Lời giải:

a)

\(\bullet \overrightarrow{IM}=\frac{1}{2}\overrightarrow{BM}=\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{AM})=\frac{1}{2}(\overrightarrow{BA}+\frac{1}{2}\overrightarrow{AC})\)

\(=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}\)

\(\bullet \overrightarrow{AI}=\overrightarrow{AM}+\overrightarrow{MI}=\frac{1}{2}\overrightarrow{AC}-\overrightarrow{IM}=\frac{1}{2}\overrightarrow{AC}-(-\frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC})\)

\(=\frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}\)

b)

Để \(\overline{A,I,K}\) thì tồn tại \(m\in\mathbb{R}|\overrightarrow{AI}=m\overrightarrow{AK}\)

\(\Leftrightarrow \overrightarrow{AI}=m(\overrightarrow{AB}+\overrightarrow{BK})\)

\(\Leftrightarrow \overrightarrow{AI}=m(\overrightarrow{AB}+x\overrightarrow{BC})\)

\(\Leftrightarrow \overrightarrow{AI}=m\overrightarrow{AB}+mx(\overrightarrow{BA}+\overrightarrow{AC})\)

\(\Leftrightarrow \frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}=(m-mx)\overrightarrow{AB}+mx\overrightarrow{AC}\)

\(\Rightarrow m-mx=\frac{1}{2}; mx=\frac{1}{4}\Rightarrow m=\frac{3}{4}; x=\frac{1}{3}\)

b) giả sử ta có A, I, K thẳng hàng=> ta có tỉ lệ \(\dfrac{AI}{AK}\)(1)

AK= AB+ BK

AK= AB+ xBC

AK= AB+ xBA+ x AC

AK= (1-x) AB+ xAC(2)

mà từ câu a) ta đã tìm được AI= 1/2AB+ 1/4AC(3)

từ (1), (2) và (3)=> \(\dfrac{1}{2-2x}=\dfrac{1}{4x}\)=> x=1/3

\(\overrightarrow{NB}=-3\overrightarrow{NM}\Rightarrow\frac{\overrightarrow{NB}}{\overrightarrow{NM}}=-3\)

\(\overrightarrow{MA}=2\overrightarrow{MC}\Rightarrow\overrightarrow{MA}=-2\overrightarrow{AC}\Rightarrow\frac{\overrightarrow{MA}}{\overrightarrow{AC}}=-2\)

Áp dụng định lý Menelaus cho tam giác BCM:

\(\frac{\overrightarrow{NB}}{\overrightarrow{NM}}.\frac{\overrightarrow{MA}}{\overrightarrow{AC}}.\frac{\overrightarrow{CP}}{\overrightarrow{PB}}=1\Leftrightarrow\left(-3\right).\left(-2\right).\frac{\overrightarrow{CP}}{\overrightarrow{PB}}=1\)

\(\Leftrightarrow\overrightarrow{PB}=6\overrightarrow{CP}\Rightarrow\overrightarrow{PB}=-6\overrightarrow{PC}\Rightarrow k=-6\)

\(\overrightarrow{KA}=-\overrightarrow{AK}=-\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=-\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)\)

\(=-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)

\(\overrightarrow{KD}=\overrightarrow{AD}-\overrightarrow{AK}=\overrightarrow{AD}+\overrightarrow{KA}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)

\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)

a/ \(\overrightarrow{AB}\)=AB=6cm,\(\overrightarrow{BC}\)=BC

Áp dụng định lý Pythagore cho \(\Delta\)ABC vuông ở A:

BC²=AB²+AC²=6²+8²=100cm=>BC=10cm

Vậy \(\overrightarrow{AB}\)=6cm,\(\overrightarrow{BC}\)10cm

b/Ta có:\(\overrightarrow{BM}\)=\(\overrightarrow{BA}\)+\(\overrightarrow{AM}\)=6+4=10cm

Vậy \(\overrightarrow{BM}\)=10cm

c/Vẽ MN//AE (N thuộc BC), vì M là trung điểm AC nên MN là đường trung bình \(\Delta\)ACE

=>MN=1/2.AE

Mặt khác I là trung điểm BM nên IE là đường trung bình \(\Delta\)BMN

=>IE=1/2.MN

Từ đó suy ra IE=1/4.AE=>AI=3/4AE

\(\Delta\)ABM vuông tại A có AI là đường trung tuyến cho nên AI=1/2BM=5cm

=>AE=20/3 cm=>\(\overrightarrow{AE}\)=20/3 cm

=>\(\overrightarrow{BE}\)=\(\overrightarrow{BA}\)+\(\overrightarrow{AE}\)=6+20/3=38/3 cm

Vậy \(\overrightarrow{BE}\)=38/3 cm

P/s: em mới học vector nên có nhầm lẫn xin mn bỏ qua

câu b) đáp an p là 2\(\sqrt{13}\)chứ bạn

BM=\(\sqrt{AB^2+AM^2}\)

= 6\(^2\)+4\(^2\)

a/ \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{BF}+\overrightarrow{FE}+\overrightarrow{CD}+\overrightarrow{DF}\)

\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{ED}+\overrightarrow{DF}+\overrightarrow{FE}\)

\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{EF}+\overrightarrow{FE}\)

\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}\)

b/ Theo tính chất trung tuyến:

\(\left\{{}\begin{matrix}\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AK}\\\overrightarrow{BA}+\overrightarrow{BC}=2\overrightarrow{BM}\end{matrix}\right.\) \(\Rightarrow\overrightarrow{AC}+\overrightarrow{BC}=2\overrightarrow{AK}+2\overrightarrow{BM}\)

\(\overrightarrow{AC}=\overrightarrow{AK}+\overrightarrow{KC}=\overrightarrow{AK}+\frac{1}{2}\overrightarrow{BC}\)

\(\Rightarrow\overrightarrow{BC}=\overrightarrow{AK}+2\overrightarrow{BM}-\frac{1}{2}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\frac{2}{3}\overrightarrow{AK}+\frac{4}{3}\overrightarrow{BM}\)

\(\Rightarrow\overrightarrow{AC}=\overrightarrow{AK}+\frac{1}{2}\left(\frac{3}{2}\overrightarrow{AK}+\frac{4}{3}\overrightarrow{BM}\right)=...\)

\(\overrightarrow{AB}=\overrightarrow{AC}-\overrightarrow{BC}=...\)

a) ta có : \(\left\{{}\begin{matrix}x_{AB}=x_B-x_A=0-1=-1\\y_{AB}=y_B-y_A=4-2=2\end{matrix}\right.\) \(\Rightarrow\overrightarrow{AB}\left(-1;2\right)\)

ta có : \(\left\{{}\begin{matrix}x_{BC}=x_C-x_B=3-0=3\\y_{BC}=y_C-y_B=2-4=-2\end{matrix}\right.\) \(\Rightarrow\overrightarrow{BC}\left(3;-2\right)\)

ta có : \(\left\{{}\begin{matrix}x_{AC}=x_C-x_A=3-1=2\\y_{AC}=y_C-y_A=2-2=0\end{matrix}\right.\) \(\Rightarrow\overrightarrow{AC}\left(2;0\right)\)

b) độ dài : \(\left\{{}\begin{matrix}AB=\sqrt{\left(x_{AB}\right)^2+\left(y_{AB}\right)^2}=\sqrt{\left(-1\right)^2+2^2}=\sqrt{5}\\AC=\sqrt{\left(x_{AC}\right)^2+\left(y_{AC}\right)^2}=\sqrt{2^2+0^2}=2\\BC=\sqrt{\left(x_{BC}\right)^2+\left(y_{BC}\right)^2}=\sqrt{3^2+\left(-2\right)^2}=\sqrt{13}\end{matrix}\right.\)

c) tọa độ trung điểm I của AB là \(\left\{{}\begin{matrix}x_I=\dfrac{x_a+x_b}{2}=\dfrac{1+0}{2}=\dfrac{1}{2}\\y_I=\dfrac{y_a+y_b}{2}=\dfrac{2+4}{2}=3\end{matrix}\right.\)

\(\Rightarrow I\left(\dfrac{1}{2};3\right)\)