Мешовит производ вектора – примери 2

Текст задатака објашњених у видео лекцији.

Пр.4)   Одреди висину која одговара основи $\left( {\overrightarrow {a,} \overrightarrow b } \right)$ паралелопипеда конструисаног над векторима: $\overrightarrow a  = 3\overrightarrow i  + 2\overrightarrow j  - \overrightarrow k $, $\overrightarrow b  = 2\overrightarrow i  + \overrightarrow j  - \overrightarrow k $, $\overrightarrow c  = \overrightarrow i  - 2\overrightarrow j  + 3\overrightarrow k $.

Пр.5)   Тачке $A\left( {2,3,1} \right)$, $B\left( {4,1, - 2} \right)$, $C\left( {6,3,7} \right)$, $D\left( { - 5, - 4,8} \right)$ су темена пирамиде. Израчунати запремину те пирамиде и висину која

одговара темену $D$.

Пр.4)  

\[\begin{gathered}
\overrightarrow a = \left( {3,2, - 1} \right) \hfill \\
\overrightarrow b = \left( {2,1, - 1} \right) \hfill \\
\overrightarrow c = \left( {1, - 2,3} \right) \hfill \\
V = B \cdot H \hfill \\
V = \left| {\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right]} \right| = \left| {\left| {\begin{array}{*{20}{c}}
3&2&{ - 1} \\
2&1&{ - 1} \\
1&{ - 2}&3
\end{array}} \right|\begin{array}{*{20}{c}}
3 \\
2 \\
1
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
2 \\
1 \\
{ - 2}
\end{array}} \right| = \left| {9 - 2 + 4 + 1 - 6 - 12} \right| = \left| { - 6} \right| = 6 \hfill \\
B = \left| {\overrightarrow a \times \overrightarrow b } \right| \hfill \\
\overrightarrow a \times \overrightarrow b = \left| {\begin{array}{*{20}{c}}
{\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k } \\
3&2&{ - 1} \\
2&1&{ - 1}
\end{array}} \right|\begin{array}{*{20}{c}}
{\overrightarrow i } \\
3 \\
2
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
{\overrightarrow j } \\
2 \\
1
\end{array} = - 2\overrightarrow i - 2\overrightarrow j + 3\overrightarrow k - 4\overrightarrow k + \overrightarrow i + 3\overrightarrow j = - \overrightarrow i + \overrightarrow j - \overrightarrow k = \left( { - 1,1, - 1} \right) \hfill \\
B = \left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{{\left( { - 1} \right)}^2} + {1^2} + {{\left( { - 1} \right)}^2}} = \sqrt 3 \hfill \\
6 = \sqrt 3 H \Rightarrow H = \frac{6}{{\sqrt 3 }} \cdot \frac{{\sqrt 3 }}{{\sqrt 3 }} = 2\sqrt 3 \hfill \\
\end{gathered} \]

Пр.5) 

\[\begin{gathered}
\overrightarrow {AB} = B - A = \left( {2, - 2, - 3} \right) \hfill \\
\overrightarrow {AC} = C - A = \left( {4,0,6} \right) \hfill \\
\overrightarrow {AD} = D - A = \left( { - 7, - 7,7} \right) \hfill \\
\left[ {\overrightarrow {AB} ,\overrightarrow {AC} ,\overrightarrow {AD} } \right] = \left| {\begin{array}{*{20}{c}}
2&{ - 2}&{ - 3} \\
4&0&6 \\
{ - 7}&{ - 7}&7
\end{array}} \right|\begin{array}{*{20}{c}}
2 \\
4 \\
{ - 7}
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
{ - 2} \\
0 \\
{ - 7}
\end{array} = 0 + 84 + 84 - 0 + 84 + 56 = 308 \hfill \\
V = \frac{1}{6}\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {AC} ,\overrightarrow {AD} } \right]} \right| = \frac{1}{6} \cdot 308 = \frac{{154}}{3} \hfill \\
\hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {{{\left( { - 12} \right)}^2} + {{\left( { - 24} \right)}^2} + {8^2}} = 28 \hfill \\
B = \frac{1}{2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = 14 \hfill \\
V = \frac{1}{3}BH \hfill \\
\frac{{154}}{3} = \frac{1}{3} \cdot 14 \cdot H \Rightarrow H = 11 \hfill \\
\end{gathered} \]