Áp dụng địnhlý hàm cos:

\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosBAC}=\sqrt{19}\)

\(\Rightarrow cosB=\dfrac{AB^2+BC^2-AC^2}{2AB.BC}=\dfrac{\sqrt{19}}{38}\)

\(BM=2MC\Rightarrow BM=\dfrac{2}{3}BC=\dfrac{2\sqrt{19}}{3}\)

\(\Rightarrow AM=\sqrt{AB^2+BM^2-2AB.BM.cosB}=\dfrac{\sqrt{139}}{3}\)

a: \(\overrightarrow{AE}=\dfrac{2}{3}\overrightarrow{EC}\)

=>E nằm giữa A và C và AE=2/3EC

Ta có: AE+EC=AC(E nằm giữa A và C)

=>\(AC=\dfrac{2}{3}EC+EC=\dfrac{5}{3}EC\)

=>\(\dfrac{AE}{AC}=\dfrac{\dfrac{2}{3}EC}{\dfrac{5}{3}EC}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{5}\)

=>\(AE=\dfrac{2}{5}AC\)

=>\(\overrightarrow{AE}=\dfrac{2}{5}\cdot\overrightarrow{AC}\)

\(\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}\)

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

b: \(\left|\overrightarrow{IA}+\overrightarrow{IG}\right|=\left|\overrightarrow{IA}-\overrightarrow{IG}\right|\)

=>\(\left[{}\begin{matrix}\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IA}-\overrightarrow{IG}\\\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IG}-\overrightarrow{IA}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2\cdot\overrightarrow{IG}=\overrightarrow{0}\\2\cdot\overrightarrow{IA}=\overrightarrow{0}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}I\equiv G\\I\equiv A\end{matrix}\right.\)

\(\cos ABC=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)

\(\Leftrightarrow89a^2-AC^2=2\cdot5a\cdot8a\cdot\dfrac{1}{2}=40a^2\)

=>AC=7a

\(AM=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{25a^2+49a^2}{2}-\dfrac{64a^2}{4}=37a^2-16a^2=21a^2\)

hay \(AM=a\sqrt{21}\left(cm\right)\)

a) \(\overrightarrow {AB} .\overrightarrow {AC}  = 2.3.\cos \widehat {BAC} = 6.\cos {60^o} = 3\)

b)

Ta có: \(\overrightarrow {AB}  + \overrightarrow {AC}  = 2\overrightarrow {AM} \)(do M là trung điểm của BC)

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

+) \(\overrightarrow {BD}  = \overrightarrow {AD}  - \overrightarrow {AB}  = \frac{7}{{12}}\overrightarrow {AC}  - \overrightarrow {AB} \)

c) Ta có:

 \(\begin{array}{l}\overrightarrow {AM} .\overrightarrow {BD}  = \left( {\frac{1}{2}\overrightarrow {AB}  + \frac{1}{2}\overrightarrow {AC} } \right)\left( {\frac{7}{{12}}\overrightarrow {AC}  - \overrightarrow {AB} } \right)\\ = \frac{7}{{24}}\overrightarrow {AB} .\overrightarrow {AC}  - \frac{1}{2}{\overrightarrow {AB} ^2} + \frac{7}{{24}}{\overrightarrow {AC} ^2} - \frac{1}{2}\overrightarrow {AC} .\overrightarrow {AB} \\ =  - \frac{1}{2}A{B^2} + \frac{7}{{24}}A{C^2} - \frac{5}{{24}}\overrightarrow {AB} .\overrightarrow {AC} \\ =  - \frac{1}{2}{.2^2} + \frac{7}{{24}}{.3^2} - \frac{5}{{24}}.3\\ = 0\end{array}\)

\( \Rightarrow AM \bot BD\)

Câu 2:

Ta có: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=\left(2a\right)^2+\left(2a\sqrt{3}\right)^2=16a^2\)

=>BC=4a

Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}=\dfrac{1}{2}\)

nên \(\widehat{ABC}=30^0\)

ΔABC vuông tại A

=>\(\widehat{ABC}+\widehat{ACB}=90^0\)

=>\(\widehat{ACB}=60^0\)

Lấy điểm E sao cho \(\overrightarrow{AB}=\overrightarrow{BE}\)

=>B là trung điểm của AE

=>\(\widehat{CBE}+\widehat{CBA}=180^0\)(hai góc kề bù)

=>\(\widehat{CBE}=180^0-30^0=150^0\)

\(\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{BE}\cdot\overrightarrow{BC}\)

\(=BE\cdot BC\cdot cos\left(\overrightarrow{BE};\overrightarrow{BC}\right)\)

\(=2a\sqrt{3}\cdot4a\cdot cos150=-12a^2\)

\(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{AB}+\overrightarrow{CA}\right|=\left|\overrightarrow{CB}\right|=CB=4a\)

Đáp án C

Đáp án A