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Solve the following.
Find the area of the parallelogram whose adjacent sides are given by the vectors and .
Find the area of the parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{\imath} + \hat{\jmath} + 4\hat{k}$ and $\vec{b} = \hat{\imath} - \hat{\jmath} + \hat{k}$.
Answer
Answer
AIArea $= |\vec{a} \times \vec{b}|$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = 5\hat{\imath} + \hat{\jmath} - 4\hat{k}$
$|\vec{a} \times \vec{b}| = \sqrt{25 + 1 + 16} = \sqrt{42}$.
Hence the area is $\sqrt{42}$ square units.
Final answer: $\sqrt{42}$ square units
From ISC 2018 Specimen Mathematics Paper 1, question 15(a).
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