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Câu hỏi: Trong không gian $Oxyz$, cho tứ diện ABCD với $A\left( 2;-1;6 \right), B\left( -3;-1;-4 \right), $ $C\left( 5;-1;0 \right)$ và $D\left( 1;2;1 \right)$. Độ dài chiều cao của tứ diện ABCD kẻ từ đỉnh A bằng
A. $3.$
B. $\dfrac{\sqrt{3}}{2}.$
C. $\dfrac{3}{2}.$
D. $5.$
A. $3.$
B. $\dfrac{\sqrt{3}}{2}.$
C. $\dfrac{3}{2}.$
D. $5.$
$\overrightarrow{BC}=\left( 8;0;4 \right);\overrightarrow{BD}=\left( 4;3;5 \right);\overrightarrow{BA}=\left( 5;0;10 \right)$
$\Rightarrow \overrightarrow{BD}\wedge \overrightarrow{BC}=\left( 12;24;-24 \right)$
${{V}_{ABCD}}=\dfrac{1}{6}\left| \left( \overrightarrow{BD}\wedge \overrightarrow{BC} \right).\overrightarrow{BA} \right|=\dfrac{\left| 5.12+0.24-10.24 \right|}{6}=30$
${{S}_{\Delta BCD}}=\dfrac{1}{2}\left| \overrightarrow{BD}\wedge \overrightarrow{BC} \right|=\dfrac{\sqrt{{{12}^{2}}+{{2.24}^{2}}}}{2}=18$
$\Rightarrow {{d}_{\left( A,\left( BCD \right) \right)}}=\dfrac{3{{V}_{ABCD}}}{{{S}_{\Delta BCD}}}=\dfrac{30.3}{18}=5$
$\Rightarrow \overrightarrow{BD}\wedge \overrightarrow{BC}=\left( 12;24;-24 \right)$
${{V}_{ABCD}}=\dfrac{1}{6}\left| \left( \overrightarrow{BD}\wedge \overrightarrow{BC} \right).\overrightarrow{BA} \right|=\dfrac{\left| 5.12+0.24-10.24 \right|}{6}=30$
${{S}_{\Delta BCD}}=\dfrac{1}{2}\left| \overrightarrow{BD}\wedge \overrightarrow{BC} \right|=\dfrac{\sqrt{{{12}^{2}}+{{2.24}^{2}}}}{2}=18$
$\Rightarrow {{d}_{\left( A,\left( BCD \right) \right)}}=\dfrac{3{{V}_{ABCD}}}{{{S}_{\Delta BCD}}}=\dfrac{30.3}{18}=5$
Đáp án D.