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Câu hỏi: Trong không gian với hệ tọa độ Oxyz, cho tứ diện ABCD có $A\left( 0;1;-1 \right);B\left( 1;1;2 \right);$ $C\left( 1;-1;0 \right);D\left( 0;0;1 \right).$ Tính độ dài đường cao AH của hình chóp A.BCD.
A. $3\sqrt{2}$
B. $2\sqrt{2}$
C. $\dfrac{\sqrt{2}}{2}$
D. $\dfrac{3\sqrt{2}}{2}$
A. $3\sqrt{2}$
B. $2\sqrt{2}$
C. $\dfrac{\sqrt{2}}{2}$
D. $\dfrac{3\sqrt{2}}{2}$
Ta có $\overrightarrow{BA}=\left( -1;0;-3 \right);\overrightarrow{BC}=\left( 0;-2;-2 \right);\overrightarrow{BD}=\left( -1;-1;-1 \right).$
$\left[ \overrightarrow{BC},\overrightarrow{BD} \right]=\left( 0;2;-2 \right)\Rightarrow \left[ \overrightarrow{BC},\overrightarrow{BD} \right].\overrightarrow{BA}=6$
${{V}_{ABCD}}=\dfrac{1}{6}.\left[ \overrightarrow{BC},\overrightarrow{BD} \right].\overrightarrow{BA}=\dfrac{1}{6}.6=1$ (đvtt)
${{S}_{BCD}}=\dfrac{1}{2}.\left| \left[ \overrightarrow{BC},\overrightarrow{BD} \right] \right|=\dfrac{1}{2}.\sqrt{{{0}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{2}$ (đvdt)
Ta có ${{V}_{ABCD}}=\dfrac{1}{3}.AH.{{S}_{BCD}}\Rightarrow AH=\dfrac{3{{V}_{ABCD}}}{{{S}_{BCD}}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}$
$\left[ \overrightarrow{BC},\overrightarrow{BD} \right]=\left( 0;2;-2 \right)\Rightarrow \left[ \overrightarrow{BC},\overrightarrow{BD} \right].\overrightarrow{BA}=6$
${{V}_{ABCD}}=\dfrac{1}{6}.\left[ \overrightarrow{BC},\overrightarrow{BD} \right].\overrightarrow{BA}=\dfrac{1}{6}.6=1$ (đvtt)
${{S}_{BCD}}=\dfrac{1}{2}.\left| \left[ \overrightarrow{BC},\overrightarrow{BD} \right] \right|=\dfrac{1}{2}.\sqrt{{{0}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{2}$ (đvdt)
Ta có ${{V}_{ABCD}}=\dfrac{1}{3}.AH.{{S}_{BCD}}\Rightarrow AH=\dfrac{3{{V}_{ABCD}}}{{{S}_{BCD}}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}$
Đáp án D.