(a)OR(b)

If the vectors $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ are such that $\displaystyle |\overrightarrow{\mathrm{a}}|=3,|\overrightarrow{\mathrm{~b}}|=\frac{2}{3}$ and $\displaystyle \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ is a unit vector, then find the angle between $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$.

Let $\displaystyle \theta$ be the angle between $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ Since $\displaystyle \vec{\mathrm{a}} \times \vec{\mathrm{b}}$ is a unit vector, we have $\displaystyle |\vec{\mathrm{a}} \times \vec{\mathrm{b}}|=1$ $\displaystyle \Rightarrow|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \sin \theta=1$ $\displaystyle \Rightarrow \sin \theta=\frac{1}{2}$, or $\displaystyle \theta=30^{\circ}$ (or $\displaystyle \frac{\pi}{6}$ ) Here\begin{aligned} & \vec{\mathrm{a}} \times \vec{\mathrm{b}}=\left|\begin{array}{rrr} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} 1 & -1 & 3 2 & -7 & 1 \end{array}\right|=20 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-5 \hat{\mathrm{k}} & \Rightarrow|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|=\sqrt{400+25+25}=\sqrt{450} \end{aligned}Area of parallelogram $\displaystyle =|\overrightarrow{\boldsymbol{\mathrm{a}}} \times \overrightarrow{\boldsymbol{\mathrm{b}}}|=\sqrt{450}=15 \sqrt{2}$