Chương II: TÍCH VÔ HƯỚNG CỦA HAI VÉC TƠ VÀ ỨNG DỤNG

Gọi cạnh của tam giác là a

\(GB=GC=\dfrac{2}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{3}\)

Áp dụng định lý cosin cho tam giác BCG:

\(cos\widehat{BGC}=\dfrac{GB^2+GC^2-BC^2}{2GB.GC}=\dfrac{2.\left(\dfrac{a\sqrt{3}}{3}\right)^2-a^2}{2\left(\dfrac{a\sqrt{3}}{3}\right)^2}=-\dfrac{1}{2}\)

\(\Rightarrow\widehat{BGC}=120^0\)

\(4MO^2=AB^2\Leftrightarrow\left(2\overrightarrow{MO}\right)^2=\left(\overrightarrow{AM}+\overrightarrow{MB}\right)^2\)

\(\Leftrightarrow\left(\overrightarrow{MA}+\overrightarrow{MB}\right)^2=\left(\overrightarrow{AM}+\overrightarrow{MB}\right)^2\)

\(\Leftrightarrow MA^2+MB^2+2\overrightarrow{MA}.\overrightarrow{MB}=AM^2+BM^2+2\overrightarrow{AM}.\overrightarrow{MB}\)

\(\Leftrightarrow4\overrightarrow{MA}.\overrightarrow{MB}=0\Leftrightarrow\overrightarrow{MA}.\overrightarrow{MB}=0\)

\(\Leftrightarrow MA\perp MB\)

1.

\(\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AB}+\dfrac{c}{b+c}\overrightarrow{BC}=\dfrac{\left(b+c\right)\overrightarrow{AB}+c\overrightarrow{BC}}{b+c}=\dfrac{b\overrightarrow{AB}+c\overrightarrow{AC}}{b+c}\)

\(\Rightarrow AD^2=\dfrac{\left(b\overrightarrow{AB}+c\overrightarrow{AC}\right)^2}{\left(b+c\right)^2}=\dfrac{2b^2c^2+2b^2c^2.cosA}{\left(b+c\right)^2}=\dfrac{2b^2c^2\left(1+cos\alpha\right)}{\left(b+c\right)^2}\)

\(\Rightarrow AD=\dfrac{bc\sqrt{2+2cos\alpha}}{b+c}\)

2.

\(MA^2+MB^2+MC^2=\left(\overrightarrow{MG}+\overrightarrow{GA}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GB}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GC}\right)^2\)

\(=3MG^2+GA^2+GB^2+GC^2+2\overrightarrow{MG}\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)\)

\(=3MG^2+GA^2+GB^2+GC^2\)

\(=3MG^2+\dfrac{4}{9}\left(AM^2+MB^2+MC^2\right)\)

\(=3MG^2+\dfrac{4}{9}\left(\dfrac{2b^2+2c^2-a^2}{4}+\dfrac{2a^2+2c^2-b^2}{4}+\dfrac{2a^2+2b^2-c^2}{4}\right)\)

\(=3MG^2+\dfrac{4}{9}.\dfrac{3}{4}\left(a^2+b^2+c^2\right)\)

\(=3MG^2+\dfrac{1}{3}\left(a^2+b^2+c^2\right)\)

3.

Hình vẽ:

Đặt các vecto đơn vị \(\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}\) cùng hướng \(\overrightarrow{AB};\overrightarrow{BC};\overrightarrow{CA}\)

Khi đó \(\left(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right)^2=3-2\left(cosA+cosB+cosC\right)=3-2P\)

\(\Rightarrow3-2P=\left(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right)^2\ge0\Rightarrow P\le\dfrac{3}{2}\)

\(maxP=\dfrac{3}{2}\Leftrightarrow\Delta ABC\) đều

B đối xứng với A qua I \(\Leftrightarrow I\) là trung điểm AB

\(\Rightarrow\left\{{}\begin{matrix}x_B=2x_I-x_A=9\\y_B=2y_I-y_A=4\end{matrix}\right.\) \(\Rightarrow B\left(9;4\right)\)

\(\left\{{}\begin{matrix}\overrightarrow{AC}=\left(3;y-3\right)\\\overrightarrow{BC}=\left(-4;y-4\right)\end{matrix}\right.\)

\(ABC\) vuông tại C \(\Leftrightarrow\overrightarrow{AB}.\overrightarrow{AC}=0\)

\(\Leftrightarrow-12+\left(y-3\right)\left(y-4\right)=0\)

\(\Leftrightarrow...\)

Có 7 vecto thỏa mãn đề bài: \(\overrightarrow{MA};\overrightarrow{PN};\overrightarrow{NP};\overrightarrow{MB};\overrightarrow{BM};\overrightarrow{AB};\overrightarrow{BA}\)