荒原之梦 - 第102页共242页 - 专注于考研数学|高等数学|线性代数|概率统计

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $[$ $\alpha$, $\beta$, $\gamma$ $]$.

三维向量的向量积运算公式（B008）

问题

若向量 $\vec{a}$ $=$ $(x_{1}, y_{1}, z_{1})$, $\vec{b}$ $=$ $(x_{2}, y_{2}, z_{2})$, $\vec{c}$ $=$ $(x_{3}, y_{3}, z_{3})$, 向量 $\mathbf{\vec{i}}$, $\mathbf{\vec{j}}$, $\mathbf{\vec{k}}$ 分别是 $x$ 轴、$y$ 轴和 $z$ 轴上的单位向量，则向量积 $\vec{a}$ $\times$ $\vec{b}$ $=$ $?$

选项

[A]. $\vec{a}$ $\times$ $\vec{b}$ $=$ $\begin{vmatrix} y_{1} & z_{1} \\ y_{2} & z_{2} \end{vmatrix} \mathbf{i}$ $+$ $\begin{vmatrix} x_{1} & z_{1} \\ x_{2} & z_{2} \end{vmatrix} \mathbf{j}$ $+$ $\begin{vmatrix} x_{1} & y_{1} \\ x_{2} & y_{2} \end{vmatrix} \mathbf{k}$

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

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

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

答案

$\textcolor{red}{\vec{a}}$ $\textcolor{green}{\times}$ $\textcolor{red}{\vec{b}}$ $=$

$\begin{vmatrix}\mathbf{\textcolor{orange}{i}} & \mathbf{\textcolor{orange}{j}} & \mathbf{\textcolor{orange}{k}} \\ x_{\textcolor{cyan}{1}} & y_{\textcolor{cyan}{1}} & z_{\textcolor{cyan}{1}} \\ x_{\textcolor{blue}{2}} & y_{\textcolor{blue}{2}} & z_{\textcolor{blue}{2}} \end{vmatrix}$ $=$

$\begin{vmatrix} y_{\textcolor{cyan}{1}} & z_{\textcolor{cyan}{1}} \\ y_{\textcolor{blue}{2}} & z_{\textcolor{blue}{2}} \end{vmatrix} \mathbf{\textcolor{orange}{i}}$ $\textcolor{green}{-}$ $\begin{vmatrix} x_{\textcolor{cyan}{1}} & z_{\textcolor{cyan}{1}} \\ x_{\textcolor{blue}{2}} & z_{\textcolor{blue}{2}} \end{vmatrix} \mathbf{\textcolor{orange}{j}}$ $\textcolor{green}{+}$ $\begin{vmatrix} x_{\textcolor{cyan}{1}} & y_{\textcolor{cyan}{1}} \\ x_{\textcolor{blue}{2}} & y_{\textcolor{blue}{2}} \end{vmatrix} \mathbf{\textcolor{orange}{k}}$

二维向量的向量积运算公式（B008）

问题

若向量 $\vec{a}$ $=$ $(x_{1}, y_{1})$, 向量 $\vec{b}$ $=$ $(x_{2}, y_{2})$, 向量 $\vec{i}$ 是 $x$ 轴上的单位向量，向量 $\vec{j}$ 是 $y$ 轴上的单位向量，则向量积 $\vec{a}$ $\times$ $\vec{b}$ $=$ $?$

选项

[A]. $\vec{a}$ $\times$ $\vec{b}$ $=$ $\frac{(x_{1} \cdot x_{2})}{\mathbf{i}}$ $+$ $\frac{(y_{1} \cdot y_{2})}{\mathbf{j}}$

[B]. $\vec{a}$ $\times$ $\vec{b}$ $=$ $(x_{1} \cdot x_{2}) + \mathbf{i}$ $+$ $(y_{1} \cdot y_{2}) + \mathbf{j}$

[C]. $\vec{a}$ $\times$ $\vec{b}$ $=$ $(x_{1} \cdot x_{2}) \mathbf{i}$ $+$ $(y_{1} \cdot y_{2}) \mathbf{j}$

[D]. $\vec{a}$ $\times$ $\vec{b}$ $=$ $(x_{1} \div x_{2}) \mathbf{i}$ $+$ $(y_{1} \div y_{2}) \mathbf{j}$

答案

$\textcolor{red}{\vec{a}}$ $\textcolor{green}{\times}$ $\textcolor{red}{\vec{b}}$ $=$ $(\textcolor{orange}{x}_{\textcolor{cyan}{1}} \textcolor{green}{\cdot} \textcolor{orange}{x}_{\textcolor{cyan}{2}}) \mathbf{\textcolor{yellow}{i}}$ $\textcolor{green}{+}$ $(\textcolor{orange}{y}_{\textcolor{cyan}{1}} \textcolor{green}{\cdot} \textcolor{orange}{y}_{\textcolor{cyan}{2}}) \mathbf{\textcolor{yellow}{j}}$

什么是向量积/叉积/外积？（B008）

问题

已知 $\vec{a}$, $\vec{b}$ 和 $\vec{c}$ 为三个向量，$\theta$ 为向量 $\vec{a}$ 与向量 $\vec{b}$ 的夹角，则以下哪些条件可以说明向量 $\vec{c}$ 为向量 $\vec{a}$ 和向量 $\vec{b}$ 的向量积？（多选）

选项

[A]. $\vec{a}$, $\vec{b}$, $\vec{c}$ 成左手系

[B]. $|\vec{c}|$ $=$ $|\vec{a}| \cdot |\vec{b}|$

[C]. $\vec{a}$, $\vec{b}$, $\vec{c}$ 成右手系

[D]. $\vec{c}$ $\perp$ $\vec{a}$, $\vec{c}$ $\perp$ $\vec{b}$

[E]. $\vec{c}$ $\perp$ $\vec{a}$, $\vec{a}$ $\perp$ $\vec{b}$

[F]. $|\vec{c}|$ $=$ $|\vec{a}| \cdot |\vec{b}| \cdot \sin \theta$

答案

若以下三个条件全部满足：

Ⅰ. $|\vec{c}|$ $=$ $|\vec{a}| \cdot |\vec{b}| \cdot \sin \theta$;
Ⅱ. $\vec{c}$ $\perp$ $\vec{a}$, $\vec{c}$ $\perp$ $\vec{b}$（即 $\vec{c}$ 垂直于 $\vec{a}$ 与 $\vec{b}$ 所形成的平面）；
Ⅲ. $\vec{a}$, $\vec{b}$, $\vec{c}$ 成右手系.

则称向量 $\vec{c}$ 为向量 $\vec{a}$ 和向量 $\vec{b}$ 的向量积.

向量的数量积/点积/内积（B008）

问题

已知向量 $\vec{a}$ $=$ $(x_{1}, y_{1}, z_{1})$, 向量 $\vec{b}$ $=$ $(x_{2}, y_{2}, z_{2})$, 且向量 $\vec{a}$ 与向量 $\vec{b}$ 之间的夹角为 $\theta$, 则向量 $\vec{a}$ 与向量 $\vec{b}$ 的数量积（数量积也被称为“点积”或者“内积”）$\vec{a}$ $\cdot$ $\vec{b}$ $=$ $?$

选项

[A]. $\vec{a}$ $\cdot$ $\vec{b}$ $=$ $(x_{1} + x_{2})$ $\cdot$ $(y_{1} + y_{2})$ $\cdot$ $(z_{1} + z_{2})$

[B]. $\vec{a}$ $\cdot$ $\vec{b}$ $=$ $x_{1} x_{2}$ $-$ $y_{1} y_{2}$ $-$ $z_{1} z_{2}$

[C]. $\vec{a}$ $\cdot$ $\vec{b}$ $=$ $x_{1} x_{2}$ $+$ $y_{1} y_{2}$ $+$ $z_{1} z_{2}$

[D]. $\vec{a}$ $\cdot$ $\vec{b}$ $=$ $x_{1} x_{2}$ $\times$ $y_{1} y_{2}$ $\times$ $z_{1} z_{2}$

答案

$\textcolor{red}{\vec{a}}$ $\textcolor{green}{\cdot}$ $\textcolor{red}{\vec{b}}$ $=$ $\textcolor{orange}{|\vec{a}|}$ $\textcolor{green}{\cdot}$ $\textcolor{orange}{|\vec{b}|}$ $\textcolor{green}{\cdot}$ $\textcolor{orange}{\cos \theta}$

$\textcolor{red}{\vec{a}}$ $\textcolor{green}{\cdot}$ $\textcolor{red}{\vec{b}}$ $=$ $\textcolor{orange}{x}_{\textcolor{cyan}{1}} \textcolor{orange}{x}_{\textcolor{cyan}{2}}$ $\textcolor{green}{+}$ $\textcolor{orange}{y}_{\textcolor{cyan}{1}} \textcolor{orange}{y}_{\textcolor{cyan}{2}}$ $\textcolor{green}{+}$ $\textcolor{orange}{z}_{\textcolor{cyan}{1}} \textcolor{orange}{z}_{\textcolor{cyan}{2}}$

向量的数乘运算（B008）

问题

若向量 $\vec{a}$ $=$ $(x_{1}, y_{2}, z_{1})$, $\lambda$ 为常数，则向量 $\vec{a}$ 的数乘 $\lambda \vec{a}$ $=$ $?$

选项

[A]. $\lambda \vec{a}$ $=$ $($ $\lambda x$, $\lambda y$, $\lambda z$ $)$

[B]. $\lambda \vec{a}$ $=$ $($ $\frac{1}{\lambda} x$, $\frac{1}{\lambda} y$, $\frac{1}{\lambda} z$ $)$

[C]. $\lambda \vec{a}$ $=$ $($ $\lambda x$, $y$, $z$ $)$

[D]. $\lambda \vec{a}$ $=$ $($ $\lambda + x$, $\lambda + y$, $\lambda + z$ $)$

答案

$\textcolor{cyan}{\lambda}$ $\textcolor{green}{\cdot}$ $\textcolor{red}{\vec{a}}$ $=$ $($ $\textcolor{cyan}{\lambda} \textcolor{green}{\cdot} \textcolor{orange}{x}$, $\textcolor{cyan}{\lambda} \textcolor{green}{\cdot} \textcolor{orange}{y}$, $\textcolor{cyan}{\lambda} \textcolor{green}{\cdot} \textcolor{orange}{z}$ $)$

向量的减法运算法则（B008）

问题

若向量 $\vec{a}$ $=$ $(x_{1}, y_{1}, z_{1})$, 向量 $\vec{b}$ $=$ $(x_{2}, y_{2}, z_{2})$, 则 $\vec{a}$ $-$ $\vec{b}$ $=$ $?$

选项

[A]. $\vec{a}$ $-$ $\vec{b}$ $=$ $($ $x_{2} + x_{1}$, $y_{2} + y_{1}$, $z_{2} + z_{1}$ $)$

[B]. $\vec{a}$ $-$ $\vec{b}$ $=$ $($ $x_{1} + x_{2}$, $y_{1} + y_{2}$, $z_{1} + z_{2}$ $)$

[C]. $\vec{a}$ $-$ $\vec{b}$ $=$ $($ $x_{1} – x_{2}$, $y_{1} – y_{2}$, $z_{1} – z_{2}$ $)$

[D]. $\vec{a}$ $-$ $\vec{b}$ $=$ $($ $\frac{x_{1}}{x_{2}}$, $\frac{y_{1}}{y_{2}}$, $\frac{z_{1}}{z_{2}}$ $)$

答案

$\textcolor{red}{\vec{a}}$ $\textcolor{green}{-}$ $\textcolor{red}{\vec{b}}$ $=$ $($ $\textcolor{orange}{x}_{\textcolor{cyan}{1}} \textcolor{green}{-} \textcolor{orange}{x}_{\textcolor{cyan}{2}}$, $\textcolor{orange}{y}_{\textcolor{cyan}{1}} \textcolor{green}{-} \textcolor{orange}{y}_{\textcolor{cyan}{2}}$, $\textcolor{orange}{z}_{\textcolor{cyan}{1}} \textcolor{green}{-} \textcolor{orange}{z}_{\textcolor{cyan}{2}}$ $)$