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A B C K I
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:
\(\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}.\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\).
A B C D I M
a)
\(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)=\dfrac{1}{2}\left(\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\right)=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\).
b)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+x\overrightarrow{BC}\)\(=\overrightarrow{AB}+x\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\left(1-x\right)\overrightarrow{AB}+x\overrightarrow{AC}\).
c) A, M, I thẳng hàng khi và chỉ khi hai véc tơ \(\overrightarrow{AM};\overrightarrow{AI}\) cùng phương
hay \(\dfrac{1-x}{\dfrac{1}{2}}=\dfrac{x}{\dfrac{3}{8}}\Leftrightarrow\dfrac{3}{8}\left(1-x\right)=\dfrac{1}{2}x\)
\(\Leftrightarrow\dfrac{7}{8}x=\dfrac{3}{8}\)\(\Leftrightarrow x=\dfrac{3}{7}\).
Thay vì \(\alpha;\beta;\gamma\) khó gõ kí tự, mình chuyển thành \(a,b,c\) cho dễ, bạn tự thay lại.
Do ABCD là hbh \(\Rightarrow\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}\)
- Chứng minh chiều thuận: I, F, K thẳng hàng \(\Rightarrow\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\)
Do I, F, K thẳng hàng \(\Rightarrow\) tồn tại một số \(k\ne0\) để \(\overrightarrow{KF}=k.\overrightarrow{KI}\)
\(\Rightarrow\left(\overrightarrow{KA}+\overrightarrow{AF}\right)=k.\left(\overrightarrow{KA}+\overrightarrow{AI}\right)\Rightarrow\left(-c.\overrightarrow{AD}+b.\overrightarrow{AC}\right)=k\left(-c.\overrightarrow{AD}+a.\overrightarrow{AB}\right)\)
\(\Rightarrow\overrightarrow{AD}\left(ck-c\right)=k.a.\overrightarrow{AB}-b.\overrightarrow{AC}=ka.\overrightarrow{AB}-b.\overrightarrow{AB}-b.\overrightarrow{AD}\)
\(\Rightarrow\overrightarrow{AD}\left(ck-c+b\right)=\overrightarrow{AB}\left(ka-b\right)\) (1)
Do \(\overrightarrow{AD};\overrightarrow{AB}\) không cùng phương \(\Rightarrow\left(1\right)\) xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}ck-c+b=0\\ka-b=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=\dfrac{c-b}{c}\\k=\dfrac{b}{a}\end{matrix}\right.\)
\(\Rightarrow\dfrac{c-b}{c}=\dfrac{b}{a}\Rightarrow1=\dfrac{b}{a}+\dfrac{b}{c}\Rightarrow\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\) (đpcm)
- Chứng minh chiều nghịch: \(\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\Rightarrow\) I, F, K thẳng hàng
\(\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\Rightarrow b=\dfrac{ac}{a+c}\)
\(\overrightarrow{FI}=\overrightarrow{FA}+\overrightarrow{AI}=-b.\overrightarrow{AC}+a.\overrightarrow{AB}=-b\left(\overrightarrow{AB}+\overrightarrow{AD}\right)+a.\overrightarrow{AB}\)
\(\Rightarrow\overrightarrow{FI}=-\dfrac{ac}{a+c}\overrightarrow{AB}-\dfrac{ac}{a+c}\overrightarrow{AD}+a.\overrightarrow{AB}=\dfrac{a^2}{a+c}\overrightarrow{AB}-\dfrac{ac}{a+c}\overrightarrow{AD}\)
\(\Rightarrow\overrightarrow{FI}=\dfrac{a}{a+c}\left(a.\overrightarrow{AB}-c.\overrightarrow{AD}\right)\) (1)
Lại có \(\overrightarrow{KI}=\overrightarrow{KA}+\overrightarrow{AI}=-c.\overrightarrow{AD}+a.\overrightarrow{AB}=a.\overrightarrow{AB}-c.\overrightarrow{AD}\) (2)
Từ (1), (2) \(\Rightarrow\overrightarrow{FI}=\dfrac{a}{a+c}\overrightarrow{KI}\) ; mà \(\dfrac{a}{a+c}\) là hằng số \(\ne0\)
\(\Rightarrow F,I,K\) thẳng hàng (đpcm)
Vậy F, I, K thẳng hàng khi và chỉ khi \(\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\)