Cho 2 vecto \(\overrightarrow{a}\), \(\overrightarrow{b}\) vuông góc và \(\left|\overrightarrow{a}\right|=1\), \(\left|\overrightarrow{b}\right|=\sqrt{2}\). Chứng minh rằng 2 vecto sau vuông góc: \(\left(2\overrightarrow{a}-\overrightarrow{b}\right),\left(\overrightarrow{a}+\overrightarrow{b}\right)\).
\(\overrightarrow{a}\perp\overrightarrow{b}\Rightarrow\overrightarrow{a}.\overrightarrow{b}=0\)
\(\left(2\overrightarrow{a}-\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)=2a^2+2\overrightarrow{a}.\overrightarrow{b}-\overrightarrow{a}.\overrightarrow{b}-b^2\)
\(=2a^2-b^2+\overrightarrow{a}.\overrightarrow{b}\)
\(=2.1-2+0=0\)
\(\Rightarrow\left(2\overrightarrow{a}-\overrightarrow{b}\right)\perp\left(\overrightarrow{a}+\overrightarrow{b}\right)\)