Bài 24 trang 119 Sách bài tập Hình học lớp 12 Nâng cao

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 câu a) 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 \(\alpha  = (\overrightarrow a ,\overrightarrow b ))\).

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