a: \(AB=\sqrt{\left(0-2\right)^2+\left(-2-1\right)^2+\left(5+3\right)^2}=\sqrt{77}\)
\(BC=\sqrt{\left(1-0\right)^2+\left(1+2\right)^2+\left(3-5\right)^2}=\sqrt{14}\)
ABCD là hình bình hành
=>\(S_{ABCD}=AB\cdot BC=\sqrt{77}\cdot\sqrt{14}=\sqrt{1078}\)
=>Sai
b: \(\left[\overrightarrow{a};\overrightarrow{b}\right]=\left[0\cdot3-\left(-2\right)\cdot\left(-1\right);\left(-2\right)\cdot2-1\cdot3;1\cdot\left(-1\right)-0\cdot2\right]\)
=>\(\left[\overrightarrow{a};\overrightarrow{b}\right]=\left(-2;-7;-1\right)\)
=>Sai
c: \(\left[\overrightarrow{a};\overrightarrow{b}\right]=\left[3\cdot2-1\cdot1;1\cdot\left(-2\right)-2\cdot2;2\cdot1-3\cdot\left(-2\right)\right]\)
=>\(\left[\overrightarrow{a};\overrightarrow{b}\right]=\left(5;-6;8\right)\)
=>Sai
d: \(\overrightarrow{n}\cdot\overrightarrow{a}=1\cdot2+1\cdot\left(-3\right)+\left(-2\right)\cdot\left(-1\right)=2-3+2=4-3=1\)
=>\(\overrightarrow{n}\) không vuông góc với \(\overrightarrow{a}\)
=>Sai
a) \(D\left(x;y;z\right)\)
\(\overrightarrow{AB}=\left(-2;-3;8\right);\overrightarrow{DC}=\left(1-x;1-y;3-z\right)\)
\(\overrightarrow{AB}=\overrightarrow{DC}\left(hbh\right)\Rightarrow\left\{{}\begin{matrix}1-x=-2\\1-y=-3\\3-x=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=4\\z=-5\end{matrix}\right.\) \(\Rightarrow D\left(3;4;-5\right)\)
\(\overrightarrow{AD}=\left(1;3;-2\right)\)
Diện tích hình bình hành \(ABCD:\)
\(S=\left[\overrightarrow{AB}.\overrightarrow{AD}\right]=\sqrt{\left(-18\right)^2+4^2+\left(-3\right)^2}=\sqrt{324+16+9}=\sqrt{349}\left(đvdt\right)\)
\(\Rightarrow\) Câu a Đúng
