Trong không gian Oxyz , cho tứ diện ABCD trong đó...

C

- Hướng dẫn giải

Ta có

\(\begin{array}{l} {V_{ABCD}} = \frac{1}{3}{S_{ABC}}.{h_D}\\ \Rightarrow {h_D} = \frac{{3{V_{ABCD}}}}{{{S_{ABC}}}} = \frac{{3.\frac{1}{6}\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right],\overrightarrow {AD} } \right|}}{{\frac{1}{2}\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right]} \right|}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right],\overrightarrow {AD} } \right|}}{{\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right]} \right|}}\\ \overrightarrow {AB} = \left( {2; - 2; - 3} \right);\overrightarrow {AC} = \left( {4;0;6} \right);\overrightarrow {AD} = \left( { - 7; - 7;7} \right)\\ \Rightarrow \left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right] = \left( { - 12; - 24;8} \right) \Rightarrow \left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right].\overrightarrow {AD} = 308\\ \Rightarrow {h_D} = \frac{{\left| {308} \right|}}{{\sqrt {{{\left( { - 12} \right)}^2} + {{\left( { - 24} \right)}^2} + {8^2}} }} = \frac{{808}}{{28}} = 11 \end{array}\)