Az $\stackrel{\to }{a}$ és $\stackrel{\to }{b}$ vektorok közötti szög $\psi =\frac{\mathrm{\pi }}{6}$.  Számolja ki a $\stackrel{\to }{p}=\stackrel{\to }{a}+\stackrel{\to }{b}$ és $\stackrel{\to }{q}=\stackrel{\to }{a}-\stackrel{\to }{b}$ vektorok által bezárt szöget, ha $\left|\stackrel{\to }{a}\right|=\sqrt{3}$ és $\left|\stackrel{\to }{b}\right|=1$
$\stackrel{\to }{p}·\stackrel{\to }{q}=\stackrel{\to }{\left|p\right|}·\stackrel{\to }{\left|q\right|}·\mathrm{cos}\alpha \phantom{\rule{0ex}{0ex}}$ $c\mathrm{os}\alpha =\frac{\stackrel{\to }{p}·\stackrel{\to }{q}}{\stackrel{\to }{\left|p\right|}·\stackrel{\to }{\left|q\right|}}\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{q}=\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)·\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{q}=\stackrel{\to }{a}·\stackrel{\to }{a}+\stackrel{\to }{a}·\stackrel{\to }{b}-\stackrel{\to }{b}·\stackrel{\to }{a}-\stackrel{\to }{b}·\stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{q}=\stackrel{\to }{\left|a\right|}·\stackrel{\to }{\left|a\right|}·\mathrm{cos}0-\stackrel{\to }{\left|b\right|}·\stackrel{\to }{\left|b\right|}·\mathrm{cos}0\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{q}=\sqrt{3}·\sqrt{3}·1-1·1·1=3-1\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{q}=2\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{p}·\stackrel{\to }{p}=\stackrel{\to }{\left|p\right|}·\stackrel{\to }{\left|p\right|}·\mathrm{cos}0\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{p}·\stackrel{\to }{p}={\stackrel{\to }{\left|p\right|}}^{2}\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{\left|p\right|}=\sqrt{\stackrel{\to }{p}·\stackrel{\to }{p}}\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{p}=\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{p}=\stackrel{\to }{a}·\stackrel{\to }{a}+\stackrel{\to }{a}·\stackrel{\to }{b}+\stackrel{\to }{b}·\stackrel{\to }{a}+\stackrel{\to }{b}·\stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{p}=\stackrel{\to }{a}·\stackrel{\to }{a}+2\stackrel{\to }{a}·\stackrel{\to }{b}+\stackrel{\to }{b}·\stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{p}=\stackrel{\to }{\left|a\right|}·\stackrel{\to }{\left|a\right|}·\mathrm{cos}0+2·\stackrel{\to }{\left|a\right|}·\stackrel{\to }{\left|b\right|}·\mathrm{cos}\psi +\stackrel{\to }{\left|b\right|}·\stackrel{\to }{\left|b\right|}·\mathrm{cos}0\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{p}=\sqrt{3}·\sqrt{3}·1+2·\sqrt{3}·1·\frac{\sqrt{3}}{2}+1·1·1\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{p}=3+3+1\phantom{\rule{0ex}{0ex}}$ $\stackrel{\to }{p}·\stackrel{\to }{p}=7\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{\left|p\right|}=\sqrt{\left(\stackrel{\to }{p}·\stackrel{\to }{p}\right)}=\sqrt{7}\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{q}·\stackrel{\to }{q}=\stackrel{\to }{\left|q\right|}·\stackrel{\to }{\left|q\right|}·\mathrm{cos}0\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{q}·\stackrel{\to }{q}={\stackrel{\to }{\left|q\right|}}^{2}\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{\left|q\right|}=\sqrt{\left(\stackrel{\to }{q}·\stackrel{\to }{q}\right)}\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{q}·\stackrel{\to }{q}=\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{q}·\stackrel{\to }{q}=\stackrel{\to }{a}·\stackrel{\to }{a}-\stackrel{\to }{a}·\stackrel{\to }{b}-\stackrel{\to }{b}·\stackrel{\to }{a}+\stackrel{\to }{b}·\stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{q}·\stackrel{\to }{q}=\stackrel{\to }{a}·\stackrel{\to }{a}+2\stackrel{\to }{a}·\stackrel{\to }{b}+\stackrel{\to }{b}·\stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{q}·\stackrel{\to }{q}=\stackrel{\to }{\left|a\right|}·\stackrel{\to }{\left|a\right|}·\mathrm{cos}0+2·\stackrel{\to }{\left|a\right|}·\stackrel{\to }{\left|b\right|}·\mathrm{cos}\psi +\stackrel{\to }{\left|b\right|}·\stackrel{\to }{\left|b\right|}·\mathrm{cos}0\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{q}·\stackrel{\to }{q}=\sqrt{3}·\sqrt{3}·1-2·\sqrt{3}·1·\frac{\sqrt{3}}{2}+1·1·1\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{q}·\stackrel{\to }{q}=3-3+1\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{q}·\stackrel{\to }{q}=1\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$\stackrel{\to }{\left|q\right|}=\sqrt{\left(\stackrel{\to }{q}·\stackrel{\to }{q}\right)}=\sqrt{1}=1\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$c\mathrm{os}\alpha =\frac{\stackrel{\to }{p}·\stackrel{\to }{q}}{\stackrel{\to }{\left|p\right|}·\stackrel{\to }{\left|q\right|}}=\frac{2}{\sqrt{7}·1}\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$c\mathrm{os}\alpha =\frac{2}{\sqrt{7}}·\frac{\sqrt{7}}{\sqrt{7}}\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}$$c\mathrm{os}\alpha =\frac{2\sqrt{7}}{7}\phantom{\rule{0ex}{0ex}}$
$c\mathrm{os}\alpha =\frac{2\sqrt{7}}{7}$