Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors then what is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ for $\overrightarrow{a}-\sqrt{2}\overrightarrow{b}$ to be a unit vector?

Answer 1

We have,

$\begin{array}{c}\left|\overrightarrow{a}\right|=1\left|\overrightarrow{b}\right|=1\\ \left|\overrightarrow{a}-\sqrt{2}\overrightarrow{b}\right|=1\\ {\left|\overrightarrow{a}-\sqrt{2}\overrightarrow{b}\right|}^2=1\\ \left(\overrightarrow{a}-\sqrt{2}\overrightarrow{b}\right).\left(\overrightarrow{a}-\sqrt{2}\overrightarrow{b}\right)=1\\ {\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2-2\sqrt{2}\overrightarrow{a}.\overrightarrow{b}=1\\ \frac{1}{\sqrt{2}}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ \cos \theta =\frac{1}{\sqrt{2}}\\ \theta =\frac{\pi }{4}={45}^0\end{array}$