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a)
\(\overrightarrow{AK}=\overrightarrow{AI}+\overrightarrow{IK}=\overrightarrow{AI}+\dfrac{1}{2}\overrightarrow{IB}=\overrightarrow{AI}+\dfrac{1}{2}\left(\overrightarrow{IA}+\overrightarrow{AB}\right)\)
\(=\overrightarrow{AI}+\dfrac{1}{2}\overrightarrow{IA}+\dfrac{1}{2}\overrightarrow{AB}\)\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}\).
b) Theo câu a:
\(\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}.\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\).
Lời giải:
\(\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BN}=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}\)
\(=\overrightarrow{MB}+\overrightarrow{BC}+2\overrightarrow{BC}=\overrightarrow{MB}+3\overrightarrow{BC}\)
\(=\overrightarrow{MA}+\overrightarrow{AB}+3(\overrightarrow{BA}+\overrightarrow{AC})\)
\(=-\overrightarrow{AM}+\overrightarrow{AB}-3\overrightarrow{AB}+3\overrightarrow{AC}\)
\(=-\frac{1}{3}\overrightarrow{AB}+\overrightarrow {AB}-3\overrightarrow{AB}+3\overrightarrow{AC}\)
\(=\frac{-7}{3}\overrightarrow{AB}+3\overrightarrow{AC}\)
Ta có đpcm.
Câu 3:
\(\left|\overrightarrow{AC}+\overrightarrow{AH}\right|=\sqrt{AC^2+AH^2+2\cdot AC\cdot AH\cdot cos30}\)
\(=\sqrt{a^2+\left(\dfrac{a\sqrt{3}}{2}\right)^2+2\cdot a\cdot\dfrac{a\sqrt{3}}{2}\cdot\dfrac{\sqrt{3}}{2}}\)
\(=\sqrt{a^2+\dfrac{3}{4}a^2+\dfrac{3a^2}{4}}=\dfrac{\sqrt{7}}{2}a\)
Xét \(\Delta ABC\) có:
\(M\) là trung điểm \(AB\)
\(D\) là trung điểm \(BC\)
\(\Rightarrow\) \(MD\) là đường trung bình của \(\Delta ABC\)
\(\Rightarrow\) \(MD\)\(=\)\(\dfrac{1}{2}AC\) và \(MD\) //\(AC\)
Ta có:
\(\overrightarrow{KD}=\overrightarrow{KM}+\overrightarrow{MD}\)
\(\Rightarrow\overrightarrow{KD}=\dfrac{1}{2}\overrightarrow{NM}+\dfrac{1}{2}\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{KD}=\dfrac{1}{2}\overrightarrow{NA}+\dfrac{1}{2}\overrightarrow{AM}+\dfrac{1}{2}\overrightarrow{AC}=\dfrac{1}{6}\overrightarrow{CA}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\\ \Rightarrow\overrightarrow{KD}=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
a) \(\overrightarrow{BI}=\overrightarrow{BC}+\overrightarrow{CI}=\overrightarrow{BC}+\dfrac{1}{4}\overrightarrow{CA}=\overrightarrow{BA}+\overrightarrow{AC}+\dfrac{1}{4}\overrightarrow{CA}\)
\(=\overrightarrow{BA}+\overrightarrow{AC}-\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AC}+\overrightarrow{BA}=\dfrac{3}{4}\overrightarrow{AC}-\overrightarrow{AB}\).
b) Có \(\overrightarrow{BJ}=\dfrac{1}{2}\overrightarrow{AC}-\dfrac{2}{3}\overrightarrow{AB}=\dfrac{3}{2}\left(\dfrac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}\right)=\dfrac{3}{2}\overrightarrow{BI}\).
Vì vậy 3 điểm B, I, J thẳng hàng.
c)
Trên cạnh AC lấy điểm K sao cho \(\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AC}\).
Tại điểm K dựng điểm T sao cho \(\overrightarrow{KT}=-\dfrac{3}{2}\overrightarrow{AB}=\dfrac{3}{2}\overrightarrow{BA}\).
\(\overrightarrow{BJ}=\dfrac{1}{2}\overrightarrow{AC}-\dfrac{3}{2}\overrightarrow{AB}=\overrightarrow{AK}+\overrightarrow{KT}=\overrightarrow{AT}\).
Dựng điểm T sao cho \(\overrightarrow{BJ}=\overrightarrow{AT}\).
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
a)Ta có:
\(\overrightarrow{OA}+\overrightarrow{OM}+\overrightarrow{ON}=\overrightarrow{CO}+\dfrac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OC}+\overrightarrow{OD}\right)\)
\(=\overrightarrow{CO}+\dfrac{1}{2}.2\overrightarrow{OC}\)
\(=\overrightarrow{0}\)
\(\RightarrowĐPCM\)
b) Ta có:
\(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AD}+2\overrightarrow{AB}\right)\)
\(\Rightarrow2\overrightarrow{AM}=\overrightarrow{AD}+2\overrightarrow{AB}\) (1)
Mà \(2\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{AC}\)(2)
Từ (1)(2) =>\(\overrightarrow{AD}+2\overrightarrow{AB}=\overrightarrow{AB}+\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AC}+\overrightarrow{AB}=\overrightarrow{AB}+\overrightarrow{AC}\)
\(\RightarrowĐPCM\)
\(\overrightarrow{AD}=2\overrightarrow{DB}\Rightarrow\overrightarrow{AD}=\dfrac{2}{3}\overrightarrow{AB}\) ; \(\overrightarrow{CE}=3\overrightarrow{EA}\Rightarrow\overrightarrow{AE}=\dfrac{1}{4}\overrightarrow{AC}\)
Lại có M là trung điểm DE
\(\Rightarrow\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{AE}\right)=\dfrac{1}{2}\left(\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{8}\overrightarrow{AC}\)
I là trung điểm BC \(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{MI}=\overrightarrow{MA}+\overrightarrow{AI}=\overrightarrow{AI}-\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{8}\overrightarrow{AC}=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)