{2} + B_{1}^{2} + C_{1}^{2}) \times (A_{2}^{2} + B_{2}^{2} + C_{2}^{2})}$

[B]. $\cos \theta$ $=$ $\frac{A_{1}A_{2} + B_{1}B_{2} + C_{1} C_{2}}{\sqrt{A_{1}^{2} + B_{1}^{2} + C_{1}^{2}} \times \sqrt{A_{2}^{2} + B_{2}^{2} + C_{2}^{2}}}$

[C]. $\cos \theta$ $=$ $\frac{|A_{1}A_{2} + B_{1}B_{2} + C_{1} C_{2}|}{\sqrt{A_{1}^{2} + B_{1}^{2} + C_{1}^{2}} \times \sqrt{A_{2}^{2} + B_{2}^{2} + C_{2}^{2}}}$

[D]. $\cos \theta$ $=$ $\frac{|A_{1}A_{2} – B_{1}B_{2} – C_{1} C_{2}|}{(A_{1}^{2} + B_{1}^{2} + C_{1}^{2}) \times (A_{2}^{2} + B_{2}^{2} + C_{2}^{2})}$

答案

$\textcolor{red}{\cos \theta}$ $=$ $\frac{|\vec{n_{\textcolor{orange}{1}}} \cdot \vec{n_{\textcolor{cyan}{2}}}|}{|\vec{n_{\textcolor{orange}{1}}}| |\vec{n_{\textcolor{cyan}{2}}}|}$ $=$ $\frac{|A_{\textcolor{orange}{1}}A_{\textcolor{cyan}{2}} + B_{\textcolor{orange}{1}}B_{\textcolor{cyan}{2}} + C_{\textcolor{orange}{1}} C_{\textcolor{cyan}{2}}|}{\sqrt{A_{\textcolor{orange}{1}}^{2} + B_{\textcolor{orange}{1}}^{2} + C_{\textcolor{orange}{1}}^{2}} \times \sqrt{A_{\textcolor{cyan}{2}}^{2} + B_{\textcolor{cyan}{2}}^{2} + C_{\textcolor{cyan}{2}}^{2}}}$

两个垂直平面间的性质（B009）

问题

那么，若 $\pi_{1}$ $\perp$ $\pi_{2}$, 则可以引申出来哪些性质？

选项

[A]. $\pi_{1}$ $\perp$ $\pi_{2}$ $\Leftrightarrow$ $\vec{n_{1}}$ $\perp$ $\vec{n_{2}}$ $\Leftrightarrow$ $\frac{A_{1}}{A_{2}}$ $+$ $\frac{B_{1}}{B_{2}}$ $+$ $\frac{C_{1}}{C_{2}}$ $=$ $0$ $\Leftrightarrow$ $\vec{n_{1}}$ $\cdot$ $\vec{n_{2}}$ $=$ $0$

[B]. $\pi_{1}$ $\perp$ $\pi_{2}$ $\Leftrightarrow$ $\vec{n_{1}}$ $//$ $\vec{n_{2}}$ $\Leftrightarrow$ $A_{1} A_{2}$ $=$ $B_{1} B_{2}$ $=$ $C_{1} C_{2}$ $=$ $0$ $\Leftrightarrow$ $\vec{n_{1}}$ $\cdot$ $\vec{n_{2}}$ $=$ $0$

[C]. $\pi_{1}$ $\perp$ $\pi_{2}$ $\Leftrightarrow$ $\vec{n_{1}}$ $\perp$ $\vec{n_{2}}$ $\Leftrightarrow$ $A_{1} A_{2}$ $+$ $B_{1} B_{2}$ $+$ $C_{1} C_{2}$ $=$ $1$ $\Leftrightarrow$ $\vec{n_{1}}$ $\times$ $\vec{n_{2}}$ $=$ $1$

[D]. $\pi_{1}$ $\perp$ $\pi_{2}$ $\Leftrightarrow$ $\vec{n_{1}}$ $\perp$ $\vec{n_{2}}$ $\Leftrightarrow$ $A_{1} A_{2}$ $+$ $B_{1} B_{2}$ $+$ $C_{1} C_{2}$ $=$ $0$ $\Leftrightarrow$ $\vec{n_{1}}$ $\cdot$ $\vec{n_{2}}$ $=$ $0$

答案

$\pi_{\textcolor{orange}{1}}$ $\textcolor{red}{\perp}$ $\pi_{\textcolor{cyan}{2}}$ $\Leftrightarrow$ $\vec{n_{\textcolor{orange}{1}}}$ $\textcolor{red}{\perp}$ $\vec{n_{\textcolor{cyan}{2}}}$ $\Leftrightarrow$ $A_{\textcolor{orange}{1}}$ $\textcolor{red}{\cdot}$ $A_{\textcolor{cyan}{2}}$ $\textcolor{yellow}{+}$ $B_{\textcolor{orange}{1}}$ $\textcolor{red}{\cdot}$ $B_{\textcolor{cyan}{2}}$ $\textcolor{yellow}{+}$ $C_{\textcolor{orange}{1}}$ $\textcolor{red}{\cdot}$ $C_{\textcolor{cyan}{2}}$ $=$ $0$ $\Leftrightarrow$ $\vec{n_{\textcolor{orange}{1}}}$ $\textcolor{red}{\cdot}$ $\vec{n_{\textcolor{cyan}{2}}}$ $=$ $\textcolor{red}{0}$

两个平行平面间的性质（B009）

问题

那么，若 $\pi_{1}$ $//$ $\pi_{2}$, 则可以引申出来哪些性质？

选项

[A]. $\pi_{1}$ $//$ $\pi_{2}$ $\Leftrightarrow$ $\vec{n_{1}}$ $//$ $\vec{n_{2}}$ $\Leftrightarrow$ $\frac{A_{1}}{A_{2}}$ $=$ $\frac{B_{1}}{B_{2}}$ $=$ $\frac{C_{1}}{C_{2}}$ $\Leftrightarrow$ $\vec{n_{1}}$ $\cdot$ $\vec{n_{2}}$ $=$ $1$

[B]. $\pi_{1}$ $//$ $\pi_{2}$ $\Leftrightarrow$ $\vec{n_{1}}$ $//$ $\vec{n_{2}}$ $\Leftrightarrow$ $\frac{A_{1}}{A_{2}}$ $=$ $\frac{B_{1}}{B_{2}}$ $=$ $\frac{C_{1}}{C_{2}}$ $\Leftrightarrow$ $\vec{n_{1}}$ $\times$ $\vec{n_{2}}$ $=$ $0$

[C]. $\pi_{1}$ $//$ $\pi_{2}$ $\Leftrightarrow$ $\vec{n_{1}}$ $//$ $\vec{n_{2}}$ $\Leftrightarrow$ $A_{1} A_{2}$ $=$ $B_{1} B_{2}$ $=$ $C_{1} C_{2}$ $\Leftrightarrow$ $\vec{n_{1}}$ $\times$ $\vec{n_{2}}$ $=$ $0$

[D]. $\pi_{1}$ $//$ $\pi_{2}$ $\Leftrightarrow$ $\vec{n_{1}}$ $\perp$ $\vec{n_{2}}$ $\Leftrightarrow$ $\frac{A_{1}}{A_{2}}$ $=$ $\frac{B_{1}}{B_{2}}$ $=$ $\frac{C_{1}}{C_{2}}$ $\Leftrightarrow$ $\vec{n_{1}}$ $\times$ $\vec{n_{2}}$ $=$ $0$

答案

$\pi_{\textcolor{orange}{1}}$ $\textcolor{red}{//}$ $\pi_{\textcolor{cyan}{2}}$ $\Leftrightarrow$ $\vec{n_{\textcolor{orange}{1}}}$ $\textcolor{red}{//}$ $\vec{n_{\textcolor{cyan}{2}}}$ $\Leftrightarrow$ $\frac{A_{\textcolor{orange}{1}}}{A_{\textcolor{cyan}{2}}}$ $=$ $\frac{B_{\textcolor{orange}{1}}}{B_{\textcolor{cyan}{2}}}$ $=$ $\frac{C_{\textcolor{orange}{1}}}{C_{\textcolor{cyan}{2}}}$ $\Leftrightarrow$ $\vec{n_{\textcolor{orange}{1}}}$ $\textcolor{yellow}{\times}$ $\vec{n_{\textcolor{cyan}{2}}}$ $=$ $\textcolor{yellow}{0}$

空间直线方程的两点式（B009）

问题

若空间直线方程过点 $(x_{1}, y_{1}, z_{1})$ 和 $(x_{2}, y_{2}, z_{2})$, 则如何使用 [两点式方程] 表示该直线？

选项

[A]. $\frac{x – x_{1}}{x_{2} – x_{1}}$ $=$ $\frac{y – y_{1}}{y_{2} – y_{1}}$ $=$ $\frac{z – z_{1}}{z_{2} – z_{1}}$

[B]. $\frac{x – x_{2}}{x_{2} – x_{1}}$ $=$ $\frac{y – y_{2}}{y_{2} – y_{1}}$ $=$ $\frac{z – z_{2}}{z_{2} – z_{1}}$

[C]. $\frac{x – x_{1}}{x – x_{2}}$ $=$ $\frac{y – y_{1}}{y – y_{2}}$ $=$ $\frac{z – z_{1}}{z – z_{2}}$

[D]. $\frac{x + x_{1}}{x_{2} + x_{1}}$ $=$ $\frac{y + y_{1}}{y_{2} + y_{1}}$ $=$ $\frac{z + z_{1}}{z_{2} + z_{1}}$

答案

$\frac{\textcolor{red}{x} – \textcolor{cyan}{x_{1}}}{\textcolor{orange}{x_{2}} – \textcolor{cyan}{x_{1}}}$ $=$ $\frac{\textcolor{red}{y} – \textcolor{cyan}{y_{1}}}{\textcolor{orange}{y_{2}} – \textcolor{cyan}{y_{1}}}$ $=$ $\frac{\textcolor{red}{z} – \textcolor{cyan}{z_{1}}}{\textcolor{orange}{z_{2}} – \textcolor{cyan}{z_{1}}}$

空间直线方程的标准式/对称式（B009）

问题

若空间直线方程过点 $(x_{0}, y_{0}, z_{0})$, 且该直线的方向向量 $\vec{s}$ $=$ $(A, B, C)$, 则如何使用 [标准式方程] 或者说 [对称式方程] 表示该直线？

选项

[A]. $\frac{x – A}{x_{0}}$ $=$ $\frac{y – B}{y_{0}}$ $=$ $\frac{z – C}{z_{0}}$

[B]. $A(x – x_{0})$ $=$ $B(y – y_{0})$ $=$ $C(z – z_{0})$

[C]. $\frac{x + x_{0}}{A}$ $=$ $\frac{y + y_{0}}{B}$ $=$ $\frac{z + z_{0}}{C}$

[D]. $\frac{x – x_{0}}{A}$ $=$ $\frac{y – y_{0}}{B}$ $=$ $\frac{z – z_{0}}{C}$

答案

$\frac{\textcolor{red}{x} – \textcolor{cyan}{x_{0}}}{\textcolor{orange}{A}}$ $=$ $\frac{\textcolor{red}{y} – \textcolor{cyan}{y_{0}}}{\textcolor{orange}{B}}$ $=$ $\frac{\textcolor{red}{z} – \textcolor{cyan}{z_{0}}}{\textcolor{orange}{C}}$

空间直线方程的参数式（B009）

问题

若空间直线方程过点 $(x_{0}, y_{0}, z_{0})$, 且该直线的方向向量 $\vec{s}$ $=$ $(A, B, C)$, 则如何使用参数式方程表示该直线？

选项

[A]. $\left\{\begin{matrix} x = x_{0} + \frac{A}{t},\\ y = y_{0} + \frac{B}{t},\\ z = z_{0} + \frac{C}{t}.\end{matrix}\right.$

[B]. $\left\{\begin{matrix} x = Ax_{0},\\ y = By_{0},\\ z = Cz_{0}.\end{matrix}\right.$

[C]. $\left\{\begin{matrix} x = x_{0} + A,\\ y = y_{0} + B,\\ z = z_{0} + C.\end{matrix}\right.$

[D]. $\left\{\begin{matrix} x = x_{0} + At,\\ y = y_{0} + Bt,\\ z = z_{0} + Ct.\end{matrix}\right.$

答案

$\left\{\begin{matrix} \textcolor{red}{x} = \textcolor{orange}{x_{0}} + \textcolor{cyan}{A} \textcolor{yellow}{t},\\ \textcolor{red}{y} = \textcolor{orange}{y_{0}} + \textcolor{cyan}{B} \textcolor{yellow}{t},\\ \textcolor{red}{z} = \textcolor{orange}{z_{0}} + \textcolor{cyan}{C} \textcolor{yellow}{t}.\end{matrix}\right.$

其中，$t$ 为参数.

空间直线方程的方向向量（B009）

问题

已知一条空间直线由两平面相交形成，且这两个平面的法向量分别为 $\vec{n_{1}}$ $=$ $(A_{1}, B_{1}, C_{1})$ 和 $\vec{n_{2}}$ $=$ $(A_{2}, B_{2}, C_{2})$, 则该直线的方向向量 $\vec{s}$ $=$ $?$

选项

[A]. $\vec{s}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_{1} & B_{1} & C_{1} \end{vmatrix}$

[B]. $\vec{s}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_{1} & B_{1} & C_{1} \\ A_{2} & B_{2} & C_{2} \end{vmatrix}$

[C]. $\vec{s}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ C_{1} & B_{1} & A_{1} \\ C_{2} & B_{2} & A_{2} \end{vmatrix}$

[D]. $\vec{s}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{1}{A_{1}} & \frac{1}{B_{1}} & \frac{1}{C_{1}} \\ \frac{1}{A_{2}} & \frac{1}{B_{2}} & \frac{1}{C_{2}} \end{vmatrix}$

答案

$\textcolor{yellow}{\vec{s}}$ $=$ $\begin{vmatrix} \mathbf{\textcolor{orange}{i}} & \mathbf{\textcolor{orange}{j}} & \mathbf{\textcolor{orange}{k}} \\ A_{\textcolor{red}{1}} & B_{\textcolor{red}{1}} & C_{\textcolor{red}{1}} \\ A_{\textcolor{cyan}{2}} & B_{\textcolor{cyan}{2}} & C_{\textcolor{cyan}{2}} \end{vmatrix}$

其中，向量 $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ 分别为 $x$ 轴、$y$ 轴和 $z$ 轴上的单位向量.

空间直线方程的一般式/交面式（B009）

问题

若两个平面的法向量分别为 $\vec{n_{1}}$ $=$ $(A_{1}, B_{1}, C_{1})$, $\vec{n_{2}}$ $=$ $(A_{2}, B_{2}, C_{2})$, 此外还有常数 $D_{1}$ 和 $D_{2}$, 则这两个平面相交所形成的直线如何表示？

选项

[A]. $\left\{\begin{matrix} \frac{1}{A_{1}}x + \frac{1}{B_{1}}y + \frac{1}{C_{1}}z + \frac{1}{D_{1}} = 0,\\ \frac{1}{A_{2}}x + \frac{1}{B_{2}}y + \frac{1}{C_{2}}z + \frac{1}{D_{2}} = 0.\end{matrix}\right.$

[B]. $\left\{\begin{matrix} A_{1}x + B_{1}y + C_{1}z = 0,\\ A_{2}x + B_{2}y + C_{2}z = 0.\end{matrix}\right.$

[C]. $\left\{\begin{matrix} A_{1}x + B_{1}y + C_{1}z + D_{1} = 0,\\ A_{2}x + B_{2}y + C_{2}z + D_{2} = 0.\end{matrix}\right.$

[D]. $\left\{\begin{matrix} \frac{A_{1}}{x} + \frac{B_{1}}{y} + \frac{C_{1}}{z} + D_{1} = 0,\\ \frac{A_{2}}{x} + \frac{B_{2}}{y} + \frac{C_{2}}{z} + D_{2} = 0.\end{matrix}\right.$

答案

$\left\{\begin{matrix} A_{\textcolor{red}{1}}\textcolor{yellow}{x} + B_{\textcolor{red}{1}}\textcolor{yellow}{y} + C_{\textcolor{red}{1}}\textcolor{yellow}{z} + D_{\textcolor{red}{1}} = 0,\\ A_{\textcolor{cyan}{2}}\textcolor{yellow}{x} + B_{\textcolor{cyan}{2}}\textcolor{yellow}{y} + C_{\textcolor{cyan}{2}}\textcolor{yellow}{z} + D_{\textcolor{cyan}{2}} = 0.\end{matrix}\right.$

空间直角坐标系下平面的法向量（B009）

问题

若平面过 $(x_{1}, y_{1}, z_{1})$, $(x_{2}, y_{2}, z_{2})$ 和 $(x_{3}, y_{3}, z_{3})$ 这三个点，则在空间直角坐标系下平面的 [法向量] $\vec{n}$ $=$ $?$

选项

[A]. $\vec{n}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\ x_{1}-x_{3} & y_{1}-y_{3} & z_{1}-z_{3} \end{vmatrix}$

[B]. $\vec{n}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_{2}+x_{1} & y_{2}+y_{1} & z_{2}+z_{1} \\ x_{3}+x_{1} & y_{3}+y_{1} & z_{3}+z_{1} \end{vmatrix}$

[C]. $\vec{n}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_{2}-x_{3} & y_{2}-y_{3} & z_{2}-z_{3} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}$

[D]. $\vec{n}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}$

答案

$\textcolor{red}{\vec{n}}$ $=$ $\begin{vmatrix} \textcolor{red}{\mathbf{i}} & \textcolor{red}{\mathbf{j}} & \textcolor{red}{\mathbf{k}} \\ x_{\textcolor{orange}{2}}-x_{\textcolor{cyan}{1}} & y_{\textcolor{orange}{2}}-y_{\textcolor{cyan}{1}} & z_{\textcolor{orange}{2}}-z_{\textcolor{cyan}{1}} \\ x_{\textcolor{yellow}{3}}-x_{\textcolor{cyan}{1}} & y_{\textcolor{yellow}{3}}-y_{\textcolor{cyan}{1}} & z_{\textcolor{yellow}{3}}-z_{\textcolor{cyan}{1}} \end{vmatrix}$

空间直角坐标系下平面方程的三点式（B009）

问题

若平面过 $(x_{1}, y_{1}, z_{1})$, $(x_{2}, y_{2}, z_{2})$ 和 $(x_{3}, y_{3}, z_{3})$ 这三个点，则空间直角坐标系下平面方程的三点式如何表示？

选项

[A]. $\begin{vmatrix} x_{1}-x & y_{1}-y & z_{1}-z \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{2} & y_{3}-y_{2} & z_{3}-z_{2} \end{vmatrix}$ $=$ $0$

[B]. $\begin{vmatrix} x+x_{1} & y+y_{1} & z+z_{1} \\ x_{2}+x_{1} & y_{2}+y_{1} & z_{2}+z_{1} \\ x_{3}+x_{1} & y_{3}+y_{1} & z_{3}+z_{1} \end{vmatrix}$ $=$ $0$

[C]. $\begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}$ $=$ $1$

[D]. $\begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}$ $=$ $0$

答案

$\begin{vmatrix} \textcolor{red}{x}-x_{\textcolor{cyan}{1}} & \textcolor{red}{y}-y_{\textcolor{cyan}{1}} & \textcolor{red}{z}-z_{\textcolor{cyan}{1}} \\ x_{\textcolor{orange}{2}}-x_{\textcolor{cyan}{1}} & y_{\textcolor{orange}{2}}-y_{\textcolor{cyan}{1}} & z_{\textcolor{orange}{2}}-z_{\textcolor{cyan}{1}} \\ x_{\textcolor{yellow}{3}}-x_{\textcolor{cyan}{1}} & y_{\textcolor{yellow}{3}}-y_{\textcolor{cyan}{1}} & z_{\textcolor{yellow}{3}}-z_{\textcolor{cyan}{1}} \end{vmatrix}$ $=$ $\textcolor{red}{0}$

空间直角坐标系下平面方程的截距式（B009）

问题

若 $a$, $b$, $c$ 分别为平面在 $x$ 轴、$y$ 轴和 $z$ 轴上的截距，则空间直角坐标系下平面方程的 [截距式] 如何表示?

选项

[A]. $\frac{x}{a}$ $+$ $\frac{y}{b}$ $+$ $\frac{z}{c}$ $=$ $0$

[B]. $\frac{x}{a}$ $-$ $\frac{y}{b}$ $-$ $\frac{z}{c}$ $=$ $1$

[C]. $\frac{x}{a}$ $+$ $\frac{y}{b}$ $+$ $\frac{z}{c}$ $=$ $1$

[D]. $\frac{a}{x}$ $+$ $\frac{b}{y}$ $+$ $\frac{c}{z}$ $=$ $1$

答案

$\frac{\textcolor{red}{x}}{\textcolor{cyan}{a}}$ $\textcolor{yellow}{+}$ $\frac{\textcolor{red}{y}}{\textcolor{cyan}{b}}$ $\textcolor{yellow}{+}$ $\frac{\textcolor{red}{z}}{\textcolor{cyan}{c}}$ $=$ $\textcolor{orange}{1}$

空间直角坐标系下平面方程的点法式（B009）

问题

若平面过点 $(x_{0}, y_{0}, z_{0})$, 且平面的法向量为 $\vec{n}$ $=$ $(A, B, C)$, 则空间直角坐标系下平面方程的 [点法式] 如何表示？

选项

[A]. $A$ $(x – x_{0})$ $+$ $B$ $(y – y_{0})$ $+$ $C$ $(z – z_{0})$ $=$ $0$

[B]. $A$ $(x – x_{0})$ $-$ $B$ $(y – y_{0})$ $-$ $C$ $(z – z_{0})$ $=$ $0$

[C]. $A$ $(x \div x_{0})$ $+$ $B$ $(y \div y_{0})$ $+$ $C$ $(z \div z_{0})$ $=$ $0$

[D]. $A$ $(x + x_{0})$ $+$ $B$ $(y + y_{0})$ $+$ $C$ $(z + z_{0})$ $=$ $0$

答案

$\textcolor{red}{A}$ $\cdot$ $(\textcolor{cyan}{x} – \textcolor{orange}{x_{0}})$ $+$ $\textcolor{red}{B}$ $\cdot$ $(\textcolor{cyan}{y} – \textcolor{orange}{y_{0}})$ $+$ $\textcolor{red}{C}$ $\cdot$ $(\textcolor{cyan}{z} – \textcolor{orange}{z_{0}})$ $=$ $0$

空间直角坐标系下平面方程的一般式（B009）

问题

若 $A$, $B$, $C$, $D$ 为常数，则空间直角坐标系下平面方程的一般式如何表示？

选项

[A]. $A x$ $+$ $B y$ $+$ $C z$ $+$ $D$ $=$ $0$

[B]. $A x$ $+$ $B y$ $+$ $C z$ $+$ $\frac{x + y +z}{D}$ $=$ $0$

[C]. $\frac{A}{x}$ $+$ $\frac{B}{y}$ $+$ $\frac{C}{z}$ $+$ $D$ $=$ $0$

[D]. $(A – x)$ $+$ $(B – y)$ $+$ $(C – z)$ $+$ $D$ $=$ $0$

答案

$\textcolor{red}{A} \textcolor{yellow}{\cdot} \textcolor{cyan}{x}$ $\textcolor{green}{+}$ $\textcolor{red}{B} \textcolor{yellow}{\cdot} \textcolor{cyan}{y}$ $\textcolor{green}{+}$ $\textcolor{red}{C} \textcolor{yellow}{\cdot} \textcolor{cyan}{z}$ $\textcolor{green}{+}$ $\textcolor{red}{D}$ $=$ $0$

向量的混合积（B008）

问题

若有三个向量 $\alpha$ $=$ $(x_{1}, y_{1}, z_{1})$, $\beta$ $=$ $(x_{2}, y_{2}, z_{2})$, $\gamma$ $=$ $(x_{3}, y_{3}, z_{3})$, 且 $\times$ 表示向量的向量积，$\cdot$ 表示向量的数量积，则混合积 $($ $\alpha$ $\times$ $\beta$ $)$ $\cdot$ $\gamma$ $=$ $?$

选项

[A]. $($ $\alpha$ $\times$ $\beta$ $)$ $\cdot$ $\gamma$ $=$ $\begin{vmatrix} x_{2} & y_{2} & z_{2} \\ x_{1} & y_{1} & z_{1} \\ x_{3} & y_{3} & z_{3} \end{vmatrix}$

[B]. $($ $\alpha$ $\times$ $\beta$ $)$ $\cdot$ $\gamma$ $=$ $\begin{vmatrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{vmatrix}$

[C]. $($ $\alpha$ $\times$ $\beta$ $)$ $\cdot$ $\gamma$ $=$ $\begin{vmatrix} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{vmatrix}$

[D]. $($ $\alpha$ $\times$ $\beta$ $)$ $\cdot$ $\gamma$ $=$ $\begin{vmatrix} x_{1} & y_{1} & z_{1} \\ x_{3} & y_{3} & z_{3} \\ x_{1} & y_{1} & z_{1} \end{vmatrix}$

答案

$($ $\alpha$ $\textcolor{red}{\times}$ $\beta$ $)$ $\textcolor{red}{\cdot}$ $\gamma$ $=$ $\begin{vmatrix} x_{\textcolor{yellow}{1}} & y_{\textcolor{yellow}{1}} & z_{\textcolor{yellow}{1}} \\ x_{\textcolor{orange}{2}} & y_{\textcolor{orange}{2}} & z_{\textcolor{orange}{2}} \\ x_{\textcolor{cyan}{3}} & y_{\textcolor{cyan}{3}} & z_{\textcolor{cyan}{3}} \end{vmatrix}$

注意：$($ $\alpha$ $\times$ $\beta$ $)$ $\cdot$ $\gamma$ 也可以记作 $[$ $\alpha$, $\beta$, $\gamma$ $]$.