The area of the parallelogram having diagonals $\overrightarrow{a}=3\hat{i}+\hat{j}-2\hat{k}$ and $\overrightarrow{b}=\hat{i}-3\hat{j}+4\hat{k}$ is?

The correct option is B

$5\sqrt{3}$

Explanation for Correct Answer:

Finding the area of a parallelogram:

The diagonals of a parallelogram are

$\overrightarrow{a}=3\hat{i}+\hat{j}-2\hat{k}$

$\overrightarrow{b}=\hat{i}-3\hat{j}+4\hat{k}$

First, we will calculate the cross product of two diagonals hence

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& -2\\ 1& -3& 4\end{array}\right|$

$\begin{array}{l}⟹\overrightarrow{a}\times \overrightarrow{b}=\hat{i}(4-6)-\hat{j}(12+2)+\hat{k}(-9-1)\\ =-2\hat{i}-14\hat{j}-10\hat{k}\\ ⟹|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{(-2{)}^2+(-14{)}^2+(-10{)}^2}\\ =\sqrt{4+196+100}\\ =\sqrt{300}\\ =10\sqrt{3}\end{array}$

We know, the area of the parallelogram is given as:

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{parallelogram}=\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ =\frac{1}{2}\times 10\sqrt{3}\left[\because \left|\overrightarrow{a}\times \overrightarrow{b}\right|=10\sqrt{3}\right] \\ =5\sqrt{3} \end{gathered}$$

The area of the parallelogram has diagonals $\overrightarrow{a}$ and $\overrightarrow{b}$ is equal to $5\sqrt{3}\mathrm{square}\mathrm{units}$

Hence, option (B) is correct.