《2025年暑假生活北京师范大学出版社高二数学人教版》

1. 空间向量的正交分解及其坐标表示
设$e_{1},e_{2},e_{3}为有公共起点O$的三个两两垂直的单位向量(我们称它们为单位正交基底).
以$e_{1},e_{2},e_{3}的公共起点O$为原点,分别以$\boldsymbol{e}_{1},\boldsymbol{e}_{2},\boldsymbol{e}_{3}$的方向为$x$轴、$y$轴、$z轴的正方向建立空间直角坐标系Oxyz$.
对于空间任意一个向量$p$,一定可以把它平移,使它的起点与原点$O$重合,得到向量$\overrightarrow {OP}= p$,由空间向量基本定理可知,存在唯一的有序实数组$(x,y,z)$,使得$p= $$x\boldsymbol{e}_{1}+y\boldsymbol{e}_{2}+z\boldsymbol{e}_{3}$. 我们把$(x,y,z)$称作向量$p在单位正交基底\{ e_{1},e_{2},e_{3}\} $下的坐标,记作$p= $$(x,y,z)$.

答案： $\boldsymbol{e}_{1},\boldsymbol{e}_{2},\boldsymbol{e}_{3}$ 起点 $x\boldsymbol{e}_{1}+y\boldsymbol{e}_{2}+z\boldsymbol{e}_{3}$ $(x,y,z)$ $(x,y,z)$

2. 空间向量运算的坐标表示
(1)空间向量的线性运算及数量积的坐标表示:
设$a= (a_{1},a_{2},a_{3}),b= (b_{1},b_{2},b_{3})$,则
①$a+b= (a_{1}+b_{1},a_{2}+b_{2},a_{3}+b_{3})$;
②$a-b= (a_{1}-b_{1},a_{2}-b_{2},a_{3}-b_{3})$;
③$\lambda a= (\lambda a_{1},\lambda a_{2},\lambda a_{3})(\lambda ∈R)$;
④$a\cdot b= $$a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$.
(2)向量平行、垂直,向量的模、夹角的坐标表示:
设$a= (a_{1},a_{2},a_{3}),b= (b_{1},b_{2},b_{3})$,则
①若$a// b(b≠0)$,则$\left\{\begin{array}{l} a_{1}= \lambda b_{1},\\ a_{2}= \lambda b_{2},(\lambda ∈R);\\ a_{3}= \lambda b_{3}\end{array} \right. $
②若$a⊥b$,则$a\cdot b= a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}= 0$;
③$|a|= \sqrt {a\cdot a}= $$\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}$;
④$cos\langle a,b\rangle =\frac {a\cdot b}{|a||b|}= \frac {a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}}{\sqrt {a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}\cdot \sqrt {b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}.$

答案： （1）④$a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$ （2）③$\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}$

3. 向量的坐标及两点间的距离公式
设$A(x_{1},y_{1},z_{1}),B(x_{2},y_{2},z_{2})$,则$\overrightarrow {AB}=$$(x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1})$,$d_{AB}= |\overrightarrow {AB}|=$$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$.

答案： $(x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1})$ $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

【典例1】如图2-1,在棱长为1的正方体$ABCD-A'B'C'D'$中,$E,F,G分别为棱DD',D'C',BC$的中点,以$\{ \overrightarrow {AB},\overrightarrow {AD},\overrightarrow {AA'}\} $为基底,求下列向量的坐标.

(1)$\overrightarrow {AE},\overrightarrow {AG},\overrightarrow {AF}$;
(2)$\overrightarrow {EF},\overrightarrow {EG},\overrightarrow {DG}$.

答案： 解:
(1)$\overrightarrow {AE}= \overrightarrow {AD}+\overrightarrow {DE}= \overrightarrow {AD}+\frac {1}{2}\overrightarrow {DD'}= \overrightarrow {AD}+\frac {1}{2}\overrightarrow {AA'}= (0,1,\frac {1}{2}).$
$\overrightarrow {AG}= \overrightarrow {AB}+\overrightarrow {BG}= \overrightarrow {AB}+\frac {1}{2}\overrightarrow {AD}= (1,\frac {1}{2},0).$
$\overrightarrow {AF}= \overrightarrow {AA'}+\overrightarrow {A'D'}+\overrightarrow {D'F}= \frac {1}{2}\overrightarrow {AB}+\overrightarrow {AD}+\overrightarrow {AA'}= (\frac {1}{2},1,1).$
(2)$\overrightarrow {EF}= \overrightarrow {AF}-\overrightarrow {AE}= \overrightarrow {AA'}+\overrightarrow {AD}+\frac {1}{2}\overrightarrow {AB}-(\overrightarrow {AD}+\frac {1}{2}\overrightarrow {AA'})= \frac {1}{2}\overrightarrow {AB}+\frac {1}{2}\overrightarrow {AA'}= (\frac {1}{2},0,\frac {1}{2}).$
$\overrightarrow {EG}= \overrightarrow {AG}-\overrightarrow {AE}= \overrightarrow {AB}+\frac {1}{2}\overrightarrow {AD}-(\overrightarrow {AD}+\frac {1}{2}\overrightarrow {AA'})= \overrightarrow {AB}-\frac {1}{2}\overrightarrow {AD}-\frac {1}{2}\overrightarrow {AA'}= (1,-\frac {1}{2},-\frac {1}{2}).$
$\overrightarrow {DG}= \overrightarrow {AG}-\overrightarrow {AD}= \overrightarrow {AB}+\frac {1}{2}\overrightarrow {AD}-\overrightarrow {AD}= \overrightarrow {AB}-\frac {1}{2}\overrightarrow {AD}= (1,-\frac {1}{2},0).$