Први разред средње школе

Вектори - примери 3

Вектори, решени задаци.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.7   У паралелограму $ABCD$ докажи да је $\overrightarrow {AC}  + \overrightarrow {BC}  = 2\overrightarrow {BC} $.

Пр.8   Нека су $M$, $N$ и $P$ средишта страница $BC$, $CA$ и $AB$ троугла $ABC$. Докажи да је $\overrightarrow {AM}  + \overrightarrow {BN}  + \overrightarrow {CP}  = \overrightarrow 0 $.

Пр.9   Дат је правилан шестооугао $ABCDEF$. Нека су $M$, $N$, $P$, $Q$, $R$ и $S$ редом средишта страница $AB$, $CD$, $DE$, $EF$ и $FA$. Докажи да је тада $\overrightarrow {MN}  + \overrightarrow {PQ}  + \overrightarrow {RS}  = \overrightarrow 0 $.

Пр.7

63

\[\begin{gathered}
\left. \begin{gathered}
\overrightarrow {BC} = \overrightarrow {BA} + \overrightarrow {AC} \hfill \\
\overrightarrow {BC} = \overrightarrow {BD} + \overrightarrow {DC} \hfill \\
\end{gathered} \right| + \hfill \\
2\overrightarrow {BC} = \overrightarrow {BA} + \overrightarrow {AC} + \overrightarrow {BD} + \overrightarrow {DC} \hfill \\
\overrightarrow {BA} + \overrightarrow {DC} = \overrightarrow 0 \hfill \\
2\overrightarrow {BC} = \overrightarrow {AC} + \overrightarrow {BD} \hfill \\
\end{gathered} \]


Пр.8

64

\[\begin{gathered}
\left. \begin{gathered}
\overrightarrow {AM} = \overrightarrow {AB} + \overrightarrow {BM} \hfill \\
\overrightarrow {BN} = \overrightarrow {BC} + \overrightarrow {CN} \hfill \\
\overrightarrow {CP} = \overrightarrow {CA} + \overrightarrow {AP} \hfill \\
\end{gathered} \right| + \hfill \\
\overrightarrow {AM} + \overrightarrow {BN} + \overrightarrow {CP} = \overrightarrow {AB} + \overrightarrow {BM} + \overrightarrow {BC} + \overrightarrow {CN} + \overrightarrow {CA} + \overrightarrow {AP} \hfill \\
\overrightarrow {BM} = \frac{1}{2}\overrightarrow {BC} \hfill \\
\overrightarrow {CN} = \frac{1}{2}\overrightarrow {CA} \hfill \\
\overrightarrow {AP} = \frac{1}{2}\overrightarrow {AB} \hfill \\
\overrightarrow {AM} + \overrightarrow {BN} + \overrightarrow {CP} = \frac{3}{2}\overrightarrow {BC} + \frac{3}{2}\overrightarrow {CA} + \frac{3}{2}\overrightarrow {AB} = \frac{3}{2}\left( {\overrightarrow {BC} + \overrightarrow {CA} + \overrightarrow {AB} } \right) \hfill \\
\overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC} \hfill \\
\overrightarrow {AM} + \overrightarrow {BN} + \overrightarrow {CP} = \frac{3}{2}\overrightarrow 0 = \overrightarrow 0 \hfill \\
\end{gathered} \]

Пр.9

65

\[\begin{gathered}
\left. \begin{gathered}
\overrightarrow {MN} = \overrightarrow {NB} + \overrightarrow {BN} \hfill \\
\overrightarrow {PQ} = \overrightarrow {PD} + \overrightarrow {DQ} \hfill \\
\overrightarrow {RS} = \overrightarrow {RF} + \overrightarrow {FS} \hfill \\
\end{gathered} \right| + \hfill \\
\overrightarrow {MN} + \overrightarrow {PQ} + \overrightarrow {RS} = \overrightarrow {NB} + \overrightarrow {BN} + \overrightarrow {PD} + \overrightarrow {DQ} + \overrightarrow {RF} + \overrightarrow {FS} = \hfill \\
= \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {BC} + \frac{1}{2}\overrightarrow {CD} + \frac{1}{2}\overrightarrow {DE} + \frac{1}{2}\overrightarrow {EF} + \frac{1}{2}\overrightarrow {FA} = \hfill \\
= \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} + \overrightarrow {DE} + \overrightarrow {EF} + \overrightarrow {FA} } \right) \hfill \\
\overrightarrow {BC} + \overrightarrow {CD} + \overrightarrow {DE} + \overrightarrow {EF} + \overrightarrow {FA} = \overrightarrow {BA} \hfill \\
\overrightarrow {MN} + \overrightarrow {PQ} + \overrightarrow {RS} = \frac{1}{2}\overrightarrow 0 = \overrightarrow 0 \hfill \\
\end{gathered} \]