Cho tam giác ABC có trọng tâm G. Gọi H đối xứng của B qua G. Cminh:
\(\overrightarrow{AH}=\frac{2}{3}\overrightarrow{AC}-\frac{1}{3}\overrightarrow{AB}\)
\(\overrightarrow{CH}=\frac{-1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 1 2021
H đối xứng B qua G \(\Rightarrow\overrightarrow{BH}=2\overrightarrow{BG}=2\left(\dfrac{1}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\right)=-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(\overrightarrow{AH}=\overrightarrow{AB}+\overrightarrow{BH}=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}=\dfrac{2}{3}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}\)
\(\overrightarrow{CH}=\overrightarrow{CA}+\overrightarrow{AH}=-\overrightarrow{AC}+\dfrac{2}{3}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}=-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{MH}=\overrightarrow{MA}+\overrightarrow{AH}=-\dfrac{1}{2}\overrightarrow{AB}-\dfrac{1}{2}\overrightarrow{AC}+\dfrac{2}{3}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}\)
\(=-\dfrac{5}{6}\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AC}\)
10 tháng 8 2019
\(\text{a) }\overrightarrow{AH}=\overrightarrow{AG}+\overrightarrow{GH}=\overrightarrow{AG}+\overrightarrow{BG}=\frac{1}{3}\left(3\overrightarrow{AG}+3\overrightarrow{BG}\right)\\ =\frac{1}{3}\left(\overrightarrow{AA}+\overrightarrow{AC}+\overrightarrow{AB}+\overrightarrow{BA}+\overrightarrow{BC}+\overrightarrow{BB}\right)\\ =\frac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{BC}\right)=\frac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{BA}+\overrightarrow{AC}\right)\\ =\frac{1}{3}\left(2\overrightarrow{AC}-\overrightarrow{AB}\right)=\frac{2}{3}\overrightarrow{AC}-\frac{1}{3}\overrightarrow{AB}\)
\(\text{b) }\overrightarrow{CH}=\overrightarrow{CA}+\overrightarrow{AH}=-\overrightarrow{AC}+\frac{2}{3}\overrightarrow{AC}-\frac{1}{3}\overrightarrow{AB}\\ =-\frac{1}{3}\overrightarrow{AC}-\frac{1}{3}\overrightarrow{AB}=-\frac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{AB}\right)\)
\(\text{c) }\overrightarrow{MH}=\overrightarrow{MC}+\overrightarrow{CH}=\frac{1}{2}\overrightarrow{BC}-\frac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{AB}\right)\\ =\frac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)-\frac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{AB}\right)\\ =-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}-\frac{1}{3}\overrightarrow{AC}-\frac{1}{3}\overrightarrow{AB}\\ =\frac{1}{6}\overrightarrow{AB}-\frac{5}{6}\overrightarrow{AB}\)
HQ
Hà Quang Minh
Giáo viên
24 tháng 9 2023
Dễ thấy: \(\overrightarrow {BC} = \overrightarrow {BA} + \overrightarrow {AC} = - \overrightarrow {AB} + \overrightarrow {AC} \)
Ta có:
+) \(\overrightarrow {AD} = \overrightarrow {AB} + \overrightarrow {BD} \). Mà \(\overrightarrow {BD} = - \overrightarrow {DB} = - \frac{1}{3}\overrightarrow {BC} \)
\( \Rightarrow \overrightarrow {AD} = \overrightarrow {AB} + \left( { - \frac{1}{3}} \right)( - \overrightarrow {AB} + \overrightarrow {AC} ) = \frac{4}{3}\overrightarrow {AB} - \frac{1}{3}\overrightarrow {AC} \)
+) \(\overrightarrow {DH} = \overrightarrow {DA} + \overrightarrow {AH} = - \overrightarrow {AD} + \overrightarrow {AH} \).
Mà \(\overrightarrow {AD} = \frac{4}{3}\overrightarrow {AB} - \frac{1}{3}\overrightarrow {AC} ;\;\;\overrightarrow {AH} = \frac{2}{3}\overrightarrow {AB} .\)
\( \Rightarrow \overrightarrow {DH} = - \left( {\frac{4}{3}\overrightarrow {AB} - \frac{1}{3}\overrightarrow {AC} } \right) + \frac{2}{3}\overrightarrow {AB} = - \frac{2}{3}\overrightarrow {AB} + \frac{1}{3}\overrightarrow {AC} .\)
+) \(\overrightarrow {HE} = \overrightarrow {HA} + \overrightarrow {AE} = - \overrightarrow {AH} + \overrightarrow {AE} \)
Mà \(\overrightarrow {AH} = \frac{2}{3}\overrightarrow {AB} ;\;\overrightarrow {AE} = \frac{1}{3}\overrightarrow {AC} \)
\( \Rightarrow \overrightarrow {HE} = - \frac{2}{3}\overrightarrow {AB} + \frac{1}{3}\overrightarrow {AC} .\)
b)
Theo câu a, ta có: \(\overrightarrow {DH} = \overrightarrow {HE} = - \frac{2}{3}\overrightarrow {AB} + \frac{1}{3}\overrightarrow {AC} \)
\( \Rightarrow \) Hai vecto \(\overrightarrow {DH} ,\overrightarrow {HE} \) cùng phương.
\( \Leftrightarrow \)D, E, H thẳng hàng