A) \[5\sqrt{70}\]
B) \[50\,\sqrt{7}\]
C) \[\,\sqrt{70}\]
D) \[5\,\sqrt{7}\]
Correct Answer: A
Solution :
The area of the parallelogram whose conterminal edges are a and b, is \[|a\times b|\]. Given, conterminal edges of the parallelogram are \[a=3\hat{i}-3\hat{j}+\hat{k}\] and \[b=4\hat{i}+9\hat{j}+2\hat{k}\] Then, \[a\times b\left| \begin{matrix} {\hat{i}} & {\hat{j}} & k \\ 3 & -3 & 1 \\ 4 & 9 & 2 \\ \end{matrix} \right|\] \[=\hat{i}(-6-9)-\hat{j}(6-4)+\hat{k}(27+12)\] \[=\hat{i}(-15)-\hat{j}(2)+\hat{k}(39)\] \[=-15-2\hat{j}+39\hat{k}\] \[\therefore \] Area of parallelogram \[=|a\times b|\] \[=\sqrt{{{(-15)}^{2}}+{{(-2)}^{2}}+{{(39)}^{2}}}\] \[=\sqrt{225+4+1521}\] \[=\sqrt{1750}=5\sqrt{70}\,\,sq\,units\]
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