Question

If $\overrightarrow{a},\overrightarrow{b}$ are two vectors such that $\left|\overrightarrow{a}\right|=1,\left|\overrightarrow{b}\right|=2\mathrm{and}\left|\overrightarrow{a}\times \stackrel{\rightharpoonup }{b}\right|=\sqrt{2}$ then the angle between $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ is _____________.

Answer 1

$$\begin{gathered} \mathrm{For}\mathrm{given}\mathrm{vectors}\overrightarrow{a}\mathrm{and}\overrightarrow{b} \\ \left|\overrightarrow{a}\right|=1,\left|\overrightarrow{b}\right|=2\mathrm{and}\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{2} \\ \therefore \mathrm{By}\mathrm{lagrange}\text{'}\mathrm{s}\mathrm{identity}, \\ {\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2 \\ \mathrm{i}.\mathrm{e}.2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2=1\times 4 \\ \mathrm{i}.\mathrm{e}.{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2=2 \\ \mathrm{i}.\mathrm{e}.\overrightarrow{a}.\overrightarrow{b}=\sqrt{2} \\ \therefore \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta =\sqrt{2}\left(\mathrm{where}\theta \mathrm{is}\mathrm{angle}\mathrm{between}a\mathrm{and}b\right) \\ \mathrm{i}.\mathrm{e}.1\times 2\cos \theta =\sqrt{2} \\ \mathrm{i}.\mathrm{e}.\cos \theta =\frac{1}{\sqrt{2}} \\ \mathrm{i}.\mathrm{e}.\theta =\frac{\mathrm{\pi }}{4}\mathrm{or}\frac{3\mathrm{\pi }}{4}\mathrm{i}.\mathrm{e}\mathrm{angle}\mathrm{between}\overrightarrow{a}\mathrm{and}\overrightarrow{b}\mathrm{is}\frac{\mathrm{\pi }}{4}\mathrm{or}\frac{3\mathrm{\pi }}{4} \end{gathered}$$