Аналитчка геометрија, вектори – примери 3

Текст задатака објашњених у видео лекцији.

Пр.1)   Одредити линеарну зависност јединичних вектора $\overrightarrow i $, $\overrightarrow j $ и $\overrightarrow k $ у правцу $x$, $y$ и $z$ - осе редом.

Пр.2)   Одредити линеарну зависност вектора: $\overrightarrow a \left( {5,4,3} \right),\overrightarrow b \left( {3,3,2} \right),\overrightarrow c \left( {8,1,3} \right)$.

Пр.1) 

\[\begin{gathered}
\alpha \overrightarrow i + \beta \overrightarrow j + \gamma \overrightarrow k = \overrightarrow 0 \hfill \\
\alpha \left( {1;0;0} \right) + \beta \left( {0;1;0} \right) + \gamma \left( {0;0;1} \right) = \left( {0;0;0} \right) \hfill \\
\left( {\alpha ;0;0} \right) + \left( {0;\beta ;0} \right) + \left( {0;0;\gamma } \right) = \left( {0;0;0} \right) \hfill \\
\left( {\alpha ;\beta ;\gamma } \right) = \left( {0;0;0} \right) \hfill \\
\hfill \\
\alpha = 0 \hfill \\
\beta = 0 \hfill \\
\gamma = 0 \hfill \\
\end{gathered} \]

$\overrightarrow i ,\overrightarrow j ,\overrightarrow k$ линеарне независне

Пр.2) 

\[\begin{gathered}
\alpha \overrightarrow a + \beta \overrightarrow b + \gamma \overrightarrow c = \overrightarrow 0 \hfill \\
\alpha \left( {5;4;3} \right) + \beta \left( {3;3;2} \right) + \gamma \left( {8;1;3} \right) = \left( {0;0;0} \right) \hfill \\
\left( {5\alpha ;4\alpha ;3\alpha } \right) + \left( {3\beta ;3\beta ;2\beta } \right) + \left( {8\gamma ;\gamma ;3\gamma } \right) = \left( {0;0;0} \right) \hfill \\
\left( {5\alpha + 3\beta + 8\gamma ;4\alpha + 3\beta + \gamma ;3\alpha + 2\beta + 3\gamma } \right) = \left( {0;0;0} \right) \hfill \\
5\alpha + 3\beta + 8\gamma = 0 \hfill \\
4\alpha + 3\beta + \gamma = 0 \hfill \\
\underline {3\alpha + 2\beta + 3\gamma = 0} \hfill \\
4\alpha + 3\beta + \gamma = 0 \hfill \\
- 27\alpha - 21\beta = 0 \hfill \\
\underline { - 9\alpha - 7\beta = 0} \hfill \\
4\alpha + 3\beta + \gamma = 0 \hfill \\
\underline { - 9\alpha - 7\beta = 0} \hfill \\
\alpha = - \frac{{7\beta }}{9} \hfill \\
\underline {4\left( { - \frac{{7\beta }}{9}} \right) + 3\beta + \gamma = 0} \hfill \\
\alpha = - \frac{{7\beta }}{9} \hfill \\
\gamma = - 3\beta + \frac{{28\beta }}{9} = \frac{1}{9}\beta \hfill \\
\left( {\alpha ;\beta ;\gamma } \right) = \left( { - \frac{{7\beta }}{9};\beta ;\frac{1}{9}\beta } \right) \hfill \\
\end{gathered} \]

$\overrightarrow a ;\overrightarrow b ;\overrightarrow c$ линеарно зависне.