Први разред средње школе

Вектори - примери 4

Вектори, решени задаци.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.10   Дат је паралелограм $ABCD$ и на правој $BC$ дата је тачка $E$ таква да је $B$- sred $CE$ . Ако је вектор $\overrightarrow {AB}  = \overrightarrow m $ и $\overrightarrow {AC}  = \overrightarrow n $, изрази векторе $\overrightarrow {BD} $, $\overrightarrow {CD} $, $\overrightarrow {DA} $, $\overrightarrow {BD} $, $\overrightarrow {AE} $, $\overrightarrow {CE} $, $\overrightarrow {DE} $ векторима $\overrightarrow m $ и $\overrightarrow n $.

Пр.11   Ако су ${A_1}$, ${B_1}$, и ${C_1}$ редом средишта страница $BC$, $CA$, $AB$ троугла $ABC$ и вектори $\overrightarrow {AB}  = \overrightarrow m $, $\overrightarrow {AC}  = \overrightarrow n $ изрази векторе $\overrightarrow {A{A_1}} $ и $\overrightarrow {B{B_1}} $ векторима $\overrightarrow m $ и $\overrightarrow n $.

Пр.10

66

\[\begin{gathered}
\overrightarrow {BC} = \overrightarrow {BA} + \overrightarrow {AC} = - \overrightarrow {AB} + \overrightarrow {AC} = \overrightarrow m + \overrightarrow n \hfill \\
\overrightarrow {CD} = \overrightarrow {BA} = - \overrightarrow {AB} = - \overrightarrow m \hfill \\
\overrightarrow {DA} = \overrightarrow {CB} = - \overrightarrow {BC} = - \left( { - \overrightarrow m + \overrightarrow n } \right) = \overrightarrow m - \overrightarrow n \hfill \\
\overrightarrow {BD} = \overrightarrow {BA} + \overrightarrow {AD} = - \overrightarrow m - \overrightarrow {DA} = - \overrightarrow m - \left( {\overrightarrow m - \overrightarrow n } \right) = - 2\overrightarrow m + \overrightarrow n \hfill \\
\overrightarrow {AE} = \overrightarrow {AB} + \overrightarrow {BE} = \overrightarrow m + \overrightarrow {CB} = \overrightarrow m - \overrightarrow {BC} = \overrightarrow m - \left( { - \overrightarrow m + \overrightarrow n } \right) = 2\overrightarrow m - \overrightarrow n \hfill \\
\overrightarrow {CE} = 2\overrightarrow {CB} = - 2\overrightarrow {BC} = - 2\left( { - \overrightarrow m + \overrightarrow n } \right) = 2\overrightarrow m - 2\overrightarrow n \hfill \\
\overrightarrow {DE} = \overrightarrow {DA} + \overrightarrow {AE} = \overrightarrow m - \overrightarrow n + 2\overrightarrow m - \overrightarrow n = 3\overrightarrow m - 2\overrightarrow n \hfill \\
\end{gathered} \]


Пр.11

67

\[\begin{gathered}
\left. \begin{gathered}
\overrightarrow {A{A_1}} = \overrightarrow {AB} + \overrightarrow {B{A_1}} \hfill \\
\overrightarrow {A{A_1}} = \overrightarrow {AC} + \overrightarrow {C{A_1}} \hfill \\
\end{gathered} \right| + \hfill \\
2\overrightarrow {A{A_1}} = \overrightarrow {AB} + \overrightarrow {B{A_1}} + \overrightarrow {AC} + \overrightarrow {C{A_1}} \hfill \\
\overrightarrow {AB} = \overrightarrow m \hfill \\
\overrightarrow {AB} = \overrightarrow n \hfill \\
\overrightarrow {B{A_1}} + \overrightarrow {C{A_1}} = \overrightarrow 0 \hfill \\
\left. {2\overrightarrow {A{A_1}} = \overrightarrow m + \overrightarrow n } \right| \div 2 \hfill \\
\overrightarrow {A{A_1}} = \frac{1}{2}\left( {\overrightarrow m + \overrightarrow n } \right) \hfill \\
\end{gathered} \]

 

\[\begin{gathered}
\left. \begin{gathered}
\overrightarrow {B{B_1}} = \overrightarrow {BA} + \overrightarrow {A{B_1}} \hfill \\
\overrightarrow {B{B_1}} = \overrightarrow {BC} + \overrightarrow {C{B_1}} \hfill \\
\end{gathered} \right| + \hfill \\
2\overrightarrow {B{B_1}} = \overrightarrow {BA} + \overrightarrow {A{B_1}} + \overrightarrow {BC} + \overrightarrow {C{B_1}} \hfill \\
\overrightarrow {A{B_1}} + \overrightarrow {C{B_1}} = \overrightarrow 0 \hfill \\
2\overrightarrow {B{B_1}} = - \overrightarrow {AB} + \overrightarrow {BA} + \overrightarrow {AC} = - 2\overrightarrow m + \overrightarrow n \hfill \\
\left. {2\overrightarrow {B{B_1}} = - 2\overrightarrow m + \overrightarrow n } \right| \div 2 \hfill \\
\overrightarrow {B{B_1}} = - \overrightarrow m + \frac{1}{2}\overrightarrow n \hfill \\
\end{gathered} \]