Question

If $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are two non-zero non-collinear vectors such that $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=1\mathrm{and}\left|\overrightarrow{a}.\overrightarrow{b}\right|=\sqrt{3}$, then the angle between $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ is _____________.

Answer 1

$$\begin{gathered} \mathrm{For}\mathrm{non}‐\mathrm{zero}\mathrm{non}\mathrm{collinear}\mathrm{vectors}\overrightarrow{a}\mathrm{and}\overrightarrow{b} \\ \mathrm{Given}\left|\overrightarrow{a}\times \overrightarrow{b}\right|=1 \\ \left|\overrightarrow{a}·\overrightarrow{b}\right|=\sqrt{3} \\ \mathrm{Since}\mathrm{by}\mathrm{lagrange}\text{'}\mathrm{s}\mathrm{identity}, \\ {\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2 \\ \mathrm{i}.\mathrm{e}.1+3={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2 \\ \mathrm{i}.\mathrm{e}.4={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2 \\ \mathrm{i}.\mathrm{e}.\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|=2 \\ \therefore \left|\overrightarrow{a}.\overrightarrow{b}\right|=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \mathrm{where}\theta \mathrm{is}\mathrm{angle}\mathrm{between}\overrightarrow{a}\mathrm{and}\overrightarrow{b} \\ \sqrt{3}=2\cos \theta \\ \mathrm{i}.\mathrm{e}.\cos \theta =\frac{\sqrt{3}}{2} \\ \mathrm{i}.\mathrm{e}.\theta =\frac{\mathrm{\pi }}{6} \end{gathered}$$