Bài 3. TÍCH CỦA VECTO VỚI MỘT SỐ - Hỏi đáp

Hình vẽ:

Lời giải:

a) Kéo dài $AG$ cắt $BC$ tại trung điểm $M$. Hiển nhiên $\overrightarrow{BM}, \overrightarrow{CM}$ là vecto đối nên tổng bằng vecto không.

Theo tính chất trọng tâm ta có:

$\overrightarrow{AI}=\frac{1}{2}\overrightarrow{AG}=\frac{1}{2}.\frac{2}{3}\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AM}$

$=\frac{1}{6}(\overrightarrow{AM}+\overrightarrow{AM})=\frac{1}{6}(\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{AC}+\overrightarrow{CM})$

$=\frac{1}{6}(\overrightarrow{AB}+\overrightarrow{AC})$

$=\frac{1}{6}(\overrightarrow{AC}+\overrightarrow{CB}+\overrightarrow{AC})$

$=\frac{1}{6}(2\overrightarrow{AC}+\overrightarrow{CB})$

$=\frac{-1}{3}\overrightarrow{CA}+\frac{1}{6}\overrightarrow{CB}$

$=\frac{-1}{3}\overrightarrow{a}+\frac{1}{6}\overrightarrow{b}$

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$\overrightarrow{AK}=\frac{1}{5}\overrightarrow{AB}=\frac{1}{5}(\overrightarrow{AC}+\overrightarrow{CB})=\frac{1}{5}(-\overrightarrow{CA}+\overrightarrow{CB})$

$=\frac{-1}{5}\overrightarrow{a}+\frac{1}{5}\overrightarrow{b}$

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$\overrightarrow{CI}=\overrightarrow{CA}+\overrightarrow{AI}=\overrightarrow{a}-\frac{1}{3}\overrightarrow{a}+\frac{1}{6}\overrightarrow{b}$

$=\frac{2}{3}\overrightarrow{a}+\frac{1}{6}\overrightarrow{b}$

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$\overrightarrow{CK}=\overrightarrow{CA}+\overrightarrow{AK}=\overrightarrow{a}-\frac{1}{5}\overrightarrow{a}+\frac{1}{5}\overrightarrow{b}=\frac{4}{5}\overrightarrow{a}+\frac{1}{5}\overrightarrow{b}$

b)

Từ phần a ta thấy: $\overrightarrow{CI}=\frac{5}{6}\overrightarrow{CK}$ nên $C,I,K$ thẳng hàng.

Đề yêu cầu gì bạn!?

Gọi H là chân đường cao hạ từ A xuống BC \(\Rightarrow\Delta ABH\) vuông cân tại H (do \(\widehat{B}=45^0\))

\(\Rightarrow BH=AH=2a\Rightarrow HC=BH+AD=4a\)

\(\Rightarrow AC=\sqrt{AH^2+HC^2}=2a\sqrt{5}\)

Vậy:

\(\left|\overrightarrow{CB}-\overrightarrow{AD}+\overrightarrow{AC}\right|=\left|\overrightarrow{CB}+\overrightarrow{DA}+\overrightarrow{AC}\right|=\left|\overrightarrow{CB}+\overrightarrow{DC}\right|=\left|\overrightarrow{DB}\right|=AC=2a\sqrt{5}\)

Ủa m đâu bạn? và làm gì có điểm K nào ở đây?

\(\overrightarrow{IJ}=\overrightarrow{AI}+\overrightarrow{AJ}=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)\)

\(\overrightarrow{IJ}=-\frac{1}{6}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\frac{1}{2}\overrightarrow{AB}+3\overrightarrow{IJ}\)

\(\overrightarrow{AK}=\overrightarrow{AI}+\overrightarrow{IK}=\frac{1}{2}\overrightarrow{AB}+m.\overrightarrow{IJ}\)

\(\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{2}\left(\frac{1}{2}\overrightarrow{B}+3\overrightarrow{IJ}\right)\)

\(\overrightarrow{AD}=\frac{5}{4}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{IJ}=\frac{5}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{3}{5}\overrightarrow{IJ}\right)\)

Vậy để A;K;D thẳng hàng \(\Leftrightarrow m=\frac{3}{5}\)

\(\overrightarrow{AG}=\frac{\overrightarrow{AB}+\overrightarrow{AM}+\overrightarrow{AN}}{3}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AM}+\frac{1}{3}\overrightarrow{AN}\)

\(=\frac{1}{3}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AB}+\frac{1}{3}.\frac{1}{2}.\left(\overrightarrow{AC}+\overrightarrow{AD}\text{​​}\right)\)

\(=\frac{4}{9}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}+\frac{1}{6}\overrightarrow{BC}\text{​​}\)

\(=\frac{4}{9}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}+\frac{1}{6}\left(\overrightarrow{AC}-\text{​​}\overrightarrow{AB}\right)\)

\(=\frac{4}{9}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}+\frac{1}{6}\overrightarrow{AC}-\text{​​}\frac{1}{6}\overrightarrow{AB}\)

\(=\frac{5}{18}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=\frac{5}{18}\text{​​}\text{​​}\overrightarrow{a}\text{​​}+\frac{1}{3}\overrightarrow{b}\)

Chưa đủ dữ kiện đề bài để chứng minh đẳng thức. Bạn xem lại đề.

\(\overrightarrow{KD}=\overrightarrow{KA}+\overrightarrow{AD}=-\overrightarrow{AK}+\overrightarrow{AD}\)

\(=-\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)+\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

\(=-\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\right)+\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\)

\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\)

H đối xứng G qua B \(\Rightarrow\) B là trung điểm của HG

\(\Rightarrow\overrightarrow{AG}+\overrightarrow{AH}=2\overrightarrow{AB}\Rightarrow\overrightarrow{AB}=\frac{1}{2}\overrightarrow{AG}+\frac{1}{2}\overrightarrow{AH}\)

Lại có: \(\overrightarrow{AB}+\overrightarrow{AC}=3\overrightarrow{AG}\Rightarrow\overrightarrow{AC}=3\overrightarrow{AG}-\overrightarrow{AB}\)

\(\Rightarrow\overrightarrow{AC}=3\overrightarrow{AG}-\left(\frac{1}{2}\overrightarrow{AG}+\frac{1}{2}\overrightarrow{AH}\right)\)

\(\Rightarrow\overrightarrow{AC}=\frac{5}{2}\overrightarrow{AG}-\frac{1}{2}\overrightarrow{AH}\)