《2025年教材帮高中数学选择性必修第二册苏教版》

例11 (2025·河北省唐山市第十一中学月考)已知 O 是坐标原点,且 A,B,C 三点的坐标分别是$(2,-1,2),(4,5,-1),(-2,2,3)$,求适合下列条件的点 P 的坐标:
(1)$\overrightarrow {OP}=\frac {1}{2}(\overrightarrow {AB}-\overrightarrow {AC})$;(2)$\overrightarrow {AP}=\frac {1}{2}(\overrightarrow {AB}-\overrightarrow {AC}).$

答案： 解析▶$\overrightarrow {AB}=(2,6,-3),\overrightarrow {AC}=(-4,3,1).$
(1)$\overrightarrow {OP}=\frac {1}{2}(\overrightarrow {AB}-\overrightarrow {AC})=\frac {1}{2}(6,3,-4)=(3,\frac {3}{2},-2),$则点 P 的坐标为$(3,\frac {3}{2},-2).$
(2)设$P(x,y,z)$,则$\overrightarrow {AP}=(x-2,y+1,z-2).$
$\because \overrightarrow {AP}=\frac {1}{2}(\overrightarrow {AB}-\overrightarrow {AC})=(3,\frac {3}{2},-2),$
$\therefore \left\{\begin{array}{l} x-2=3,\\ y+1=\frac {3}{2},\\ z-2=-2,\end{array}\right. $解得$\left\{\begin{array}{l} x=5,\\ y=\frac {1}{2},\\ z=0,\end{array}\right. $
则点 P 的坐标为$(5,\frac {1}{2},0).$

3. (1)若向量$\boldsymbol{a}=(1,1,x),\boldsymbol{b}=(1,2,1),\boldsymbol{c}=(1,1,$1)满足条件$(\boldsymbol{c}-\boldsymbol{a})· (2\boldsymbol{b})=-2$,则$x=$2.
(2)已知空间四点$A(4,1,3),B(2,3,1),C(3,7,$$-5),D(x,-1,3)$共面,则 x 的值为11.

答案： 3.
(1)$2c - a = (0,0,1 - x)$，$(c - a)·(2b) = 2(0,0,1 - x)·(1,2,1) = 2(1 - x) = -2$，解得$x = 2$。
(2)$\overrightarrow{AB} = (-2,2,-2)$，$\overrightarrow{AC} = (-1,6,-8)$，$\overrightarrow{AD} = (x - 4,-2,0)$。
$\because A$,$B$,$C$,$D$共面，$\therefore\overrightarrow{AB}$,$\overrightarrow{AC}$,$\overrightarrow{AD}$共面，
$\therefore$存在实数$\lambda$,$\mu$，使$\overrightarrow{AD} = \lambda\overrightarrow{AB} + \mu\overrightarrow{AC}$，
即$(x - 4,-2,0) = (-2\lambda - \mu,2\lambda + 6\mu,-2\lambda - 8\mu)$，
$\begin{cases}x - 4 = -2\lambda - \mu,\\-2 = 2\lambda + 6\mu,\\0 = -2\lambda - 8\mu,\end{cases}$解得$\begin{cases}\lambda = -4,\\\mu = 1,\\x = 11.\end{cases}$

例12 已知向量$\boldsymbol{a}=(1,2,\frac {1}{2}),\boldsymbol{b}=(\frac {1}{2},-\frac {1}{2},$$1),\boldsymbol{c}=(-2,3,-\frac {1}{2}),\boldsymbol{d}=(1,-\frac {3}{2},\frac {1}{4})$. 求证:$\boldsymbol{a}⊥\boldsymbol{b},\boldsymbol{c}// \boldsymbol{d}.$

答案： 解析▶$\because \boldsymbol{a}=(1,2,\frac {1}{2}),\boldsymbol{b}=(\frac {1}{2},-\frac {1}{2},1),$
$\therefore \boldsymbol{a}· \boldsymbol{b}=1×\frac {1}{2}+2×(-\frac {1}{2})+\frac {1}{2}×1=0,\therefore \boldsymbol{a}⊥\boldsymbol{b}.$
（证明$\boldsymbol{a}⊥\boldsymbol{b}$,就是要证明$\boldsymbol{a}· \boldsymbol{b}=0$）
$\because \boldsymbol{c}=(-2,3,-\frac {1}{2}),\boldsymbol{d}=(1,-\frac {3}{2},\frac {1}{4}),$
$\therefore \boldsymbol{c}=-2(1,-\frac {3}{2},\frac {1}{4})=-2\boldsymbol{d},\therefore \boldsymbol{c}// \boldsymbol{d}.$
（证明$\boldsymbol{c}// \boldsymbol{d}$,需证明$\boldsymbol{c}=λ\boldsymbol{d}$）

例13 已知$\boldsymbol{a}=(1,5,-1),\boldsymbol{b}=(-2,3,5).$
(1)$(k\boldsymbol{a}+\boldsymbol{b})// (\boldsymbol{a}-3\boldsymbol{b})$,求 k 的值;
(2)$(k\boldsymbol{a}+\boldsymbol{b})⊥(\boldsymbol{a}-3\boldsymbol{b})$,求 k 的值.

答案： 解析▶$k\boldsymbol{a}+\boldsymbol{b}=(k-2,5k+3,-k+5),$
$\boldsymbol{a}-3\boldsymbol{b}=(7,-4,-16).$
(1)若$(k\boldsymbol{a}+\boldsymbol{b})// (\boldsymbol{a}-3\boldsymbol{b}),$
则$\frac {k-2}{7}=\frac {5k+3}{-4}=\frac {-k+5}{-16}$,解得$k=-\frac {1}{3}.$
（只有$\boldsymbol{a}-3\boldsymbol{b}$坐标中不含0时才可使用此公式）
(2)若$(k\boldsymbol{a}+\boldsymbol{b})⊥(\boldsymbol{a}-3\boldsymbol{b})$,则$(k-2)×7+(5k+$$3)×(-4)+(-k+5)×(-16)=0$,解得$k=\frac {106}{3}.$
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