Câu hỏi:

Ta có: $AB = 3AM \Rightarrow \overrightarrow{AM} = \frac{1}{3} \overrightarrow{AB} = \frac{1}{3} \overrightarrow{a}$
$CN = 2BC \Rightarrow \overrightarrow{CN} = 2\overrightarrow{BC} = 2(\overrightarrow{AC} - \overrightarrow{AB}) = 2(\overrightarrow{b} - \overrightarrow{a}) \Rightarrow \overrightarrow{AN} = \overrightarrow{AC} + \overrightarrow{CN} = \overrightarrow{b} + 2(\overrightarrow{b} - \overrightarrow{a}) = 3\overrightarrow{b} - 2\overrightarrow{a} = -2\overrightarrow{a} + 3\overrightarrow{b}$

Ta có: $\overrightarrow{y} = 2\overrightarrow{a} + 6\overrightarrow{b} = 2(\overrightarrow{a} + 3\overrightarrow{b}) = -2(\overrightarrow{a} - (-3)\overrightarrow{b}) = -2(\overrightarrow{a} - 3\overrightarrow{b}) = -2\overrightarrow{x}$.
Vì $\overrightarrow{y} = -2\overrightarrow{x}$ nên $\overrightarrow{y}$ và $\overrightarrow{x}$ cùng phương, ngược hướng.

Ta có $\overrightarrow{BI} = \frac{3}{4}\overrightarrow{AC} - \overrightarrow{AB} = \frac{3}{4}(\overrightarrow{BC} - \overrightarrow{BA}) - \overrightarrow{AB} = \frac{3}{4}\overrightarrow{BC} - \frac{3}{4}\overrightarrow{BA} - \overrightarrow{AB} = \frac{3}{4}\overrightarrow{BC} - \frac{1}{4}\overrightarrow{AB}$.
Suy ra $\overrightarrow{BI} = \frac{3}{4}\overrightarrow{BC} + \frac{1}{4}\overrightarrow{BA}$.
$\overrightarrow{BJ} = \frac{1}{2}\overrightarrow{AC} - \frac{2}{3}\overrightarrow{AB} = \frac{1}{2}(\overrightarrow{BC} - \overrightarrow{BA}) - \frac{2}{3}\overrightarrow{AB} = \frac{1}{2}\overrightarrow{BC} - \frac{1}{2}\overrightarrow{BA} - \frac{2}{3}\overrightarrow{AB} = \frac{1}{2}\overrightarrow{BC} - \frac{1}{6}\overrightarrow{AB}$.
Suy ra $\overrightarrow{BJ} = \frac{1}{2}\overrightarrow{BC} + \frac{1}{6}\overrightarrow{BA}$.
$\overrightarrow{BJ} = x \overrightarrow{BI} \Leftrightarrow \frac{1}{2}\overrightarrow{BC} + \frac{1}{6}\overrightarrow{BA} = x(\frac{3}{4}\overrightarrow{BC} + \frac{1}{4}\overrightarrow{BA}) \Leftrightarrow \frac{1}{2} = \frac{3}{4}x \land \frac{1}{6} = \frac{1}{4}x \Leftrightarrow x = \frac{2}{3}$.
Do đó $\overrightarrow{BJ} = \frac{2}{3}\overrightarrow{BI}$.
Vậy B, I, J thẳng hàng.
Tuy nhiên, đáp án này không đúng. Kiểm tra lại đề bài.
$\overrightarrow{BI} = \frac{3}{4}\overrightarrow{AC} - \overrightarrow{AB}$
$\overrightarrow{BJ} = \frac{1}{2}\overrightarrow{AC} - \frac{2}{3}\overrightarrow{AB}$
$\overrightarrow{IJ} = \overrightarrow{BJ} - \overrightarrow{BI} = (\frac{1}{2}\overrightarrow{AC} - \frac{2}{3}\overrightarrow{AB}) - (\frac{3}{4}\overrightarrow{AC} - \overrightarrow{AB}) = -\frac{1}{4}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB}$
$\overrightarrow{IC} = \overrightarrow{AC} - \overrightarrow{AI}$. Ta cần tìm hệ số $k$ sao cho $\overrightarrow{AI} = k \overrightarrow{AC}$
$\overrightarrow{BI} = \frac{3}{4}\overrightarrow{AC} - \overrightarrow{AB} = \overrightarrow{AI} - \overrightarrow{AB} = k\overrightarrow{AC} - \overrightarrow{AB}$
$\Rightarrow k = \frac{3}{4} \Rightarrow \overrightarrow{AI} = \frac{3}{4}\overrightarrow{AC}$
$\overrightarrow{IC} = \overrightarrow{AC} - \frac{3}{4}\overrightarrow{AC} = \frac{1}{4}\overrightarrow{AC}$
$\overrightarrow{IJ} = -\frac{1}{4}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB} = -\overrightarrow{IC} + \frac{1}{3}\overrightarrow{AB}$
$\Rightarrow \overrightarrow{CJ} = \overrightarrow{CI} + \overrightarrow{IJ} = -\frac{1}{4}\overrightarrow{AC} - \frac{1}{4}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB} = -\frac{1}{2}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB} = -(\frac{1}{2}\overrightarrow{AC} - \frac{1}{3}\overrightarrow{AB}) = -\overrightarrow{BJ}$
Vậy J là trung điểm BC.
$\overrightarrow{IJ} = -\frac{1}{4}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB} = k \overrightarrow{IC} = k \frac{1}{4}\overrightarrow{AC}$
$\overrightarrow{IJ} = x \overrightarrow{JC}$
Vậy I, J, C thẳng hàng.

Tam giác ABC vuông tại A, ta có:

• $AC = \sqrt{BC^2 - AB^2} = \sqrt{8^2 - 4^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3}$

Xét tam giác ABC vuông tại A, ta có:

• $\sin{\widehat{ABC}} = \frac{AC}{BC} = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}$

Suy ra $\widehat{ABC} = 60^\circ$. Ta có:

• $(\overrightarrow{CB}, \overrightarrow{CA}) = 180^\circ - \widehat{BCA} = \widehat{ABC} = 60^\circ$

Vậy góc giữa hai vector \(\overrightarrow {CB} \) và \(\overrightarrow {CA} \) bằng 60 độ.

Ta có công thức tích vô hướng của hai vector:
$\overrightarrow a .\overrightarrow b = |\overrightarrow a |.|\overrightarrow b |.cos(\overrightarrow a , \overrightarrow b )$
Suy ra:
$\sqrt 3 = |\overrightarrow a |.2.cos(30^\circ) = |\overrightarrow a |.2.\frac{\sqrt 3}{2} = |\overrightarrow a |. \sqrt 3$
Do đó:
$|\overrightarrow a | = \frac{\sqrt 3}{\sqrt 3} = 1$