cho 2 vecto \(\overrightarrow{a}=\left(3;-6\right);\overrightarrow{b}=\left(4;2\right)\) và \(\overrightarrow{c}=\left(8;m+1\right)\)
a) Tính \(\left|\overrightarrow{a}\right|\)
b) Tính \(\left|\overrightarrow{b}\right|\)
c) Tính \(\overrightarrow{a}.\overrightarrow{b}\)
d) chứng minh \(\overrightarrow{a}\perp\overrightarrow{b}\)
e) tìm m để \(\overrightarrow{a}\perp\overrightarrow{c}\)
f) tìm m để \(\overrightarrow{b}\) cùng phương với \(\overrightarrow{c}\)
g) tìm m để \(\left|\overrightarrow{c}\right|=5\)
a: \(\left|\overrightarrow{a}\right|=\sqrt{3^2+\left(-6\right)^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}\)
b: \(\left|\overrightarrow{b}\right|=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\)
c: \(\overrightarrow{a}\cdot\overrightarrow{b}=3\cdot4+\left(-6\right)\cdot2=12-12=0\)
d: \(\overrightarrow{a}\cdot\overrightarrow{b}=0\)
=>\(\overrightarrow{a}\perp\overrightarrow{b}\)
e: \(\overrightarrow{a}\perp\overrightarrow{c}\)
=>\(3\cdot8+\left(-6\right)\left(m+1\right)=0\)
=>6(m+1)=24
=>m+1=4
=>m=3
f: Để \(\overrightarrow{b};\overrightarrow{c}\) cùng phương thì \(\dfrac{4}{8}=\dfrac{2}{m+1}\)
=>\(m+1=2\cdot\dfrac{8}{4}=2\cdot2=4\)
=>m=4-1=3
g: \(\left|\overrightarrow{c}\right|=5\)
=>\(\sqrt{8^2+\left(m+1\right)^2}=5\)
=>\(\left(m+1\right)^2+64=25\)
=>\(\left(m+1\right)^2=-39< 0\left(loại\right)\)
=>\(m\in\varnothing\)
