Ôn tập chương I

Do 2 vecto có giá vuông góc nên tích vô hướng của chúng bằng 0

Hay \(\overrightarrow{a}.\overrightarrow{b}=0\)

Đặt \(S=\left|\overrightarrow{a}-\overrightarrow{b}\right|\)

\(\Rightarrow S^2=a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}=6^2+8^2-2.0=100\)

\(\Rightarrow S=10\)

1.

Do tung độ của 2 vecto cùng dấu nên 2 vecto cùng hướng khi tọa độ của chúng tương ứng tỉ lệ, hay:

\(\dfrac{m}{1}=\dfrac{6}{2}\Rightarrow m=3\)

Do \(3\in\left(2;4\right)\) nên B là đáp án đúng

2.

\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(2;2\right)\\\overrightarrow{AC}=\left(-6;m-2\right)\end{matrix}\right.\)

3 điểm A,B,C thẳng hàng khi hai vecto trên cùng phương hay tọa độ của chúng tương ứng tỉ lệ:

\(\dfrac{-6}{2}=\dfrac{m-2}{2}\Rightarrow m-2=-6\Rightarrow m=-4\in\left(-5;-2\right)\)

5.

Gọi M là trung điểm BC

\(\Rightarrow\left\{{}\begin{matrix}x_M=\dfrac{x_B+x_C}{2}=\dfrac{2+\left(-1\right)}{2}=\dfrac{1}{2}\\y_M=\dfrac{y_B+y_C}{2}=\dfrac{1+\left(-2\right)}{2}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow M\left(\dfrac{1}{2};-\dfrac{1}{2}\right)\)

Gọi G là trọng tâm tam giác

\(\Rightarrow\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{4}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=1\end{matrix}\right.\) \(\Rightarrow G\left(\dfrac{4}{3};1\right)\)

b.

Gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-1;-3\right)\\\overrightarrow{DC}=\left(-1-x;-2-y\right)\end{matrix}\right.\)

ABCD là hbh khi

\(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}-1-x=-1\\-2-y=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

\(\Rightarrow D\left(0;1\right)\)

6.

\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{1}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=0\end{matrix}\right.\)

\(\Rightarrow G\left(\dfrac{1}{3};0\right)\)

b.

Gọi \(D\left(x;y\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{BG}=\left(-\dfrac{26}{3};10\right)\\\overrightarrow{DC}=\left(-5-x;4-y\right)\end{matrix}\right.\)

BGCD là hình bình hành khi:

\(\overrightarrow{BG}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}-5-x=-\dfrac{26}{3}\\4-y=10\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=-6\end{matrix}\right.\)

\(\Rightarrow D\left(\dfrac{11}{3}=-6\right)\)

Câu 3: 

\(\Leftrightarrow\overrightarrow{AB}=2\overrightarrow{AM}\)

=>M là trung điểm của AB

Do M là trung điểm BC nên: \(\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)

Tương tự: \(\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{BC}\) ; \(\overrightarrow{CP}=\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)

Cộng vế:

\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)

\(=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BA}\right)+\dfrac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)+\dfrac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CB}\right)=\overrightarrow{0}\)

b. Từ câu a ta có:

\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{AO}+\overrightarrow{OM}+\overrightarrow{BO}+\overrightarrow{ON}+\overrightarrow{CO}+\overrightarrow{OP}=\overrightarrow{0}\)

\(\Leftrightarrow-\overrightarrow{OA}+\overrightarrow{OM}-\overrightarrow{OB}+\overrightarrow{ON}-\overrightarrow{OC}+\overrightarrow{OP}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OM}+\overrightarrow{ON}+\overrightarrow{OP}\) (đpcm)

Do G là trọng tâm tam giác 

\(\Rightarrow\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AD}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=\dfrac{1}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{AC}\)

\(=\dfrac{2}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}=-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)

Do I là trung điểm AG

\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AG}=\dfrac{1}{2}\left(-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\right)=-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)

\(\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{AB}=\dfrac{1}{5}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)=-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)

\(\overrightarrow{CI}=\overrightarrow{CA}+\overrightarrow{AI}=\overrightarrow{CA}-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)

\(\overrightarrow{CK}=\overrightarrow{CA}+\overrightarrow{AK}=\overrightarrow{CA}-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}=\dfrac{4}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)

a: \(\overrightarrow{CN}=\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)

\(=\dfrac{1}{2}\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{CB}\)

\(=\dfrac{1}{2}\overrightarrow{u}-\overrightarrow{v}\)

\(\left|3\overrightarrow{AB}+\overrightarrow{AC}\right|=4a\)