Câu hỏi:
Cho tam giác ABC có điểm I nằm trên cạnh AC sao cho \(\overrightarrow {BI} = \frac{3}{4}\overrightarrow {AC} - \overrightarrow {AB} \), J là điểm thỏa mãn \(\overrightarrow {BJ} = \frac{1}{2}\overrightarrow {AC} - \frac{2}{3}\overrightarrow {AB} \). Ba điểm nào sau đây thẳng hàng ?
C. I, A, B;
D. I, G, B.
Trả lời:
Đáp án đúng: A
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.
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.
Câu hỏi này thuộc đề thi trắc nghiệm dưới đây, bấm vào Bắt đầu thi để làm toàn bài
