Копланарност вектора, разлагање вектора. Решени задаци.
Текст задатака објашњених у видео лекцији.
Пр.1) Доказати да су вектори $\overrightarrow a $, $\overrightarrow b $ и $\overrightarrow c $ компланарни, а затим вектор $\overrightarrow c $ разложи по правцима $\overrightarrow a $ и $\overrightarrow b $ ако је: $\overrightarrow a = \left( {3, - 2,} \right)$, $\overrightarrow b = \left( { - 2,1} \right)$, $\overrightarrow c = \left( {7, - 4} \right)$.
Пр.1)
\[\begin{gathered}
\alpha \overrightarrow a + \beta \overrightarrow b + \gamma \overrightarrow c = \overrightarrow 0 \hfill \\
\alpha \left( {3; - 2} \right) + \beta \left( { - 2;1} \right) + \gamma \left( {7; - 4} \right) = \left( {0;0} \right) \hfill \\
\left( {3\alpha ; - 2\alpha } \right) + \left( { - 2\beta ;\beta } \right) + \left( {7\gamma ; - 4\gamma } \right) = \left( {0;0} \right) \hfill \\
\left( {3\alpha - 2\beta + 7\gamma ; - 2\alpha + \beta - 4\gamma } \right) = \left( {0;0} \right) \hfill \\
\hfill \\
3\alpha - 2\beta + 7\gamma = 0 \hfill \\
\underline { - 2\alpha + \beta - 4\gamma = 0} \hfill \\
- 2\alpha + \beta - 4\gamma = 0 \hfill \\
\underline { - \alpha - \gamma = 0} \hfill \\
\alpha = - \gamma \hfill \\
2\gamma + \beta - 4\gamma = 0 \hfill \\
\beta = 2\gamma \hfill \\
\left( {\alpha ;\beta ;\gamma } \right) = \left( { - \gamma ;2\gamma ;\gamma } \right) \hfill \\
\end{gathered} \]
$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ - линеарно зависне.
\[\begin{gathered}
\overrightarrow c = \alpha \overrightarrow a + \beta \overrightarrow b \hfill \\
\left( {7; - 4} \right) = \alpha \left( {3; - 2} \right) + \beta \left( { - 2;1} \right) = \hfill \\
= \left( {3\alpha ; - 2\alpha } \right) + \left( { - 2\beta ;\beta } \right) = \left( {3\alpha - 2\beta ; - 2\alpha + \beta } \right) \hfill \\
\hfill \\
3\alpha - 2\beta = 7 \hfill \\
\underline { - 2\alpha + \beta = - 4} \hfill \\
- 2\alpha + \beta = - 4 \hfill \\
\underline { - \alpha = - 1} \hfill \\
- 2\alpha + \beta = - 4 \hfill \\
\alpha = 1 \hfill \\
\beta = - 2 \hfill \\
\hfill \\
\overrightarrow c = \overrightarrow a - 2\overrightarrow b \hfill \\
\end{gathered} \]