Question

Area of a parallelogram, whose diagonals are $3\hat{i}+\hat{j}-2\hat{k}$ and $\hat{i}-3\hat{j}+4\hat{k}$ will be

• A. $14unit$
• B. $5\sqrt{3}unit$
• C. $10\sqrt{3}unit$
• D. $20\sqrt{3}unit$.

Answer 1

The correct option is B $5\sqrt{3}unit$

Given diagonals are, $\overrightarrow{AC}=3\hat{i}+\hat{j}-2\hat{k},\overrightarrow{DB}=\hat{i}-3\hat{j}+4\hat{k}$
therefore, $\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=\frac{\overrightarrow{AC}}{2}+\frac{\overrightarrow{DB}}{2}=\frac{1}{2}(3\hat{i}+\hat{j}-2\hat{k})+\frac{1}{2}(\hat{i}-3\hat{j}+4\hat{k})=2\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{AC}\times \overrightarrow{AB}=(3\hat{i}+\hat{j}-2\hat{k})\times (2\hat{i}-\hat{j}+\hat{k})=-\hat{i}-7\hat{j}-5\hat{k}$
thus, area of a parallelogram$=|\overrightarrow{AC}\times \overrightarrow{AB}|=\sqrt{1^2+7^2+5^2}=\sqrt{75}=5\sqrt{3}unit$