Google
Câu hỏi: Cho tứ diện ABCD có $AB=AC=AD$ và $\widehat{BAC}=\widehat{BAD}={{60}^{o}}$. Hãy xác định góc giữa cặp vecto $\overrightarrow{AB}$ và $\overrightarrow{CD}?$
A. ${{60}^{o}}.$
B. ${{45}^{o}}.$
C. ${{120}^{o}}.$
D. ${{90}^{o}}.$
Ta có $\overrightarrow{AB}.\overrightarrow{CD}=\overrightarrow{AB}.\left( \overrightarrow{AD}-\overrightarrow{AC} \right)=\overrightarrow{AB}.\overrightarrow{AD}-\overrightarrow{AB}.\overrightarrow{AC}$
$\begin{aligned}
& =\left| \overrightarrow{AB} \right|.\left| \overrightarrow{AD} \right|.\cos \left( \overrightarrow{AB}.\overrightarrow{AD} \right)-\left| \overrightarrow{AB} \right|.\left| \overrightarrow{AC} \right|.\cos \left( \overrightarrow{AB}.\overrightarrow{AC} \right) \\
& =\left| \overrightarrow{AB} \right|.\left| \overrightarrow{AD} \right|.\cos {{60}^{o}}-\left| \overrightarrow{AB} \right|.\left| \overrightarrow{AC} \right|.\cos {{60}^{o}} \\
\end{aligned}$
Mà $AC=AD\Rightarrow \overrightarrow{AB}.\overrightarrow{CD}=0\Rightarrow \left( \overrightarrow{AB},\overrightarrow{CD} \right)={{90}^{o}}$
A. ${{60}^{o}}.$
B. ${{45}^{o}}.$
C. ${{120}^{o}}.$
D. ${{90}^{o}}.$
Ta có $\overrightarrow{AB}.\overrightarrow{CD}=\overrightarrow{AB}.\left( \overrightarrow{AD}-\overrightarrow{AC} \right)=\overrightarrow{AB}.\overrightarrow{AD}-\overrightarrow{AB}.\overrightarrow{AC}$
$\begin{aligned}
& =\left| \overrightarrow{AB} \right|.\left| \overrightarrow{AD} \right|.\cos \left( \overrightarrow{AB}.\overrightarrow{AD} \right)-\left| \overrightarrow{AB} \right|.\left| \overrightarrow{AC} \right|.\cos \left( \overrightarrow{AB}.\overrightarrow{AC} \right) \\
& =\left| \overrightarrow{AB} \right|.\left| \overrightarrow{AD} \right|.\cos {{60}^{o}}-\left| \overrightarrow{AB} \right|.\left| \overrightarrow{AC} \right|.\cos {{60}^{o}} \\
\end{aligned}$
Mà $AC=AD\Rightarrow \overrightarrow{AB}.\overrightarrow{CD}=0\Rightarrow \left( \overrightarrow{AB},\overrightarrow{CD} \right)={{90}^{o}}$
Đáp án D.