question_answer

What is the area of the parallelogram having diagonals \[3\hat{i}+\hat{j}-2\hat{k}\] and \[\hat{i}-3\hat{j}+4\hat{k}\]?

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 units