ABCD is a parallelogram such that $\displaystyle \overrightarrow{\mathrm{AC}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\displaystyle \overrightarrow{\mathrm{BD}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$. Find $\displaystyle \overrightarrow{\mathrm{AB}}$ and $\displaystyle \overrightarrow{\mathrm{AD}}$. Also, find the area of the parallelogram ABCD .

Let $\displaystyle \overrightarrow{AB} = \vec{a}$ and $\displaystyle \overrightarrow{AD} = \vec{b}$$\displaystyle \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{a} + \vec{b} = \hat{\imath} + \hat{\jmath}$$\displaystyle \overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD} = \vec{b} - \vec{a} = 2\hat{\imath} + \hat{\jmath} + \hat{k}$Adding we get, $\displaystyle 2\overrightarrow{AD} = \overrightarrow{AC} + \overrightarrow{BD} = 3\hat{\imath} + 2\hat{\jmath} + \hat{k}$$\displaystyle \Rightarrow \overrightarrow{AD} = \dfrac{3}{2}\hat{\imath} + \hat{\jmath} + \dfrac{1}{2}\hat{k}$Subtracting, we get$\displaystyle 2\overrightarrow{AB} = \overrightarrow{AC} - \overrightarrow{BD} = -\hat{\imath} - \hat{k} \Rightarrow \overrightarrow{AB} = -\dfrac{1}{2}\hat{\imath} - \dfrac{1}{2}\hat{k}$\[\left|\overrightarrow{AC} \times \overrightarrow{BD}\right| = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & 0 \\ 2 & 1 & 1 \end{vmatrix} = \hat{\imath} - \hat{\jmath} - \hat{k}\]Area $\displaystyle = \dfrac{1}{2}\left|\overrightarrow{AC} \times \overrightarrow{BD}\right|$\[= \frac{\sqrt{3}}{2}\]