《2025年成才之路高中新课程学习指导高中数学必修第二册北师大版》

例3. 设两个非零向量$\boldsymbol{a}$与$\boldsymbol{b}$不共线，
(1)若$\overrightarrow{AB} = \boldsymbol{a} + \boldsymbol{b}$,$\overrightarrow{BC} = 2\boldsymbol{a} + 8\boldsymbol{b}$,$\overrightarrow{CD} = 3(\boldsymbol{a} - \boldsymbol{b})$,求证:$A$、$B$、$D$三点共线；
(2)试确定实数$k$,使$k\boldsymbol{a} + \boldsymbol{b}$与$\boldsymbol{a} + k\boldsymbol{b}$共线.
【分析】 (1)欲证三点$A$、$B$、$D$共线，即证存在实数$\lambda$,使$\overrightarrow{AB} = \lambda\overrightarrow{BD}$,只要由已知条件找出$\lambda$即可.
(2)由两向量共线，列出关于$a$、$b$的等式，再由$\boldsymbol{a}$与$\boldsymbol{b}$不共线知，若$\lambda\boldsymbol{a} = \mu\boldsymbol{b}$,则$\lambda = \mu = 0$.
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答案： 例3:证明$: (1)\because\overrightarrow{AB} = \boldsymbol{a} + \boldsymbol{b},\overrightarrow{BC} = 2\boldsymbol{a} + 8\boldsymbol{b},$
$\therefore\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD} = 2\boldsymbol{a} + 8\boldsymbol{b} + 3(\boldsymbol{a} - \boldsymbol{b})$
$ = 2\boldsymbol{a} + 8\boldsymbol{b} + 3\boldsymbol{a} - 3\boldsymbol{b} = 5(\boldsymbol{a} + \boldsymbol{b}) = 5\overrightarrow{AB}.$
$\therefore\overrightarrow{AB}、$$\overrightarrow{BD}$共线,
又$\because$它们有公共点$B,\therefore A、$B、D三点共线.
$(2)\because ka + b$与a + kb共线,
$\therefore$存在实数$\lambda,$使$ka + b = \lambda(a + kb),$
即$ka + b = \lambda a + \lambda kb,\therefore (k - \lambda)a = (\lambda k - 1)b,$
$\because a、$b是不共线的两个非零向量,
$\therefore k - \lambda = \lambda k - 1 = 0,\therefore k^2 - 1 = 0.\therefore k = \pm 1.$

对点训练3
已知向量$\overrightarrow{AB} = \boldsymbol{a} + 5\boldsymbol{b}$,$\overrightarrow{BC} = - 2\boldsymbol{a} + 8\boldsymbol{b}$,$\overrightarrow{CD} = 3(\boldsymbol{a} - \boldsymbol{b})$,
(1)求证:$A$、$B$、$D$三点共线；
(2)求证:$\overrightarrow{CA} = x\overrightarrow{CB} + y\overrightarrow{CD}$(其中$x + y = 1$).

答案： 对点训练3:【证明$】 (1)\because\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD} = -2\boldsymbol{a} + 8\boldsymbol{b} +$
$3(\boldsymbol{a} - \boldsymbol{b}) = \boldsymbol{a} + 5\boldsymbol{b},\overrightarrow{AB} = \boldsymbol{a} + 5\boldsymbol{b},$
$\therefore\overrightarrow{AB} = \overrightarrow{BD},\therefore\overrightarrow{AB} // \overrightarrow{BD},$
又$\overrightarrow{AB}、$$\overrightarrow{BD}$有公共点B,所以A,B,D三点共线.
$(2)\because\overrightarrow{CA} = \overrightarrow{CB} + \overrightarrow{BA} = -\overrightarrow{BC} - \overrightarrow{AB}$
$ = 2\boldsymbol{a} - 8\boldsymbol{b} - \boldsymbol{a} - 5\boldsymbol{b} = \boldsymbol{a} - 13\boldsymbol{b},$
$x\overrightarrow{CB} + y\overrightarrow{CD} = x(2\boldsymbol{a} - 8\boldsymbol{b}) + 3y(\boldsymbol{a} - \boldsymbol{b})$
$ = (2x + 3y)\boldsymbol{a} + (-8x - 3y)\boldsymbol{b}.$
$\therefore \begin{cases} 2x + 3y = 1, \\ -8x - 3y = -13, \end{cases}$
所以$ \begin{cases} x = 2, \\ y = -1, \end{cases}$
$\therefore\overrightarrow{CA} = x\overrightarrow{CB} + y\overrightarrow{CD},$其中x + y = 1.