Question

Let $\overrightarrow{a},\overrightarrow{b}$ be unit vectors such that $\overrightarrow{a}-\sqrt{2}\overrightarrow{b}$ is also a unit vector. Then the angle between $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ is _____________.

Answer 1

Given:
$\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are unit vectors
$\overrightarrow{a}-\sqrt{2}\overrightarrow{b}$ is also a unit vector


$$\begin{gathered} \overrightarrow{a}-\sqrt{2}\overrightarrow{b}\mathrm{is}\mathrm{a}\mathrm{unit}\mathrm{vector} \\ \Rightarrow \left|\overrightarrow{a}-\sqrt{2}\overrightarrow{b}\right|=1 \\ \Rightarrow {\left|\overrightarrow{a}-\sqrt{2}\overrightarrow{b}\right|}^2={\left(1\right)}^2 \\ \Rightarrow \left(\overrightarrow{a}-\sqrt{2}\overrightarrow{b}\right)·\left(\overrightarrow{a}-\sqrt{2}\overrightarrow{b}\right)=1 \\ \Rightarrow \overrightarrow{a}·\overrightarrow{a}+\sqrt{2}\overrightarrow{b}·\sqrt{2}\overrightarrow{b}-2\sqrt{2}\overrightarrow{a}·\overrightarrow{b}=1 \\ \Rightarrow {\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2-2\sqrt{2}\left(\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \right)=1 \\ \Rightarrow 1^2+2{\left(1\right)}^2-2\sqrt{2}\left(\left(1\right)\left(1\right)\cos \theta \right)=1\left(\because \left|\overrightarrow{a}\right|=1,\left|\overrightarrow{b}\right|=1\right) \\ \Rightarrow 1+2-2\sqrt{2}\cos \theta =1 \\ \Rightarrow -2\sqrt{2}\cos \theta =-2 \\ \Rightarrow \cos \theta =\frac{1}{\sqrt{2}} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{4} \end{gathered}$$


Hence, the angle between $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ is $\underline{\frac{\mathrm{\pi }}{4}}$.