Show that the area of the parallelogram whose diagonals are given by $\vec{a}$ and $\vec{b}$ is $\frac{|\vec{a} \times \vec{b}|}{2}$. Also,find the area of the parallelogram whose diagonals are $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3 \hat{j}-\hat{k}$.

(N/A) Let $ABCD$ be a parallelogram such that $\overrightarrow{AB} = \vec{p}$ and $\overrightarrow{AD} = \vec{q}$. Then $\overrightarrow{BC} = \vec{q}$.
By the triangle law of vector addition,we have:
$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{p} + \vec{q} = \vec{a} \quad \dots(i)$
$\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB} = \vec{q} - \vec{p} = \vec{b} \quad \dots(ii)$
Adding $(i)$ and $(ii)$,we get $\vec{a} + \vec{b} = 2\vec{q} \Rightarrow \vec{q} = \frac{1}{2}(\vec{a} + \vec{b})$.
Subtracting $(ii)$ from $(i)$,we get $\vec{a} - \vec{b} = 2\vec{p} \Rightarrow \vec{p} = \frac{1}{2}(\vec{a} - \vec{b})$.
The area of the parallelogram is given by $|\vec{p} \times \vec{q}|$.
$|\vec{p} \times \vec{q}| = |\frac{1}{2}(\vec{a} - \vec{b}) \times \frac{1}{2}(\vec{a} + \vec{b})| = \frac{1}{4} |\vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}|$.
Since $\vec{a} \times \vec{a} = 0$ and $\vec{b} \times \vec{b} = 0$,and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$,we have:
$|\vec{p} \times \vec{q}| = \frac{1}{4} |\vec{a} \times \vec{b} + \vec{a} \times \vec{b}| = \frac{1}{4} |2(\vec{a} \times \vec{b})| = \frac{1}{2} |\vec{a} \times \vec{b}|$.
Now,for diagonals $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} = \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 + 1) = -2\hat{i} + 3\hat{j} + 7\hat{k}$.
Area $= \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} \sqrt{(-2)^2 + 3^2 + 7^2} = \frac{1}{2} \sqrt{4 + 9 + 49} = \frac{\sqrt{62}}{2} \text{ sq units}$.