《2025年新坐标同步练习高中数学A版选择性必修第一册人教版青海专用》

思考2
设平面向量$\boldsymbol{a}= (x_{1},y_{1})$,$\boldsymbol{b}= (x_{2},y_{2})$,则$\boldsymbol{a}+\boldsymbol{b}$, $\boldsymbol{a}-\boldsymbol{b}$,$\lambda\boldsymbol{a}$,$\boldsymbol{a}\cdot\boldsymbol{b}$的运算结果分别是什么？
$\boldsymbol{a}+\boldsymbol{b}=(x_{1}+x_{2},y_{1}+y_{2})$，$\boldsymbol{a}-\boldsymbol{b}=(x_{1}-x_{2},y_{1}-y_{2})$，$\lambda\boldsymbol{a}=(\lambda x_{1},\lambda y_{1})$，$\boldsymbol{a}\cdot\boldsymbol{b}=x_{1}x_{2}+y_{1}y_{2}$

答案： $a+b=(x_{1}+x_{2},y_{1}+y_{2}),a - b=(x_{1}-x_{2},y_{1}-y_{2}),λa=(λx_{1},λy_{1}),a\cdot b=x_{1}x_{2}+y_{1}y_{2}.$

思考3
有了空间向量的坐标表示,你能类比平面向量的坐标运算,得出空间向量运算的坐标表示吗？
设$a=(a_{1},a_{2},a_{3}),b=(b_{1},b_{2},b_{3})$，与平面向量运算的坐标表示一样，有$a + b=(a_{1}+b_{1},a_{2}+b_{2},a_{3}+b_{3}),a - b=(a_{1}-b_{1},a_{2}-b_{2},a_{3}-b_{3}),λa=(λa_{1},λa_{2},λa_{3}),λ∈R,a\cdot b=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}.$

答案： 设$a=(a_{1},a_{2},a_{3}),b=(b_{1},b_{2},b_{3})$，与平面向量运算的坐标表示一样，有$a + b=(a_{1}+b_{1},a_{2}+b_{2},a_{3}+b_{3}),a - b=(a_{1}-b_{1},a_{2}-b_{2},a_{3}-b_{3}),λa=(λa_{1},λa_{2},λa_{3}),λ∈R,a\cdot b=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}.$

1. 若$\boldsymbol{a}= (a_{1},a_{2},a_{3})$,$\boldsymbol{b}= (b_{1},b_{2},b_{3})$,则$\boldsymbol{a}+\boldsymbol{b}=$$(a_{1}+b_{1},a_{2}+b_{2},a_{3}+b_{3})$，$\boldsymbol{a}-\boldsymbol{b}=$$(a_{1}-b_{1},a_{2}-b_{2},a_{3}-b_{3})$，$\lambda\boldsymbol{a}=$$(λa_{1},λa_{2},λa_{3})$，$\boldsymbol{a}\cdot\boldsymbol{b}=$$a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$。

答案： 1.$(a_{1}+b_{1},a_{2}+b_{2},a_{3}+b_{3})$;2.$(a_{1}-b_{1},a_{2}-b_{2},a_{3}-b_{3})$;3.$(λa_{1},λa_{2},λa_{3})$;4.$a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$

2. 设$A(x_{1},y_{1},z_{1})$,$B(x_{2},y_{2},z_{2})$,则$\overrightarrow{AB}= ($$x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}$$)$。即一个空间向量的坐标等于表示此向量的有向线段的终点坐标减去起点坐标。

答案： 5.$x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}$;6.减去

1. (2025·福州期中)已知点$A(1,2,0)$,若向量$\overrightarrow{AB}= (2,3,-1)$,则点$B$的坐标是(D)
A.$(-1,-1,1)$
B.$(1,1,-1)$
C.$(-3,-5,1)$
D.$(3,5,-1)$

答案： D

2. 已知$\boldsymbol{a}= (-1,2,1)$,$\boldsymbol{b}= (2,0,1)$,则$(2\boldsymbol{a}+3\boldsymbol{b})\cdot(\boldsymbol{a}-\boldsymbol{b})= $$-4$。

答案： $-4$

3. (2025·北京期中)在空间直角坐标系中,已知点$A(1,m,1)$,$B(-1,1,2)$,$C(3,-2,1)$,$D(1,-3,2)$,若$A$,$B$,$C$,$D$四点共面,则$m= $2

答案： 2

二 空间向量平行、垂直的坐标表示及应用
思考
设平面向量$\boldsymbol{a}= (x_{1},y_{1})$,$\boldsymbol{b}= (x_{2},y_{2})$,则$\boldsymbol{a}//\boldsymbol{b}$与$\boldsymbol{a}\perp\boldsymbol{b}$的充要条件分别是什么？对于空间向量是不是也有类似结论？
$a// b\Leftrightarrow x_{1}y_{2}-x_{2}y_{1}=0;a\perp b\Leftrightarrow x_{1}x_{2}+y_{1}y_{2}=0$.对于空间向量也有类似结论.

答案： $a// b\Leftrightarrow x_{1}y_{2}-x_{2}y_{1}=0;a\perp b\Leftrightarrow x_{1}x_{2}+y_{1}y_{2}=0$.对于空间向量也有类似结论.

若$\boldsymbol{a}= (a_{1},a_{2},a_{3})$,$\boldsymbol{b}= (b_{1},b_{2},b_{3})$,则
1. $\boldsymbol{a}$与$\boldsymbol{b}$共线的充要条件是：存在实数$\lambda$，使得$a = λb$；
2. 用坐标表示为：$a_{1}=λb_{1},a_{2}=λb_{2},a_{3}=λb_{3}$；
3. $\boldsymbol{a}$与$\boldsymbol{b}$垂直的充要条件是：$a\cdot b = 0$；
4. 用坐标表示为：$a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}=0$。

答案： 1.$a = λb$;2.$a_{1}=λb_{1},a_{2}=λb_{2},a_{3}=λb_{3}$;3.$a\cdot b = 0$;4.$a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}=0$

角度1 证平行与垂直
[例1] (对接教材例2)在正方体$ABCD - A_{1}B_{1}C_{1}D_{1}$中,已知$E$,$F$,$G$,$H分别是CC_{1}$,$BC$,$CD$,$A_{1}C_{1}$的中点。求证：
(1)$AB_{1}// GE$,$AB_{1}\perp EH$；
(2)$A_{1}G\perp平面EFD$。
证明：如图，以A为坐标原点，以$\{\overrightarrow {AB},\overrightarrow {AD},\overrightarrow {AA_{1}}\}$为正交基底建立空间直角坐标系.设正方体的棱长为1，则$A(0,0,0),B(1,0,0),C(1,1,0),D(0,1,0),A_{1}(0,0,1),B_{1}(1,0,1),C_{1}(1,1,1).$由中点坐标公式，得$E(1,1,\frac{1}{2}),F(1,\frac{1}{2},0),G(\frac{1}{2},1,0),H(\frac{1}{2},\frac{1}{2},1).$
(1)$\overrightarrow {AB_{1}}=(1,0,1),\overrightarrow {GE}=(\frac{1}{2},0,\frac{1}{2}),\overrightarrow {EH}=(-\frac{1}{2},-\frac{1}{2},\frac{1}{2}).$因为$\overrightarrow {AB_{1}} = 2\overrightarrow {GE},\overrightarrow {AB_{1}}\cdot \overrightarrow {EH}=1×(-\frac{1}{2})+0 + 1×\frac{1}{2}=0$，所以$\overrightarrow {AB_{1}}// \overrightarrow {GE},\overrightarrow {AB_{1}}\perp\overrightarrow {EH}$，即$AB_{1}// GE,AB_{1}\perp EH.$
(2)$\overrightarrow {A_{1}G}=(\frac{1}{2},1,-1),\overrightarrow {DF}=(1,-\frac{1}{2},0),\overrightarrow {DE}=(1,0,\frac{1}{2}).$因为$\overrightarrow {A_{1}G}\cdot \overrightarrow {DF}=\frac{1}{2}-\frac{1}{2}+0 = 0,\overrightarrow {A_{1}G}\cdot\overrightarrow {DE}=\frac{1}{2}+0-\frac{1}{2}=0$，所以$\overrightarrow {A_{1}G}\perp\overrightarrow {DF},\overrightarrow {A_{1}G}\perp\overrightarrow {DE}$，即$A_{1}G\perp DF,A_{1}G\perp DE.$因为$DF\cap DE = D,DF,DE\subset$平面EFD，所以$A_{1}G\perp$平面EFD.

答案： 证明：如图，以A为坐标原点，以$\{\overrightarrow {AB},\overrightarrow {AD},\overrightarrow {AA_{1}}\}$为正交基底建立空间直角坐标系.设正方体的棱长为1，则$A(0,0,0),B(1,0,0),C(1,1,0),D(0,1,0),A_{1}(0,0,1),B_{1}(1,0,1),C_{1}(1,1,1).$由中点坐标公式，得$E(1,1,\frac{1}{2}),F(1,\frac{1}{2},0),G(\frac{1}{2},1,0),H(\frac{1}{2},\frac{1}{2},1).$
(1)$\overrightarrow {AB_{1}}=(1,0,1),\overrightarrow {GE}=(\frac{1}{2},0,\frac{1}{2}),\overrightarrow {EH}=(-\frac{1}{2},-\frac{1}{2},\frac{1}{2}).$因为$\overrightarrow {AB_{1}} = 2\overrightarrow {GE},\overrightarrow {AB_{1}}\cdot \overrightarrow {EH}=1×(-\frac{1}{2})+0 + 1×\frac{1}{2}=0$，所以$\overrightarrow {AB_{1}}// \overrightarrow {GE},\overrightarrow {AB_{1}}\perp\overrightarrow {EH}$，即$AB_{1}// GE,AB_{1}\perp EH.$
(2)$\overrightarrow {A_{1}G}=(\frac{1}{2},1,-1),\overrightarrow {DF}=(1,-\frac{1}{2},0),\overrightarrow {DE}=(1,0,\frac{1}{2}).$因为$\overrightarrow {A_{1}G}\cdot \overrightarrow {DF}=\frac{1}{2}-\frac{1}{2}+0 = 0,\overrightarrow {A_{1}G}\cdot\overrightarrow {DE}=\frac{1}{2}+0-\frac{1}{2}=0$，所以$\overrightarrow {A_{1}G}\perp\overrightarrow {DF},\overrightarrow {A_{1}G}\perp\overrightarrow {DE}$，即$A_{1}G\perp DF,A_{1}G\perp DE.$因为$DF\cap DE = D,DF,DE\subset$平面EFD，所以$A_{1}G\perp$平面EFD.