Question

Answer 1

Câu 12: Đáp án A
Ta có: \(\left|\overrightarrow{AB}+\overrightarrow{AD}-\dfrac{1}{2}\overrightarrow{AC}\right|=\left|\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{AC}\right|=\left|\dfrac{1}{2}\overrightarrow{AC}\right|=\dfrac{AC}{2}\)
Áp dụng định lý pytago trong tam giác ABC vuông tại B có:
\(AB^2+BC^2=AC^2\) 
\(\Rightarrow a^2+a^2=AC^2\)
\(\Rightarrow AC^2=2a^2\)
\(\Rightarrow AC=a\sqrt{2}\) \(\Rightarrow\dfrac{AC}{2}=\dfrac{a\sqrt{2}}{2}\)
Vậy \(\left|\overrightarrow{AB}-\dfrac{1}{2}\overrightarrow{AC}+\overrightarrow{AD}\right|=\dfrac{a\sqrt{2}}{2}\)