Cho hai véc tơ u 1 = ( x 1 ; y 1 ; z 1 ) và u 2 = ( x 2 ; y 2 ; z 2 ) . Kí hiệu u = [u1 , u2] , khi đó:

2/22

Cho hai véc tơ \[\overrightarrow {{u_1}} = \left( {{x_1};{y_1};{z_1}} \right)\]và \[\overrightarrow {{u_2}} = \left( {{x_2};{y_2};{z_2}} \right)\]. Kí hiệu \[\overrightarrow u = \left[ {\overrightarrow {{u_1}} ,\overrightarrow {{u_2}} } \right],\]khi đó:

\[\vec u = \left( {\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{y_2}}\\{{y_1}}\end{array}}&{\begin{array}{*{20}{l}}{{z_2}}\\{{z_1}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{z_2}}\\{{z_1}}\end{array}}&{\begin{array}{*{20}{l}}{{x_2}}\\{{x_1}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{x_2}}\\{{x_1}}\end{array}}&{\begin{array}{*{20}{l}}{{y_2}}\\{{y_1}}\end{array}}\end{array}} \right|} \right)\]

\[\vec u = \left( {\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}}&{\begin{array}{*{20}{l}}{{y_1}}\\{{y_2}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{y_1}}\\{{y_2}}\end{array}}&{\begin{array}{*{20}{l}}{{z_1}}\\{{z_2}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{z_1}}\\{{z_2}}\end{array}}&{\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}}\end{array}} \right|} \right)\]

\[\vec u = \left( {\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{y_1}}\\{{y_2}}\end{array}}&{\begin{array}{*{20}{l}}{{z_1}}\\{{z_2}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{z_1}}\\{{z_2}}\end{array}}&{\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}}&{\begin{array}{*{20}{l}}{{y_1}}\\{{y_2}}\end{array}}\end{array}} \right|} \right)\]

\[\vec u = \left( {\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{z_1}}\\{{z_2}}\end{array}}&{\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}}&{\begin{array}{*{20}{l}}{{y_1}}\\{{y_2}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{y_1}}\\{{y_2}}\end{array}}&{\begin{array}{*{20}{l}}{{z_1}}\\{{z_2}}\end{array}}\end{array}} \right|} \right)\]

Giải thích

Công thức xác định tọa độ tích có hướng

\[\left[ {\overrightarrow {{u_1}} ,\overrightarrow {{u_2}} } \right] = \left( {\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{y_1}}\\{{y_2}}\end{array}}&{\begin{array}{*{20}{l}}{{z_1}}\\{{z_2}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{z_1}}\\{{z_2}}\end{array}}&{\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}}&{\begin{array}{*{20}{l}}{{y_1}}\\{{y_2}}\end{array}}\end{array}} \right|} \right)\]

\[ = \left( {{y_1}{z_2} - {y_2}{z_1};{z_1}{x_2} - {z_2}{x_1};{x_1}{y_2} - {x_2}{y_1}} \right)\]

Đáp án cần chọn là: C