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![](https://rs.olm.vn/images/avt/0.png?1311)
1.
Gọi G là trọng tâm tam giác
\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{OG}=\overrightarrow{0}\)
\(\Leftrightarrow O\equiv G\)
\(\Rightarrow O\) là trọng tâm tam giác ABC
\(\Rightarrow\Delta ABC\) đều
Gọi độ dài các cạnh tam giác là a
\(\overrightarrow{BN}.\overrightarrow{AM}=\dfrac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=-\dfrac{1}{4}a^2-\dfrac{1}{8}a^2-\dfrac{1}{8}a^2+\dfrac{1}{2}a^2=0\)
Mặt khác \(\overrightarrow{BN}.\overrightarrow{AM}=BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)\)
\(\Rightarrow BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow\left(\overrightarrow{AM};\overrightarrow{BN}\right)=90^o\)
\(BD=\dfrac{AB}{cos45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)
\(\overrightarrow{BQ}.\overrightarrow{BP}=\dfrac{1}{4}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\left(\overrightarrow{BC}+\overrightarrow{BD}\right)\)
\(=\dfrac{1}{4}BA.BC.cos90^o+\dfrac{1}{4}BA.BD.cos45^o+\dfrac{1}{4}BD.BC.cos45^o+\dfrac{1}{4}BD^2\)
\(=\dfrac{1}{4}a^2+\dfrac{1}{4}a^2+\dfrac{1}{2}a^2=a^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(\left|\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OB}+\overrightarrow{OD}\right|=\left|\overrightarrow{0}+\overrightarrow{0}\right|=0\)
b/ \(\left|\overrightarrow{OA}+\overrightarrow{OB}\right|+\left|\overrightarrow{OC}+\overrightarrow{OD}\right|=a+a=2a\)
c/
\(\left|\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OB}\right|+\left|\overrightarrow{OD}\right|=\left|\overrightarrow{OB}\right|+\left|\overrightarrow{OD}\right|=2\left|\overrightarrow{OB}\right|=2\sqrt{a^2-\frac{a^2}{4}}=a\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
tam giác OAB vuông cân tại O \(\Rightarrow\)OA = OB = a.
2OA - OB = 2OA - OA = OA =a
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
Dựng hình bình hành ABCD \(\Rightarrow\left|\overrightarrow{BM}+\overrightarrow{BA}\right|=\left|\overrightarrow{MC}+\overrightarrow{CD}\right|=MD\)
Hạ ME vuông góc với CD \(\Rightarrow CE=ME=\frac{1}{2}AC\) và \(DE=CD+CE\)
\(\Delta ABC\) vuông cân tại A, theo Pytago ta có:
\(AC=\frac{\sqrt{BC^2}}{2}=a\)
\(\Rightarrow ME=\frac{a}{2}\) và \(DE=CE+CD=\frac{a}{2}+a=\frac{3a}{2}\)
\(\Delta EDM\) vuông tại E, theo Pytago ta có:
\(MD=\sqrt{ME^2+ED^2}=\sqrt{\frac{a^2}{4}+\frac{9a^2}{4}}=\frac{a\sqrt{10}}{2}\)
Câu 2:
Dựng \(\overrightarrow{OC}=\frac{11}{4}\overrightarrow{OA}\Rightarrow OC=\frac{11}{4}a\), \(\overrightarrow{OD}=\frac{3}{7}\overrightarrow{OB}\Rightarrow OD=\frac{3}{7}a\)
Ta có:
\(\left|\overrightarrow{v}\right|=\left|\frac{11}{4}\overrightarrow{OA}-\frac{3}{7}\overrightarrow{OB}\right|=\left|\overrightarrow{OC}-\overrightarrow{OD}\right|=\left|\overrightarrow{DC}\right|=DC\)
Tam giác OCD vuông tại O, theo Pytago, ta có:
\(DC=\sqrt{OD^2+OC^2}=\sqrt{\frac{9a^2}{49}+\frac{121a^2}{16}}\)\(=a\sqrt{\frac{6073}{784}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|\overrightarrow{OA}-\overrightarrow{CB}\right|=\left|\overrightarrow{OA}+\overrightarrow{BC}\right|=\left|\overrightarrow{OA}+\overrightarrow{AD}\right|=\left|\overrightarrow{OD}\right|=OD=\dfrac{1}{2}BD=\dfrac{a\sqrt{2}}{2}\)
\(\left|\overrightarrow{AB}+\overrightarrow{DC}\right|=\left|\overrightarrow{AB}+\overrightarrow{AB}\right|=2\left|\overrightarrow{AB}\right|=2AB=2a\)
\(\left|\overrightarrow{CD}-\overrightarrow{DA}\right|=\left|\overrightarrow{CD}+\overrightarrow{AD}\right|=\left|\overrightarrow{BA}+\overrightarrow{AD}\right|=\left|\overrightarrow{BD}\right|=BD=a\sqrt{2}\)