Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
\(\overrightarrow{KA}=-\overrightarrow{AK}=-\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=-\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)\)
\(=-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{KD}=\overrightarrow{AD}-\overrightarrow{AK}=\overrightarrow{AD}+\overrightarrow{KA}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
\(AC=\sqrt{AB^2+AD^2}=5\)
a.
\(\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=\left|\overrightarrow{AC}\right|=AC=5\)
b.
Đặt \(T=\left|2\overrightarrow{AB}+3\overrightarrow{AD}\right|\Rightarrow T^2=4AB^2+9AD^2+12\overrightarrow{AB}.\overrightarrow{AD}=4AB^2+9AD^2\)
\(\Rightarrow T^2=180\Rightarrow T=6\sqrt{5}\)
Vậy \(\left|2\overrightarrow{AB}+3\overrightarrow{AD}\right|=6\sqrt{5}\)
\(A^2=\left|3a+5b\right|^2=9a^2+25b^2+30ab=9.1+25.1+30.3=124\)
\(\Rightarrow A=2\sqrt{31}\)