question_answer

The area of the parallelogram whose diagonals are given by the vectors \[\mathbf{3\hat{i}}+\mathbf{\hat{J}}-\mathbf{2\hat{k}}\] and \[\mathbf{\hat{i}}-\mathbf{2}\hat{j}+\mathbf{4\hat{k}}\] is:

A)   4                   

B)  8              

C)   \[5\sqrt{3}\]                

D)  \[10\sqrt{3}\]

Correct Answer: C

Solution :

[c] \[\vec{a}=3\hat{i}+\hat{j}-2\hat{k}\] \[\vec{b}=\hat{j}-3\hat{j}+4\hat{k}\] Since, \[\vec{a}\] and \[\vec{b}\]represent the diagonal of the parallelogram. Area of parallelogram \[=\frac{1}{2}\left| \vec{a}\times \vec{b} \right|\] Now, \[=\,\sqrt{{{(-2)}^{2}}+{{(-14)}^{2}}+{{(-10)}^{2}}}=\sqrt{4+196+100}\]\[=\sqrt{300}=10\sqrt{3}\] Hence, area of parallelogram \[=\frac{1}{2}.10\sqrt{3}=5\sqrt{3}\] Hence, option [c] is correct.