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Векторски производ вектора - примери 2

Израчунавање површине применом векторског производа вектора.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.2)   Одредити висину ${h_a}$ троугла $ABC$, ако је $A\left( { - 1,0 - 1} \right),B\left( {0,2, - 3} \right)$ и $C\left( {4,4,1} \right)$.

Пр.3)   Нека је $\left| {\overrightarrow a } \right| = 10,\left| {\overrightarrow b } \right| = 2$. Ако је $\overrightarrow a  \cdot \overrightarrow b  = 12$ одредити $\left| {\overrightarrow a  \times \overrightarrow b } \right|$.

Пр.2)

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\[\begin{gathered}
P = \frac{1}{2}\left| {\overrightarrow {BA} \times \overrightarrow {BC} } \right| \hfill \\
P = \frac{{a{h_a}}}{2} = \frac{{\left| {\overrightarrow {BC} } \right|{h_a}}}{2} \hfill \\
\overrightarrow {BA} = A - B = \left( { - 1, - 2,2} \right) \hfill \\
\overrightarrow {BC} = C - B = \left( {4,2,4} \right) \hfill \\
\overrightarrow {BA} \times \overrightarrow {BC} = \left| {\begin{array}{*{20}{c}}
{\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k } \\
{ - 1}&{ - 2}&2 \\
4&2&4
\end{array}} \right|\begin{array}{*{20}{c}}
{\overrightarrow i } \\
{ - 1} \\
4
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
{\overrightarrow j } \\
{ - 2} \\
2
\end{array} = - 8\overrightarrow i + 8\overrightarrow j - 2\overrightarrow k + 8\overrightarrow k - 4\overrightarrow i + 4\overrightarrow {j = } \hfill \\
= - 12\overrightarrow i + 12\overrightarrow j + 6\overrightarrow k = \left( { - 12,12,6} \right) \hfill \\
\hfill \\
\left| {\overrightarrow {BA} \times \overrightarrow {BC} } \right| = \sqrt {{{\left( { - 12} \right)}^2} + {{12}^2} + {6^2}} = \sqrt {324} = 18 \hfill \\
P = \frac{1}{2}18 = 9 \hfill \\
\left| {\overrightarrow {BC} } \right| = \sqrt {{4^2} + {2^2} + {4^2}} = 6 \hfill \\
18 = \frac{{6{h_a}}}{2} \hfill \\
{h_a} = 6 \hfill \\
\end{gathered} \]

Пр.3)

\[\begin{gathered}
\overrightarrow a \cdot \overrightarrow b = \left| {\overrightarrow a } \right| \cdot \left| {\overrightarrow b } \right|\cos \measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) \hfill \\
12 = 10 \cdot 2\cos \measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) \hfill \\
\cos \measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{3}{5} \hfill \\
\hfill \\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1 \hfill \\
{\sin ^2}\alpha + {\left( {\frac{3}{5}} \right)^2} = 1 \hfill \\
{\sin ^2}\alpha = 1 - \frac{9}{{25}} \hfill \\
{\sin ^2}\alpha = \frac{{16}}{{25}} \hfill \\
\sin \alpha = \frac{4}{5} \hfill \\
\hfill \\
\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right| \cdot \left| {\overrightarrow b } \right|sin\measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) = 10 \cdot 2 \cdot \frac{4}{5} = 16 \hfill \\
\end{gathered} \]