Аналитчка геометрија, вектори – примери 6

Текст задатака објашњених у видео лекцији.

Пр.2)   Доказати да вектори $\overrightarrow a ,\overrightarrow b $ и $\overrightarrow c $ припадају истој равни, а затим вектор $\overrightarrow c $ разложити по правцима вектора $\overrightarrow a $ и $\overrightarrow b $ ако је: $\overrightarrow a  = \left( { - 3,0,2} \right),\overrightarrow b  = \left( {2,1, - 4} \right),\overrightarrow c  = \left( {11, - 2, - 2} \right)$

Пр.3)   Дати су вектори $\overrightarrow a  = \left( {2,3} \right),\overrightarrow b  = \left( {1, - 3} \right),\overrightarrow c  = \left( { - 1,3} \right)$. Одредити реални број $m$ тако да вектори $\overrightarrow p  = \overrightarrow a  + m\overrightarrow b $ и $\overrightarrow q  = \overrightarrow a  + 2\overrightarrow c $  буду колинеарни.

Пр.2) 

\[{D_s} = \left| {\begin{array}{*{20}{c}}
{ - 3}&0&2 \\
2&1&{ - 4} \\
{11}&{ - 2}&{ - 2}
\end{array}} \right|\begin{array}{*{20}{c}}
{ - 3} \\
2 \\
{11}
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
0 \\
1 \\
{ - 2}
\end{array} = 6 + 0 - 8 - 22 + 24 - 0 = 0\]

$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ - линеарно зависне.

\[\begin{gathered}
\overrightarrow c = \alpha \overrightarrow a + \beta \overrightarrow b \hfill \\
\left( {11; - 2; - 2} \right) = \alpha \left( { - 3;0;2} \right) + \beta \left( {2;1; - 4} \right) = \hfill \\
= \left( { - 3\alpha ;0;2\alpha } \right) + \left( {2\beta ;\beta ; - 4\beta } \right) = \left( { - 3\alpha + 2\beta ;\beta ;2\alpha - 4\beta } \right) \hfill \\
\hfill \\
- 3\alpha + 2\beta = 11 \hfill \\
\beta = - 2 \hfill \\
\underline {2\alpha - 4\beta = - 2} \hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
- 3\alpha - 4\beta = 11 \hfill \\
- 3\alpha = 15 \hfill \\
\alpha = - 5 \hfill \\
\hfill \\
\end{gathered} &{}&\begin{gathered}
2\alpha + 8 = - 2 \hfill \\
2\alpha = - 10 \hfill \\
\alpha = - 5 \hfill \\
\hfill \\
\end{gathered}
\end{array} \hfill \\
\overrightarrow c = - 5\overrightarrow a - 2\overrightarrow b \hfill \\
\end{gathered} \]

Пр.3)

\[\begin{gathered}
\overrightarrow p = \overrightarrow a + m\overrightarrow b = \left( {2,3} \right) + m\left( {1, - 3} \right) = \left( {2,3} \right) + \left( {m, - 3m} \right) = \left( {2 + m,3 - 3m} \right) \hfill \\
\overrightarrow q = \overrightarrow a + 2\overrightarrow c = \left( {2,3} \right) + 2\left( { - 1,3} \right) = \left( {2,3} \right) + \left( { - 2,6} \right) = \left( {0,9} \right) \hfill \\
\end{gathered} \]

\[\begin{gathered}
{D_s} = \left| {\begin{array}{*{20}{c}}
{2 + m}&{3 - 3m} \\
0&9
\end{array}} \right| = 0 \hfill \\
\left( {2 + m} \right) \cdot 9 - 0 = 0 \hfill \\
2 + m = 0 \hfill \\
m = - 2 \hfill \\
\end{gathered} \]