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.You need to login to perform this action.
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