a: CI+BI=CB

=>\(\dfrac{3}{2}BI+BI=CB\)

=>\(\dfrac{5}{2}BI=CB\)

=>\(BI=\dfrac{2}{5}BC\)

=>\(CI=\dfrac{3}{2}\cdot BI=\dfrac{3}{2}\cdot\dfrac{2}{5}CB=\dfrac{3}{5}CB\)

\(\overrightarrow{AI}=\overrightarrow{AB}+\overrightarrow{BI}\)

\(=\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{BC}\)

\(=\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\)

\(=\dfrac{3}{5}\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{AC}\)

JB=2/5JC mà J không nằm trong đoạn thẳng BC

nên B nằm giữa J và C

=>JB+BC=JC

=>\(BC+\dfrac{2}{5}JC=JC\)

=>\(BC=\dfrac{3}{5}JC\)

\(\dfrac{JB}{BC}=\dfrac{\dfrac{2}{5}JC}{\dfrac{3}{5}JC}=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2}{3}\)

=>\(JB=\dfrac{2}{3}BC\)

\(\overrightarrow{AJ}=\overrightarrow{AB}+\overrightarrow{BJ}\)

\(=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{BC}\)

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

\(=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{BA}-\dfrac{2}{3}\overrightarrow{AC}=\dfrac{5}{3}\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AC}\)

b:

Gọi giao điểm của AG với BC là M

G là trọng tâm của ΔABC

nên AG cắt BC tại trung điểm M của BC

=>\(AG=\dfrac{2}{3}AM\)

Xét ΔABC có AM là trung tuyến

nên \(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

=>\(\overrightarrow{AG}=\dfrac{2}{3}\cdot\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)

Đặt \(\overrightarrow{AG}=x\cdot\overrightarrow{AI}+y\cdot\overrightarrow{AJ}\)

\(\overrightarrow{AG}=\dfrac{1}{3}\cdot\overrightarrow{AB}+\dfrac{1}{3}\cdot\overrightarrow{AC};\overrightarrow{AI}=\dfrac{3}{5}\cdot\overrightarrow{AB}+\dfrac{2}{5}\cdot\overrightarrow{AC};\overrightarrow{AJ}=\dfrac{5}{3}\overrightarrow{AB}-\dfrac{2}{3}\cdot\overrightarrow{AC}\)

Ta có hệ phương trình sau:

\(\left\{{}\begin{matrix}\dfrac{1}{3}=x\cdot\dfrac{3}{5}+y\cdot\dfrac{5}{3}\\\dfrac{1}{3}=x\cdot\dfrac{2}{5}+y\cdot\dfrac{-2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\cdot\dfrac{3}{5}+y\cdot\dfrac{5}{3}=\dfrac{1}{3}\\x\cdot\dfrac{2}{5}+y\cdot\dfrac{-2}{3}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+25y=5\\6x-10y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}18x+50y=10\\18x-30y=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}80y=-5\\6x-10y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{1}{16}\\6x=10y+5=-\dfrac{5}{8}+5=\dfrac{35}{8}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{1}{16}\\x=\dfrac{35}{48}\end{matrix}\right.\)

Vậy: \(\overrightarrow{AG}=\dfrac{35}{48}\overrightarrow{AI}-\dfrac{1}{16}\overrightarrow{AJ}\)

\(\overrightarrow{BI}=-\frac{2}{7}\overrightarrow{IC}=-\frac{2}{7}\left(\overrightarrow{BC}-\overrightarrow{BI}\right)\Rightarrow\overrightarrow{BI}=-\frac{2}{5}\overrightarrow{BC}\)

\(\Rightarrow\overrightarrow{AI}=\overrightarrow{AB}+\overrightarrow{BI}=\overrightarrow{AB}-\frac{2}{5}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\frac{5}{2}\overrightarrow{AB}-\frac{5}{2}\overrightarrow{AI}\) (1)

\(\overrightarrow{BJ}=\frac{3}{2}\overrightarrow{IC}=-\frac{3}{7}\overrightarrow{BI}=\frac{6}{35}\overrightarrow{BC}\)

\(\overrightarrow{AJ}=\overrightarrow{AB}+\overrightarrow{BJ}=\overrightarrow{AB}+\frac{6}{35}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\frac{35}{6}\overrightarrow{AJ}-\frac{35}{6}\overrightarrow{AB}\) (2)

Từ (1);(2) ta có:

\(\frac{5}{2}\overrightarrow{AB}-\frac{5}{2}\overrightarrow{AI}=\frac{35}{6}\overrightarrow{AJ}-\frac{35}{6}\overrightarrow{AB}\)

\(\Rightarrow\overrightarrow{AB}=...\)

Quá trình tính có thể nhầm lẫn con số và dấu, bạn kiểm tra lại

Tham khảo

Tham khảo:

Lời giải:

Xử lý điều kiện đề bài:

\(\overrightarrow{NC}+2\overrightarrow{NB}=\overrightarrow{0}\Rightarrow \overrightarrow{BN}+\overrightarrow{NC}+2\overrightarrow{NB}=\overrightarrow{BN}\)

\(\Leftrightarrow \overrightarrow{BC}=\overrightarrow{BN}-2\overrightarrow{NB}=3\overrightarrow{BN}\)

\(\overrightarrow{IN}=2\overrightarrow{AI}\Rightarrow 3\overrightarrow{AI}=\overrightarrow{AI}+\overrightarrow{IN}=\overrightarrow{AN}\)

Áp dụng kết quả trên ta có:

a)

\(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{3}(\overrightarrow{BA}+\overrightarrow{AC})\)

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

\(\overrightarrow{AI}=\frac{1}{3}\overrightarrow{AN}=\frac{1}{3}(\frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC})=\frac{2}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}\)

\(\overrightarrow{BI}=\overrightarrow{AI}-\overrightarrow{AB}=\frac{2}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}-\overrightarrow{AB}=\frac{-7}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}\)

\(\overrightarrow{CI}=\overrightarrow{AI}-\overrightarrow{AC}=\frac{2}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}-\overrightarrow{AC}=\frac{2}{9}\overrightarrow{AB}-\frac{8}{9}\overrightarrow{AC}\)

b)

Với $K\in AC$, tồn tại $m\in\mathbb{R}$ sao cho \(\overrightarrow{AK}=m\overrightarrow{AC}\)

\(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}=-\overrightarrow{AB}+m\overrightarrow{AC}\)

\(\overrightarrow{BI}=\frac{-7}{9}\overrightarrow{AB}+\frac{1}{9}\overrightarrow{AC}\)

Để $B,K,I$ thẳng hàng: \(\frac{-1}{\frac{-7}{9}}=\frac{m}{\frac{1}{9}}\Rightarrow m=\frac{1}{7}\)

Hình vẽ:

Hình vẽ:

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 \(\overrightarrow{BJ}=\overrightarrow{AT}\).


 

a) Nối BM

Ta có AM= AB.cosMAB

=> || = ||.cos(, )

Ta có: . = ||.|| ( vì hai vectơ , cùng phương)

=> . = ||.||.cosAMB.

nhưng ||.||.cos(, ) = .

Vậy . = .

Với . = . lý luận tương tự.

b) . = .

. = .

=> . + . = ( + )

=> . + . = 4R²