question_answer

If \[3\hat{i}+\hat{j}-2\hat{k}\]and \[\hat{i}-3\hat{j}+4\hat{k}\]  are the diagonals of a parallelogram, then the area of the parallelogram is

A)  \[10\sqrt{3}\] sq unit   

B)  \[5\sqrt{3}\] sq unit

C)  \[5\sqrt{2}\]sq unit    

D)  \[10\sqrt{2}\]sq unit

Correct Answer: B

Solution :

Let \[\overrightarrow{AC}=3\hat{i}+\hat{j}-2\hat{k}\] and \[\overrightarrow{BD}=\hat{i}-3\hat{j}+4\hat{k}\] Then, area of parallelogram is given by \[=\frac{1}{2}|\overrightarrow{AC}\,\,\times \,\,\overrightarrow{BD}|\] Now, \[\overrightarrow{AC}\,\,\times \,\,\overrightarrow{BD}\] \[=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    3 & 1 & -2  \\    1 & -3 & 4  \\ \end{matrix} \right|\] \[=\hat{i}(+4-6)-\hat{j}(12+2)+\hat{k}(-9-1)\] \[=-2\hat{i}-14\hat{j}-10\hat{k}\] \[|\overrightarrow{AC}\,\,\times \,\,\overrightarrow{BD}|=\sqrt{4+196+100}\] \[=\sqrt{300}=10\sqrt{3}\] \[\therefore \] Required area \[=\frac{1}{2}\times 10\sqrt{3}=5\sqrt{3}\] sq. unit