Đáp án A sai vì:

\(\begin{array}{l} \begin{array}{*{20}{l}} {\left| {\left[ {\vec u,\vec v} \right]} \right| = \left| {\vec u} \right|\left| {\vec v} \right|.\sin \left( {\vec u,\vec v} \right)} \end{array}\\ Thật\,vậy\,,Gọi\,\overrightarrow u \left( {{u_1};{u_2};{u_3}} \right),\overrightarrow v \left( {{v_1};{v_2};{v_3}} \right)\\ Nếu\,\left[ \begin{array}{l} \overrightarrow u = \overrightarrow 0 \\ \overrightarrow v = \overrightarrow 0 \end{array} \right. \Rightarrow \left| {\left[ {\vec u,\vec v} \right]} \right| = \left| {\vec u} \right|\left| {\vec v} \right|.\sin \left( {\vec u,\vec v} \right) = 0\\ Nếu\,\left[ \begin{array}{l} \overrightarrow u \ne \overrightarrow 0 \\ \overrightarrow v \ne \overrightarrow 0 \end{array} \right. thì\,ta\,có\\ \left| {\vec u} \right|\left| {\vec v} \right|.\sin \left( {\vec u,\vec v} \right) = \left| {\vec u} \right|\left| {\vec v} \right|.\sqrt {1 - {{\cos }^2}\left( {\overrightarrow {u,} \overrightarrow v } \right)} = \left| {\vec u} \right|\left| {\vec v} \right|.\sqrt {1 - \frac{{{{\left( {\overrightarrow {u.} \overrightarrow v } \right)}^2}}}{{{{\left| {\overrightarrow u } \right|}^2}.{{\left| {\overrightarrow v } \right|}^2}}}} \\ = \sqrt {{{\overrightarrow u }^2}.{{\overrightarrow v }^2} - {{\left( {\overrightarrow u .\overrightarrow v } \right)}^2}} = \sqrt {{{\left( {{u_2}.{v_3} - {u_3}.{v_2}} \right)}^2} + {{\left( {{u_3}.{v_1} - {u_1}.{v_3}} \right)}^2} + {{\left( {{u_1}.{v_2} - {u_2}.{v_1}} \right)}^2}} \\ = \left| {\left[ {\overrightarrow u ,\overrightarrow v } \right]} \right|. \end{array}\)