$$(2\overset{\rightharpoonup}{a}+\overset{\rightharpoonup}{b})\cdot (\overset{\rightharpoonup}{a}-\overset{\rightharpoonup}{b})=$$ $$=2\overset{\rightharpoonup}{a}\cdot\overset{\rightharpoonup}{a}-2\overset{\rightharpoonup}{a}\cdot\overset{\rightharpoonup}{b}+\overset{\rightharpoonup}{b}\cdot\overset{\rightharpoonup}{a}-\overset{\rightharpoonup}{b}\cdot\overset{\rightharpoonup}{b}=$$ $$=2|\overset{\rightharpoonup}{a}|^2-\overset{\rightharpoonup}{a}\cdot\overset{\rightharpoonup}{b}-|\overset{\rightharpoonup}{b}|^2=$$ $$= 2|\overset{\rightharpoonup}{a}|^2-|\overset{\rightharpoonup}{a}||\overset{\rightharpoonup}{b}|\cos 45^\circ-|\overset{\rightharpoonup}{b}|^2=$$ $$=2\cdot 9-3\cdot 2\cdot \frac{\sqrt{2}}{2}-4=14-3\sqrt{2}$$

Dolžina vektorja je enaka kvadratnemu korenu skalarnega produkta vektorja s samim seboj:$$|4\overset{\rightharpoonup}{a}-3\overset{\rightharpoonup}{b}|=\sqrt{(4\overset{\rightharpoonup}{a}-3\overset{\rightharpoonup}{b})\cdot (4\overset{\rightharpoonup}{a}-3\overset{\rightharpoonup}{b})}.$$ Naprej ravnaj podobno kot v prejšnjem zgledu in upoštevaj, da je $|\overset{\rightharpoonup}{a}|=|\overset{\rightharpoonup}{b}|=1$.

$|4\overset{\rightharpoonup}{a}-3\overset{\rightharpoonup}{b}|=\sqrt{37}$

Pokaži, da je $\overset{\Large\rightharpoonup}{AC}\cdot \overset{\Large\rightharpoonup}{BD}=0$.

ZA vektorja $\overset{\rightharpoonup}{a}=\overset{\Large\rightharpoonup}{AB}$ in $\overset{\rightharpoonup}{b}=\overset{\Large\rightharpoonup}{AD}$ velja $|\overset{\rightharpoonup}{a}|=|\overset{\rightharpoonup}{b}|$. Iz tega izhaja: $$\overset{\Large\rightharpoonup}{AC}\cdot\overset{\Large\rightharpoonup}{BD}=(\overset{\rightharpoonup}{a}+\overset{\rightharpoonup}{b})(\overset{\rightharpoonup}{b}-\overset{\rightharpoonup}{a})=|\overset{\rightharpoonup}{b}|^2-|\overset{\rightharpoonup}{a}|^2=0$$