Трећи разред средње школе

Мешовит производ вектора - примери 1

Компланарност тачака и вектора. Решени задаци.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.1)   Приказати да тачке $A,B,C$ и $D$ припадају истој равни. $A\left( {2, - 1, - 2} \right),B\left( {1,2,1} \right)C\left( {2,3,0} \right)D\left( {5,0, - 6} \right)$

Пр.2)   Одредити $x$ тако да вектори $\overrightarrow a ,\overrightarrow b $ и $\overrightarrow c $ буду компланарни. $\overrightarrow a \left( {1n\left( {x - 2} \right), - 2,6} \right),\overrightarrow b \left( {x, - 2,5} \right),\overrightarrow c \left( {0, - 1,3} \right)$

Пр.3) Израчунати запремину паралелопипеда конструисаног над векторима: $\overrightarrow a (4,5, - 3),\overrightarrow b (1, - 2,1),\overrightarrow c (1,1,1)$.

Пр.1) 

397 png

\[\begin{gathered}
\overrightarrow {AB} = B - A = \left( { - 1,3,3} \right) \hfill \\
\overrightarrow {AC} = C - A = \left( {0,4,2} \right) \hfill \\
\overrightarrow {AD} = D - A = \left( {3,1, - 4} \right) \hfill \\
\hfill \\
\left[ {\overrightarrow {AB} ,\overrightarrow {AC} ,\overrightarrow {AD} } \right] = \left| {\begin{array}{*{20}{c}}
{ - 1}&3&3 \\
0&4&2 \\
3&1&{ - 4}
\end{array}} \right|\begin{array}{*{20}{c}}
{ - 1} \\
0 \\
3
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
3 \\
4 \\
1
\end{array} = 16 + 18 + 0 - 36 + 2 - 0 = 0 \hfill \\
\end{gathered} \]

Tачке $A,B,C$ и $D$ су компланарне.

Пр.2)

\[\begin{gathered}
\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}
{1n\left( {x - 2} \right)}&{ - 2}&6 \\
x&{ - 2}&5 \\
0&{ - 1}&3
\end{array}} \right|\begin{array}{*{20}{c}}
{1n\left( {x - 2} \right)} \\
x \\
0
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
{ - 2} \\
{ - 2} \\
{ - 1}
\end{array} = 0 \hfill \\
\hfill \\
- 61n\left( {x - 2} \right) + 0 - 6x - 0 + 51n\left( {x - 2} \right) + 6x = 0 \hfill \\
- 1n\left( {x - 2} \right) = 0 \hfill \\
x - 2 = 1 \hfill \\
x = 3 \hfill \\
\end{gathered} \]

Пр.3)

\[\begin{gathered}
V = \left| {\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right]} \right| \hfill \\
\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}
4&5&{ - 3} \\
1&{ - 2}&1 \\
1&1&1
\end{array}} \right| = 1 \cdot \left| {\begin{array}{*{20}{c}}
5&{ - 3} \\
{ - 2}&1
\end{array}} \right| - 1 \cdot \left| {\begin{array}{*{20}{c}}
4&{ - 3} \\
1&1
\end{array}} \right| + 1 \cdot \left| {\begin{array}{*{20}{c}}
4&5 \\
1&{ - 2}
\end{array}} \right| = \hfill \\
= 5 - 6 - \left( {4 + 3} \right) + \left( { - 8 - 5} \right) = - 1 - 7 - 13 = - 21 \hfill \\
V = 21 \hfill \\
\end{gathered} \]