Trong không gian với hệ tọa độ Oxyz, cho

* Hướng dẫn giải

Đáp án D

Ta có $\overrightarrow{BA}=\left(-1;0;-3\right);\overrightarrow{BC}=\left(0;-2;-2\right);\overrightarrow{BD}=\left(-1;-1;-1\right).$

$\left[\overrightarrow{BC},\overrightarrow{BD}\right]=\left(0;2;-2\right)\Rightarrow \left[\overrightarrow{BC},\overrightarrow{BD}\right].\overrightarrow{BA}=6$ 

$V_{ABCD}=\frac{1}{6}.\left[\overrightarrow{BC},\overrightarrow{BD}\right].\overrightarrow{BA}=\frac{1}{6}.6=1$ (đvtt)

$S_{BCD}=\frac{1}{2}.\left|\left[\overrightarrow{BC},\overrightarrow{BD}\right]\right|=\frac{1}{2}.\sqrt{0^2+{\left(-2\right)}^2+{\left(-2\right)}^2}=\sqrt{2}$ (đvdt)

Ta có $V_{ABCD}=\frac{1}{3}.AH.S_{BCD}\Rightarrow AH=\frac{3V_{ABCD}}{S_{BCD}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$