Ôn tập chương II

\(\overrightarrow{AB}+\overrightarrow{AC}\) vuông góc \(\overrightarrow{AB}+\overrightarrow{CA}\)

\(\Rightarrow\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\left(\overrightarrow{AB}+\overrightarrow{CA}\right)=0\)

\(\Leftrightarrow\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\left(\overrightarrow{AB}-\overrightarrow{AC}\right)=0\)

\(\Leftrightarrow\overrightarrow{AB}^2-\overrightarrow{AC}^2=0\)

\(\Leftrightarrow AB^2-AC^2=0\)

\(\Leftrightarrow AB=AC\)

Hay tam giác ABC cân tại A

Cách 2: gọi M là trung điểm BC \(\Rightarrow\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AM}\)

Lại có \(\overrightarrow{AB}+\overrightarrow{CA}=\overrightarrow{CB}\)

\(\Rightarrow2\overrightarrow{AM}\perp\overrightarrow{CB}\Rightarrow AM\perp BC\)

\(\Rightarrow\) AM là đường cao đồng thời là trung tuyến

\(\Rightarrow\Delta ABC\) cân tại A

\(3\overrightarrow{MA}+2\overrightarrow{MB}=\overrightarrow{0}\Leftrightarrow3\overrightarrow{MA}+2\left(\overrightarrow{MA}+\overrightarrow{AB}\right)=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{MA}=-\dfrac{2}{5}\overrightarrow{AB}\) \(\Rightarrow\left\{{}\begin{matrix}AM=\dfrac{2}{5}AB\\BM=\dfrac{3}{5}AB\end{matrix}\right.\)

Do CPMQ là hình bình hành \(\Rightarrow\left\{{}\begin{matrix}MP||BC\\MQ||AC\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{AP}{AC}=\dfrac{AM}{AB}=\dfrac{2}{5}\\\dfrac{BQ}{BC}=\dfrac{BM}{BA}=\dfrac{3}{5}\end{matrix}\right.\)

Do A,N,Q thẳng hàng, đặt  \(\overrightarrow{AN}=x.\overrightarrow{AQ}\)

Ta có: \(\overrightarrow{BP}=\overrightarrow{BA}+\overrightarrow{AP}=-\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{AC}\)

\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}=-\overrightarrow{AB}+x.\overrightarrow{AQ}=-\overrightarrow{AB}+x\left(\overrightarrow{AB}+\overrightarrow{BQ}\right)\)

\(=-\overrightarrow{AB}+x.\overrightarrow{AB}+x.\dfrac{3}{5}\overrightarrow{BC}=\left(x-1\right)\overrightarrow{AB}+\dfrac{3}{5}x\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)

\(=\left(\dfrac{2x}{5}-1\right)\overrightarrow{AB}+\dfrac{3x}{5}\overrightarrow{AC}\)

Do B,N,P thẳng hàng \(\Rightarrow\overrightarrow{BN};\overrightarrow{BP}\) tương ứng tỉ lệ

\(\Rightarrow\dfrac{\dfrac{2x}{5}-1}{-1}=\dfrac{\dfrac{3x}{5}}{\dfrac{2}{5}}\Rightarrow x=\dfrac{10}{19}\) 

\(\Rightarrow\overrightarrow{AN}=\dfrac{10}{19}\overrightarrow{AQ}=\dfrac{10}{19}\left(\overrightarrow{AN}+\overrightarrow{NQ}\right)\)

\(\Rightarrow9\overrightarrow{AN}=10\overrightarrow{NQ}\Rightarrow9\overrightarrow{NA}+10\overrightarrow{NQ}=\overrightarrow{0}\)

\(\Rightarrow a+b=19\)

a. Em tự giải

b. 

\(\overrightarrow{DM}=\overrightarrow{DA}+\overrightarrow{AB}+\overrightarrow{BM}=-\overrightarrow{AD}+\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AD}=\overrightarrow{AB}-\dfrac{1}{2}\overrightarrow{AD}\)

\(\overrightarrow{DN}=\overrightarrow{DA}+\overrightarrow{AN}=-\overrightarrow{AD}+x.\overrightarrow{AC}=-\overrightarrow{AD}+x\left(\overrightarrow{AB}+\overrightarrow{AD}\right)=x.\overrightarrow{AB}+\left(x-1\right)\overrightarrow{AD}\)

Do D,M,N thẳng hàng

\(\Rightarrow\dfrac{x}{1}=\dfrac{x-1}{-\dfrac{1}{2}}\Rightarrow x=\dfrac{2}{3}\)

b.

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AD}\)

\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}=-\overrightarrow{AB}+x.\overrightarrow{AC}=-\overrightarrow{AB}+x\left(\overrightarrow{AB}+\overrightarrow{AD}\right)=\left(x-1\right)\overrightarrow{AB}+x.\overrightarrow{AD}\)

\(AM\perp BN\Leftrightarrow\overrightarrow{AM}.\overrightarrow{BN}=0\)

\(\Leftrightarrow\left(\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AD}\right)\left(\left(x-1\right)\overrightarrow{AB}+x.\overrightarrow{AD}\right)=0\)

\(\Leftrightarrow\left(x-1\right).AB^2+\dfrac{1}{2}x.AD^2+x.\overrightarrow{AB}.\overrightarrow{AD}+\dfrac{1}{2}\left(x-1\right)\overrightarrow{AB}.\overrightarrow{AD}=0\)

\(\Leftrightarrow\left(x-1\right)a^2+\dfrac{1}{2}x.a^2+0+0\)

\(\Leftrightarrow x-1+\dfrac{1}{2}x=0\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

16.

a.

Theo tính chất trọng tâm ta có: \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\left(\overrightarrow{MG}+\overrightarrow{GA}\right)+\left(\overrightarrow{MG}+\overrightarrow{GB}\right)+\left(\overrightarrow{MG}+\overrightarrow{GC}\right)\)

\(=3\overrightarrow{MG}+\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)=3\overrightarrow{MG}+\overrightarrow{0}\)

\(=3\overrightarrow{MG}\) (đpcm)

b.

Câu b đề yêu cầu biểu diễn vecto \(\overrightarrow{AN}\) và vecto nào vậy nhỉ?

\(BN=2NC\Rightarrow BN=\dfrac{2}{3}BC\)

\(\Rightarrow\overrightarrow{BN}=\dfrac{2}{3}\overrightarrow{BC}\)

Từ đó: \(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)

c.

Do AH là trung tuyến \(\Rightarrow\overrightarrow{AH}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\Rightarrow\overrightarrow{AH}+\overrightarrow{AC}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\)

Đặt \(T=\left|\overrightarrow{AH}+\overrightarrow{AC}\right|=\left|\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\right|\)

\(\Rightarrow T^2=\dfrac{1}{4}AB^2+\dfrac{9}{4}AC^2+\dfrac{3}{2}\overrightarrow{AB}.\overrightarrow{AC}\)

\(=\dfrac{1}{4}a^2+\dfrac{9}{4}a^2+\dfrac{3}{2}.a.a.cos60^0=\dfrac{13a^2}{4}\)

\(\Rightarrow T=\dfrac{a\sqrt{13}}{2}\)

a) \(2\overrightarrow{CE}+\overrightarrow{EB}=\overrightarrow{0}\Leftrightarrow2\overrightarrow{AE}-2\overrightarrow{AC}+\overrightarrow{AB}-\overrightarrow{AE}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{AE}=-\overrightarrow{AB}+2\overrightarrow{AC}=\overrightarrow{AB}+2\overrightarrow{AD}\)

\(3\overrightarrow{DF}+\overrightarrow{BD}=\overrightarrow{0}\Leftrightarrow3\overrightarrow{AF}-3\overrightarrow{AD}+\overrightarrow{AD}-\overrightarrow{AB}=\overrightarrow{0}\)

\(3\overrightarrow{AF}=\overrightarrow{AB}+2\overrightarrow{AD}\)

Do đó \(\overrightarrow{AE}=3\overrightarrow{AF}\)

Vậy A, E, F thẳng hàng.

b) \(2\overrightarrow{AM}=3\overrightarrow{AF}\Leftrightarrow2\overrightarrow{AM}=\overrightarrow{AB}+2\overrightarrow{AD}\)

\(\Leftrightarrow2\overrightarrow{AM}=\overrightarrow{AC}+\overrightarrow{AD}\)

\(\Leftrightarrow2\overrightarrow{AM}=\overrightarrow{AM}+\overrightarrow{MC}+\overrightarrow{AM}+\overrightarrow{MD}\)

\(\Leftrightarrow\overrightarrow{MC}+\overrightarrow{MD}=\overrightarrow{0}\)

Vậy M là trung điểm của CD.

Chọn C

Gọi (C): x^2+y^2-2ax-2by+c=0 là phương trình đường tròn ngoại tiếp ΔABC

Theo đề, ta có hệ:

\(\left\{{}\begin{matrix}1+4-2a-4b+c=0\\16+1-8a-2b+c=0\\9+16-6a-8b+c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2a-4b+c=-5\\-8a-2b+c=-17\\-6a-8b+c=-25\end{matrix}\right.\)

=>a=11/4; b=9/4; c=19/2

=>(C): x^2+y^2-11/2x-9/2y+19/2=0

=>x^2-2*x*11/4+121/16+y^2-2*y*9/4+81/16=755/48

=>(x-11/4)^2+(y-9/4)^2=755/48

=>\(R=\sqrt{\dfrac{755}{48}}\)