Аналитчка геометрија, вектори – примери 2

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Пр.4)   Одредити темена $B$, $C$, $D$ и пресек дијагонала $T$ паралелограма

           $ABCD$, ако је $A\left( {2, - 1,5} \right)$, $\overrightarrow {AB}  = \left( {1,3,1} \right)$ и     $\overrightarrow {AD}  = \left( {1, - 5,3} \right)$

Пр.4)

\[\begin{gathered}
\overrightarrow {AB} = B - A \hfill \\
B = \overrightarrow {AB} + A \hfill \\
B = \left( {1;3;1} \right) + \left( {2; - 1;5} \right) \hfill \\
B = \left( {3;2;6} \right) \hfill \\
\overrightarrow {AD} = D - A \hfill \\
D = \overrightarrow {AD} + A \hfill \\
D = \left( {1; - 5;3} \right) + \left( {2; - 1;5} \right) \hfill \\
D = \left( {3; - 6;8} \right) \hfill \\
\overrightarrow {DC} = \overrightarrow {AB} \hfill \\
\overrightarrow {DC} = \left( {1;3;1} \right) \hfill \\
\overrightarrow {DC} = C - D \hfill \\
C = \overrightarrow {DC} + D \hfill \\
C = \left( {1;3;1} \right) + \left( {3; - 6;8} \right) \hfill \\
C = \left( {4; - 3;9} \right) \hfill \\
\overrightarrow {AC} = C - A \hfill \\
\overrightarrow {AC} = \left( {2; - 2;4} \right) \hfill \\
\overrightarrow {AT} = \frac{1}{2}\overrightarrow {AC} \hfill \\
\overrightarrow {AT} = \frac{1}{2}\left( {2; - 2;4} \right) \hfill \\
\overrightarrow {AT} = \left( {1; - 1;2} \right) \hfill \\
\overrightarrow {AT} = T - A \hfill \\
T = \overrightarrow {AT} + A \hfill \\
T = \left( {1; - 1;2} \right) + \left( {2; - 1;5} \right) \hfill \\
T = \left( {3; - 2;7} \right) \hfill \\
\end{gathered} \]