Vector Algebra Question 55
Question: 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} $ ?
Options:
A) $ 5\sqrt{5} $ square units
B) $ 4\sqrt{5} $ square units
C) $ 5\sqrt{3} $ square units
D) $ 15\sqrt{2} $ square units
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Answer:
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} \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \\ \end{vmatrix} $
$ =\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
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