Вектори – примери 2

Текст задатака објашњених у видео лекцији.

Пр.4   Дат је паралелограм $ABCD$ и произвољна тачка $M$ ван паралелограма. Ако је тачка $O$ пресек дијагонала паралелограма, докажи да је: $\overrightarrow {MO}  = \frac{1}{4}\left( {\overrightarrow {MA}  + \overrightarrow {MB}  + \overrightarrow {MC}  + \overrightarrow {MD} } \right)$.

Пр. 5  Дат је правилни шестоугао $ABCDEF$. Покажи једнакост $\overrightarrow {AB}  + \overrightarrow {AC}  + \overrightarrow {AD}  + \overrightarrow {AE}  + \overrightarrow {AF}  = 3\overrightarrow {AD} $.

Пр.6   Тачка $C$ и $D$ деле дуж $AD$ на три једнака одсечка. Тачка $O$ је произвољна тачка изван праве $AB$. $\overrightarrow {OA}  = \overrightarrow a $ и $\overrightarrow {OB}  = \overrightarrow b $, изразити векторе $\overrightarrow {OC} $ и $\overrightarrow {OD} $ помоћу вектора $\overrightarrow a $ и $\overrightarrow b $.

Пр.4

\[\begin{gathered}
\overrightarrow {MO} = \frac{1}{4}\left( {\overrightarrow {MA} + \overrightarrow {MB} + \overrightarrow {MC} + \overrightarrow {MD} } \right) \hfill \\
\left. \begin{gathered}
\overrightarrow {MO} = \overrightarrow {MA} + \overrightarrow {AO} \hfill \\
\overrightarrow {MO} = \overrightarrow {MB} + \overrightarrow {BO} \hfill \\
\overrightarrow {MO} = \overrightarrow {MC} + \overrightarrow {CO} \hfill \\
\overrightarrow {MO} = \overrightarrow {MD} + \overrightarrow {DO} \hfill \\
\end{gathered} \right| + \hfill \\
4\overrightarrow {MO} = \overrightarrow {MA} + \overrightarrow {AO} + \overrightarrow {MB} + \overrightarrow {BO} + \overrightarrow {MC} + \overrightarrow {CO} + \overrightarrow {MD} + \overrightarrow {DO} \hfill \\
\overrightarrow {AO} + \overrightarrow {CO} = \overrightarrow 0 \hfill \\
\overrightarrow {BO} + \overrightarrow {DO} = \overrightarrow 0 \hfill \\
\left. {4\overrightarrow {MO} = \overrightarrow {MA} + \overrightarrow {MB} + \overrightarrow {MC} + \overrightarrow {MD} } \right| \div 4 \hfill \\
\overrightarrow {MO} = \frac{1}{4}\left( {\overrightarrow {MA} + \overrightarrow {MB} + \overrightarrow {MC} + \overrightarrow {MD} } \right) \hfill \\
\end{gathered} \]

Пр.5

\[\begin{gathered}
\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} = 3\overrightarrow {AD} \hfill \\
\left. \begin{gathered}
\overrightarrow {AD} = \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} \hfill \\
\overrightarrow {AD} = \overrightarrow {AF} + \overrightarrow {FE} + \overrightarrow {ED} \hfill \\
\overrightarrow {AD} = \overrightarrow {AD} \hfill \\
\end{gathered} \right| + \hfill \\
3\overrightarrow {AD} = \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} + \overrightarrow {AF} + \overrightarrow {FE} + \overrightarrow {ED} + \overrightarrow {AD} \hfill \\
\overrightarrow {BC} + \overrightarrow {CD} = \overrightarrow {BD} = \overrightarrow {AE} \hfill \\
\overrightarrow {FE} + \overrightarrow {ED} = \overrightarrow {FD} = \overrightarrow {AC} \hfill \\
3\overrightarrow {AD} = \overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} \hfill \\
\end{gathered} \]

Пр.6

\[\begin{gathered}
\overrightarrow {OC} = \overrightarrow {OA} + \overrightarrow {AC} = \overrightarrow a + \frac{1}{3}\overrightarrow {AB} = \overrightarrow a + \frac{1}{3}\left( {\overrightarrow {AO} + \overrightarrow {OB} } \right) = \hfill \\
= \overrightarrow a + \frac{1}{3}\left( { - \overrightarrow {OA} + \overrightarrow b } \right) = \overrightarrow a + \frac{1}{3}\left( { - \overrightarrow a + \overrightarrow b } \right) = \overrightarrow a - \frac{1}{3}\overrightarrow a + \frac{1}{3}\overrightarrow b = \frac{2}{3}\overrightarrow a + \frac{1}{3}\overrightarrow b \hfill \\
\end{gathered} \]

\[\begin{gathered}
\overrightarrow {OD} = \overrightarrow {OB} + \overrightarrow {BD} = \overrightarrow b + \frac{1}{3}\overrightarrow {BA} = \overrightarrow b - \frac{1}{3}\overrightarrow {AB} = \overrightarrow b - \frac{1}{3}\left( { - \overrightarrow a + \overrightarrow b } \right) = \hfill \\
= \overrightarrow b + \frac{1}{3}\overrightarrow a - \frac{1}{3}\overrightarrow b = \frac{2}{3}\overrightarrow b + \frac{1}{3}\overrightarrow a \hfill \\
\end{gathered} \]