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# 复习向量的大小和方向

$\mid \mid \phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\left(a,b\right)\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\mid \mid =\sqrt{{a}^{2}+{b}^{2}}$

$\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)$

$\left(\mid \mid \phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\stackrel{\to }{u}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\mid \mid \mathrm{cos}\left(\theta \right),\mid \mid \phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\stackrel{\to }{u}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\mid \mid \mathrm{sin}\left(\theta \right)\right)$

## 练习类别 1：根据分矢量求出向量大小

$||\left(a,b\right)||=\sqrt{{a}^{2}+{b}^{2}}$

$\stackrel{\to }{u}=\left(1,7\right)$
$||\stackrel{\to }{u}||=$

## 练习类别 2：根据分矢量求出向量方向

$\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)$

### 示例 1：象限 $\text{I}$‍

${\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)\approx {53}^{\circ }$

### 示例 2： 象限 $\text{IV}$‍

${\mathrm{tan}}^{-1}\left(\frac{-4}{3}\right)\approx -{53}^{\circ }$

$-{53}^{\circ }+{360}^{\circ }={307}^{\circ }$

### 示例 3： 象限 $\text{II}$‍

${\mathrm{tan}}^{-1}\left(\frac{4}{-3}\right)\approx -{53}^{\circ }$
$-{53}^{\circ }$ 位于象限 $\text{IV}$，不是象限 $\text{II}$。因此，我们得加上 ${180}^{\circ }$ ，得到相反方向的角：
$-{53}^{\circ }+{180}^{\circ }={127}^{\circ }$

$\stackrel{\to }{u}=5\stackrel{^}{i}+8\stackrel{^}{j}$
$\theta =$
${}^{\circ }$

## 练习类别 3：根据向量的大小和方向求出分矢量

$\stackrel{\to }{u}=\left(||\stackrel{\to }{u}||\mathrm{cos}\left(\theta \right),||\stackrel{\to }{u}||\mathrm{sin}\left(\theta \right)\right)$

$\left(2\mathrm{cos}\left({30}^{\circ }\right),2\mathrm{sin}\left({30}^{\circ }\right)\right)=\left(\sqrt{3},1\right)$

$\right)$