Sửa đề: Chứng minh \(\overrightarrow{AB}+\overrightarrow{MC}=\overrightarrow{AC}+\overrightarrow{MB}\)

\(\overrightarrow{AB}-\overrightarrow{MB}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AM}\)

\(\overrightarrow{AC}-\overrightarrow{MC}=\overrightarrow{AC}+\overrightarrow{CM}=\overrightarrow{AC}\)

Do đó: \(\overrightarrow{AB}-\overrightarrow{MB}=\overrightarrow{AC}-\overrightarrow{MC}\)

=>\(\overrightarrow{AB}+\overrightarrow{MC}=\overrightarrow{AC}+\overrightarrow{MB}\)

\(\left|3\overrightarrow{AB}+\overrightarrow{AC}\right|=4a\)

Tam giác vuông cân tại C \(\Rightarrow AC=\dfrac{AB}{\sqrt{2}}=a\sqrt{2}\)

Do I là trung điểm BC \(\Rightarrow\overrightarrow{IC}=-\overrightarrow{IB}\)

Vậy:

\(\left|\overrightarrow{AI}-\overrightarrow{IB}\right|=\left|\overrightarrow{AI}+\overrightarrow{IC}\right|=\left|\overrightarrow{AC}\right|=AC=a\sqrt{2}\)

\(a,\overrightarrow{AB}=\left(2;10\right)\)

\(\overrightarrow{AC}=\left(-5;5\right)\)

\(\overrightarrow{BC}=\left(-7;-5\right)\)

\(b,\) Thiếu dữ kiện

\(c,Cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=\dfrac{\left|2\left(-5\right)+10.5\right|}{\sqrt{2^2+10^2}.\sqrt{\left(-5\right)^2+5^2}}=\dfrac{2\sqrt{13}}{13}\)

\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{AC}\right)=56^o18'\)

\(Cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)=\dfrac{\left|2\left(-7\right)+10\left(-5\right)\right|}{\sqrt{2^2+10^2}.\sqrt{\left(-7\right)^2+\left(-5\right)^2}}\)

\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=43^o9'\)

giúp em với mn ơi🥲

Lời giải:

$|\overrightarrow{BC}|=BC=\sqrt{AB^2+AC^2}=\sqrt{(3a)^2+(4a)^2}=5a$ theo định lý Pitago.