Câu 5:
a: \(\overrightarrow{AB}\cdot\overrightarrow{AC}=AB\cdot AC\cdot cos\left(\overrightarrow{AB};\overrightarrow{AC}\right)\)
\(=5\cdot7\cdot cos120=-\frac{35}{2}\)
b: Xét ΔABC có \(cosA=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(7^2+5^2-BC^2=2\cdot7\cdot5\cdot cos120=-35\)
=>\(BC^2=49+25+35=49+60=109\)
=>\(BC=\sqrt{109}\)
Xét ΔABC có AM là đường trung tuyến
nên \(AM^2=\frac{AB^2+AC^2}{2}-\frac{BC^2}{4}=\frac{7^2+5^2}{2}-\frac{109}{4}=\frac{2\left(7^2+5^2\right)-109}{4}=\frac{39}{4}\)
=>\(AM=\frac{\sqrt{39}}{2}\)
Câu 2:
a: \(\overrightarrow{a}\) ⊥\(\overrightarrow{b}\)
=>\(\overrightarrow{a}\cdot\overrightarrow{b}=0\)
=>1*(-1)+2m=0
=>2m-1=0
=>2m=1
=>m=1/2
b: \(\overrightarrow{a}=\left(1;2\right);\overrightarrow{b}=\left(-1;m\right)\)
\(\left|\overrightarrow{a}\right|=\sqrt{1^2+2^2}=\sqrt5\)
\(\left|\overrightarrow{b}\right|=\sqrt{\left(-1\right)^2+m^2}=\sqrt{m^2+1}\)
Ta có: \(\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|\)
=>\(m^2+1=5\)
=>\(m^2=4\)
=>m=2 hoặc m=-2
