Question

If magnitude A vector cross B vector = 1
Magnitude of A vector = 3/2
Magnitude of B vector = 4/3
Dot product of A vector and B vector is -root 3
Angle between A vector and B vector = 30 or 150 degrees

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{\theta }\mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{\mathrm{a}}\mathrm{and}\overrightarrow{\mathrm{b}}. \\ \mathrm{Now},\left|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right|=1 \\ \Rightarrow \left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\sin \mathrm{\theta }=1 \\ \Rightarrow \frac{3}{2}\times \frac{4}{3}\times \sin \mathrm{\theta }=1 \\ \Rightarrow 2\sin \mathrm{\theta }=1 \\ \Rightarrow \sin \mathrm{\theta }=\frac{1}{2} \\ \Rightarrow \mathrm{\theta }=30°\mathrm{or}150°......\left(1\right) \\ \mathrm{Also},\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=-\sqrt{3} \\ \Rightarrow \left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\cos \mathrm{\theta }=-\sqrt{3} \\ \Rightarrow \frac{3}{2}\times \frac{4}{3}\times \cos \mathrm{\theta }=-\sqrt{3} \\ \Rightarrow \cos \mathrm{\theta }=-\frac{\sqrt{3}}{2} \\ \Rightarrow \mathrm{\theta }=150°\mathrm{or}210°.....\left(2\right) \\ \mathrm{So},\mathrm{from}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}\mathrm{\theta }=150° \end{gathered}$$