- LG a
- LG b
- LG c
- LG d
- LG e
- LG g
Chứng minh các tính chất sau đây có tích có hướng :
LG a
\[\eqalign{\left[ {\overrightarrow a ,\overrightarrow b } \right] = - \left[ { \overrightarrow b ,\overrightarrow a } \right]\cr} \]
Lời giải chi tiết:
Giả sử \[\overrightarrow a = [{x_1};{y_1};{z_1}],\overrightarrow b = [{x_2};{y_2};{z_2}],\overrightarrow c = [{x_3};{y_3};{z_3}]\]
\[\eqalign{ &\left[ {\overrightarrow a ,\overrightarrow b } \right] = \left[ {\left| \matrix{ {y_1} \hfill \cr {y_2} \hfill \cr} \right.\left. \matrix{ {z_1} \hfill \cr {z_2} \hfill \cr} \right|;\left| \matrix{ {z_1} \hfill \cr {z_2} \hfill \cr} \right.\left. \matrix{ {x_1} \hfill \cr {x_2} \hfill \cr} \right|;\left| \matrix{ {x_1} \hfill \cr {x_2} \hfill \cr} \right.\left. \matrix{ {y_1} \hfill \cr {y_2} \hfill \cr} \right|} \right] \cr & = [{y_1}{z_2} - {y_2}{z_1};{z_1}{x_2} - {z_2}{x_1};{x_1}{y_2} - {x_2}{y_1}] \cr & = - [{y_2}{z_1} - {y_1}{z_2};{z_2}{x_1} - {z_1}{x_2};{x_2}{y_1} - {x_1}{y_2}] \cr & = - \left[ {\left| \matrix{ {y_2} \hfill \cr {y_1} \hfill \cr} \right.\left. \matrix{ {z_2} \hfill \cr {z_1} \hfill \cr} \right|;\left| \matrix{ {z_2} \hfill \cr {z_1} \hfill \cr} \right.\left. \matrix{ {x_2} \hfill \cr {x_1} \hfill \cr} \right|;\left| \matrix{ {x_2} \hfill \cr {x_1} \hfill \cr} \right.\left. \matrix{ {y_2} \hfill \cr {y_1} \hfill \cr} \right|} \right] \cr & = - \left[ {\overrightarrow b ,\overrightarrow a } \right]. \cr} \]
LG b
\[\eqalign{\left[ {\overrightarrow a ,\overrightarrow a } \right] = \overrightarrow 0\cr} \]
Lời giải chi tiết:
Từ câua]ta có \[\left[ {\overrightarrow a ,\overrightarrow a } \right] = - \left[ {\overrightarrow a ,\overrightarrow a } \right]\] , suy ra \[\left[ {\overrightarrow a ,\overrightarrow a } \right] = \overrightarrow 0 \].
LG c
\[\eqalign{\left[ {k\overrightarrow a ,\overrightarrow b } \right] = k\left[ {\overrightarrow a ,\overrightarrow b } \right] = \left[ {\overrightarrow a ,k\overrightarrow b } \right] & \cr} \]
Lời giải chi tiết:
\[\eqalign{ & k\left[ {\overrightarrow a ,\overrightarrow b } \right] = \left[ {k\left| \matrix{ {y_1} \hfill \cr {y_2} \hfill \cr} \right.\left. \matrix{ {z_1} \hfill \cr {z_2} \hfill \cr} \right|;k\left| \matrix{ {z_1} \hfill \cr {z_2} \hfill \cr} \right.\left. \matrix{ {x_1} \hfill \cr {x_2} \hfill \cr} \right|;k\left| \matrix{ {x_1} \hfill \cr {x_2} \hfill \cr} \right.\left. \matrix{ {y_1} \hfill \cr {y_2} \hfill \cr} \right|} \right] \cr & = \left[ {\left| \matrix{ k{y_1} \hfill \cr {y_2} \hfill \cr} \right.\left. \matrix{ k{z_1} \hfill \cr {z_2} \hfill \cr} \right|;\left| \matrix{ k{z_1} \hfill \cr {z_2} \hfill \cr} \right.\left. \matrix{ k{x_1} \hfill \cr {x_2} \hfill \cr} \right|;\left| \matrix{ k{x_1} \hfill \cr {x_2} \hfill \cr} \right.\left. \matrix{ k{y_1} \hfill \cr {y_2} \hfill \cr} \right|} \right] \cr & = \left[ {k\overrightarrow a ,\overrightarrow b } \right]. \cr} \]
Tương tự \[k\left[ {\overrightarrow a ,\overrightarrow b } \right] = \left[ {\overrightarrow a ,k\overrightarrow b } \right].\]
LG d
\[\eqalign{\left[ {\overrightarrow c ,\overrightarrow a + \overrightarrow b } \right] = \left[ {\overrightarrow c ,\overrightarrow a } \right] + \left[ {\overrightarrow c ,\overrightarrow b } \right]\cr} \]
Lời giải chi tiết:
LG e
\[\eqalign{\overrightarrow a \left[ {\overrightarrow b ,\overrightarrow c } \right] = \left[ {\overrightarrow a ,\overrightarrow b } \right].\overrightarrow c \cr} \]
Lời giải chi tiết:
\[\eqalign{ & \overrightarrow a .\left[ {\overrightarrow b ,\overrightarrow c } \right] \cr&= {x_1}\left[ {\left| \matrix{ {y_2} \hfill \cr {y_3} \hfill \cr} \right.\left. \matrix{ {z_2} \hfill \cr {z_3} \hfill \cr} \right| + {y_1}\left| \matrix{ {z_2} \hfill \cr {z_3} \hfill \cr} \right.\left. \matrix{ {x_2} \hfill \cr {x_3} \hfill \cr} \right| + {z_1}\left| \matrix{ {x_2} \hfill \cr {x_3} \hfill \cr} \right.\left. \matrix{ {y_2} \hfill \cr {y_3} \hfill \cr} \right|} \right] \cr & = {x_3}\left[ {\left| \matrix{ {y_1} \hfill \cr {y_2} \hfill \cr} \right.\left. \matrix{ {z_1} \hfill \cr {z_2} \hfill \cr} \right| + {y_3}\left| \matrix{ {z_1} \hfill \cr {z_2} \hfill \cr} \right.\left. \matrix{ {x_1} \hfill \cr {x_2} \hfill \cr} \right| + {z_3}\left| \matrix{ {x_1} \hfill \cr {x_2} \hfill \cr} \right.\left. \matrix{ {y_1} \hfill \cr {y_2} \hfill \cr} \right|} \right] \cr & =\left[ {\overrightarrow a ,\overrightarrow b } \right].\overrightarrow c \cr} \]
LG g
\[\eqalign{\left| {{{\left[ {\overrightarrow a ,\overrightarrow b } \right]}^2}} \right| = {\left| {\overrightarrow a } \right|^2}.{\left| {\overrightarrow b } \right|^2} - {[\overrightarrow a .\overrightarrow b ]^2}. \cr} \]
Lời giải chi tiết:
\[\eqalign{ VP &= {\left| {\overrightarrow a } \right|^2}.{\left| {\overrightarrow b } \right|^2} - {[\overrightarrow a .\overrightarrow b ]^2} \cr&= {\left| {\overrightarrow a } \right|^2}.{\left| {\overrightarrow b } \right|^2} - {\left| {\overrightarrow a } \right|^2}.{\left| {\overrightarrow b } \right|^2}\cos^2 \alpha \cr & = {\left| {\overrightarrow a } \right|^2}.{\left| {\overrightarrow b } \right|^2}[1 - {\cos ^2}\alpha ] = {\left| {\overrightarrow a } \right|^2}.{\left| {\overrightarrow a } \right|^2}.{\sin ^2}\alpha \cr} \]
\[ = {\left| {\left[ {\overrightarrow a ,\overrightarrow b } \right]} \right|^2} = VT\] [ ở đây \[\alpha = [\overrightarrow a ,\overrightarrow b ]]\].