Question

If $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are unit vectors such that $\overrightarrow{a}\times \overrightarrow{b}$ is also a unit vector, then the angle between $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ is ___________.

Answer 1

$$\begin{gathered} \mathrm{Given}\mathrm{for}\overrightarrow{a},\overrightarrow{b} \\ \overrightarrow{a},\overrightarrow{b}\mathrm{and}\overrightarrow{a}\times \overrightarrow{b}\mathrm{are}\mathrm{unit}\mathrm{vectors}. \\ \mathrm{i}.\mathrm{e}\left|\overrightarrow{a}\right|=1=\left|\overrightarrow{b}\right|=\left(\overrightarrow{a}\times \overrightarrow{b}\right) \\ \therefore \mathrm{By}\mathrm{lagrange}\text{'}\mathrm{s}\mathrm{identity}, \\ {\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2={\left|a\right|}^2{\left|b\right|}^2 \\ \mathrm{i}.\mathrm{e}{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2=1-1=0 \\ \mathrm{i}.\mathrm{e}{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2=0 \\ \mathrm{i}.\mathrm{e}\left|\overrightarrow{a}.\overrightarrow{b}\right|=0 \\ \mathrm{i}.\mathrm{e}.\overrightarrow{a}.\overrightarrow{b}=0 \\ \mathrm{i}.\mathrm{e}.\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta =0;\mathrm{where}\theta \mathrm{is}\mathrm{angle}\mathrm{between}\overrightarrow{a}\mathrm{and}\overrightarrow{b} \\ \mathrm{i}.\mathrm{e}.1\times 1\cos \theta =0 \\ \mathrm{i}.\mathrm{e}.\cos \theta =0 \\ \mathrm{i}.\mathrm{e}.\theta =\frac{\mathrm{\pi }}{2} \\ \mathrm{i}.\mathrm{e}.\mathrm{angle}\mathrm{between}\overrightarrow{a}\mathrm{and}\overrightarrow{b}\mathrm{is}\frac{\mathrm{\pi }}{2} \end{gathered}$$