A) \[5\sqrt{5}\] square units
B) \[4\sqrt{5}\] square units
C) \[5\sqrt{3}\] square units
D) \[15\sqrt{2}\] square units
Correct Answer: C
Solution :
[c] Diagonal \[{{d}_{1}},\overrightarrow{AC}=3i+j-2k\] Diagonal \[{{d}_{2}},\overrightarrow{BD}=i+3j+4k\] Area of parallelogram is \[\frac{1}{2}|{{\overset{\to }{\mathop{d}}\,}_{1}}\times {{\overset{\to }{\mathop{d}}\,}_{2}}|\] Hence area \[=\frac{1}{2}\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \\ \end{matrix} \right|\] \[=\frac{1}{2}|[\hat{i}(4-6)-\hat{j}(12+2)+\hat{k}(-9-1)]|\] \[=\frac{1}{2}|-2\hat{i}-14\hat{j}-10\hat{k}|\] \[=\frac{1}{2}\sqrt{4+196+100}\] \[=\frac{10\sqrt{3}}{2}=5\sqrt{3}\] square unitsYou need to login to perform this action.
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