Sam, v paru ali v skupini dokaži spodnjo trditev.
Če so $\overset{\rightharpoonup}{a},\overset{\rightharpoonup}{b},\overset{\rightharpoonup}{c}$ nekoplanarni vektorji in je $$m\overset{\rightharpoonup}{a}+n\overset{\rightharpoonup}{b}+p\overset{\rightharpoonup}{c}=\overset{\rightharpoonup}{0}\quad (m,n,p\in\mathbb{R}),$$ potem je $m=n=p=0.$
V kocki $ABCDEFGH$ so dani bazni vektorji $\overset{\rightharpoonup}{a}=\overset{\Large\rightharpoonup}{AB}$, $\overset{\rightharpoonup}{b}=\overset{\Large\rightharpoonup}{AD}$, $\overset{\rightharpoonup}{c}=\overset{\Large\rightharpoonup}{AE}$.
a) $\overset{\Large\rightharpoonup}{CH}$ b) $\overset{\Large\rightharpoonup}{EC}$ c) $\overset{\Large\rightharpoonup}{AG}$
č) $\overset{\Large\rightharpoonup}{DM}$, če je $M$ razpolovišče roba $CG$
d) $\overset{\Large\rightharpoonup}{ES}$, če je $S$ središče kocke
e) $\overset{\Large\rightharpoonup}{NP}$, če je $N$ središče ploskve $BCGH$, $P$ pa deli rob $DH$ v razmerju $|DP|:|PH|=2:1$
Za vektorje $\overset{\rightharpoonup}{a},\overset{\rightharpoonup}{b}$ in $\overset{\rightharpoonup}{c}$ velja $2\overset{\rightharpoonup}{a}+3\overset{\rightharpoonup}{b}-4\overset{\rightharpoonup}{c}=\overset{\rightharpoonup}{0}$. Izrazi vektor $\overset{\rightharpoonup}{c}$. Ali so vektorji $\overset{\rightharpoonup}{a},\overset{\rightharpoonup}{b},\overset{\rightharpoonup}{c}$ koplanarni ali nekoplanarni? Odgovor utemelji.
Vektorji $\overset{\rightharpoonup}{a},\overset{\rightharpoonup}{b},\overset{\rightharpoonup}{c}$ so nekoplanarni. Določi števila $m,n,t$ tako, da bo: $$(m-5)\overset{\rightharpoonup}{a}+(n+3)\overset{\rightharpoonup}{b}-(t-2)\overset{\rightharpoonup}{c}=\overset{\rightharpoonup}{0}$$