《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2. (1) 设平面向量$\boldsymbol{a} = (1,2),\boldsymbol{b} = (-2,y)$，若$\boldsymbol{a} // \boldsymbol{b}$，则$|3\boldsymbol{a} + \boldsymbol{b}|$等于 (A)
A. $\sqrt{5}$
B. $\sqrt{6}$
C. $\sqrt{17}$
D. $\sqrt{26}$
(2) 已知向量$\boldsymbol{a} = (\cos \theta,\sin \theta)$，向量$\boldsymbol{b} = (\sqrt{3},0)$，则$|2\boldsymbol{a} - \boldsymbol{b}|$的最大值为$2 + \sqrt{3}$.
►[归纳提升]

答案：
(1)A
(2)$2 + \sqrt{3}$
(1)
∵$\boldsymbol{a}//\boldsymbol{b}$,
∴$1× y-2×(-2)=0$,解得$y = - 4$,从而$3\boldsymbol{a} + \boldsymbol{b}=(1,2)$,$\vert3\boldsymbol{a} + \boldsymbol{b}\vert=\sqrt{5}$.
(2)$2\boldsymbol{a} - \boldsymbol{b}=(2\cos\theta-\sqrt{3},2\sin\theta)$,
$\vert2\boldsymbol{a} - \boldsymbol{b}\vert=\sqrt{(2\cos\theta-\sqrt{3})^{2}+(2\sin\theta)^{2}}$
$=\sqrt{4\cos^{2}\theta - 4\sqrt{3}\cos\theta + 3 + 4\sin^{2}\theta}=\sqrt{7 - 4\sqrt{3}\cos\theta}$,
当且仅当$\cos\theta=-1$时,$\vert2\boldsymbol{a} - \boldsymbol{b}\vert$取最大值$2 + \sqrt{3}$.

对点训练2
(1) 已知向量$\boldsymbol{a} = (m,1),\boldsymbol{b} = (3,m)$，若$\boldsymbol{a}$与$\boldsymbol{b}$方向相反，则$|\boldsymbol{a} - \sqrt{3}\boldsymbol{b}| =$ (B)
A. 54
B. 8
C. $3\sqrt{3}$
D. $4\sqrt{3}$
(2) 已知$|\boldsymbol{a}| = 10,\boldsymbol{b} = (1,2)$，且$\boldsymbol{a} // \boldsymbol{b}$，求$\boldsymbol{a}$的坐标.

答案：
(1)B
(2)设$\boldsymbol{a}$的坐标为$(x,y)$,由题意得$\begin{cases}2x - y = 0,\\\sqrt{x^{2}+y^{2}} = 10,\end{cases}$
解得$\begin{cases}x = 2\sqrt{5},\\y = 4\sqrt{5}\end{cases}$或$\begin{cases}x = -2\sqrt{5},\\y = -4\sqrt{5}\end{cases}$.
所以$\boldsymbol{a}=(2\sqrt{5},4\sqrt{5})$或$\boldsymbol{a}=(-2\sqrt{5},-4\sqrt{5})$.

例3. 设平面上向量$\boldsymbol{a} = (\cos \alpha,\sin \alpha)(0^{\circ} \leq \alpha \leq 90^{\circ}),\boldsymbol{b} = \left( -\frac{1}{2},\frac{\sqrt{3}}{2} \right)$.
(1) 求$\boldsymbol{a}$与$\boldsymbol{b}$的夹角$\theta$.
(2) 求证：$\boldsymbol{a} + \boldsymbol{b}$与$\boldsymbol{a} - \boldsymbol{b}$垂直.
►[归纳提升]

答案：
(1)由题意知,$\vert\boldsymbol{a}\vert = 1$,$\vert\boldsymbol{b}\vert = 1$,
$\boldsymbol{a}·\boldsymbol{b}=-\frac{1}{2}\cos\alpha+\frac{\sqrt{3}}{2}\sin\alpha$,则$\cos\theta=\frac{\boldsymbol{a}·\boldsymbol{b}}{\vert\boldsymbol{a}\vert\vert\boldsymbol{b}\vert}=-\frac{1}{2}\cos\alpha+\frac{\sqrt{3}}{2}\sin\alpha=\cos(120^{\circ}-\alpha)$.
∵$0^{\circ}\leq\alpha\leq90^{\circ}$,
∴$30^{\circ}\leq120^{\circ}-\alpha\leq120^{\circ}$.
又$0^{\circ}\leq\theta\leq180^{\circ}$,
∴$\theta = 120^{\circ}-\alpha$,
即两向量的夹角为$120^{\circ}-\alpha$.
(2)证明:
∵$(\boldsymbol{a} + \boldsymbol{b})·(\boldsymbol{a} - \boldsymbol{b})$
$=(\cos\alpha-\frac{1}{2},\sin\alpha+\frac{\sqrt{3}}{2})·(\cos\alpha+\frac{1}{2},\sin\alpha-\frac{\sqrt{3}}{2})$
$=(\cos\alpha-\frac{1}{2})(\cos\alpha+\frac{1}{2})+(\sin\alpha+\frac{\sqrt{3}}{2})(\sin\alpha-\frac{\sqrt{3}}{2})=\cos^{2}\alpha-\frac{1}{4}+\sin^{2}\alpha-\frac{3}{4}$
$=1-\frac{1}{4}-\frac{3}{4}=0$,
∴$(\boldsymbol{a} + \boldsymbol{b})\perp(\boldsymbol{a} - \boldsymbol{b})$.

对点训练3
(1) 设向量$\boldsymbol{a} = (2,2),\boldsymbol{b} = (2,-1)$. 若$2\boldsymbol{a} \perp (\boldsymbol{a} - t\boldsymbol{b})$，则实数$t =$4.
(2) 已知向量$\boldsymbol{a}$与$\boldsymbol{b}$的夹角为$60^{\circ}$，且$\boldsymbol{a} = (-2,-6),|\boldsymbol{b}| = \sqrt{10}$，则$\boldsymbol{a} · \boldsymbol{b} =$10.

答案：
(1)4
(2)10
(1)$2\boldsymbol{a}\perp(\boldsymbol{a}-t\boldsymbol{b})\Rightarrow2\boldsymbol{a}·(\boldsymbol{a}-t\boldsymbol{b})=(4,4)·(2 - 2t,2 + t)=16 - 4t=0\Rightarrow t = 4$.
(2)因为$\boldsymbol{a}=(-2,-6)$,所以$\vert\boldsymbol{a}\vert=\sqrt{(-2)^{2}+(-6)^{2}}=2\sqrt{10}$.又$\vert\boldsymbol{b}\vert=\sqrt{10}$,向量$\boldsymbol{a}$与$\boldsymbol{b}$的夹角为$60^{\circ}$,
所以$\boldsymbol{a}·\boldsymbol{b}=\vert\boldsymbol{a}\vert\vert\boldsymbol{b}\vert\cos60^{\circ}=2\sqrt{10}×\sqrt{10}×\frac{1}{2}=10$.