The diagonals of a parallelogram are given by the vectors \(\hat{\imath}+2\hat{\jmath}\) and \(\hat{\jmath}-2\hat{k}\). The area of the parallelogram is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\sqrt{21}}{2}\) square units
Let the adjacent sides be \(\vec{a}\) and \(\vec{b}\). Then the diagonals are \(\vec{d}_1 = \vec{a}+\vec{b}\) and \(\vec{d}_2 = \vec{a}-\vec{b}\), and
\(\vec{d}_1\times\vec{d}_2 = \vec{a}\times\vec{a}-\vec{a}\times\vec{b}+\vec{b}\times\vec{a}-\vec{b}\times\vec{b}\)
\( = -2(\vec{a}\times\vec{b})\)
So the area from the diagonals is
\( \displaystyle |\vec{a}\times\vec{b}| = \frac{1}{2}|\vec{d}_1\times\vec{d}_2| \)
Here
\( \displaystyle \vec{d}_1\times\vec{d}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & 0 \\ 0 & 1 & -2 \end{vmatrix} \)
\( = \hat{\imath}(-4-0)-\hat{\jmath}(-2-0)+\hat{k}(1-0)\)
\( = -4\hat{\imath}+2\hat{\jmath}+\hat{k}\)
\(|\vec{d}_1\times\vec{d}_2| = \sqrt{16+4+1} = \sqrt{21}\)
So the area is
\( \displaystyle \frac{\sqrt{21}}{2} \)
Why the other options are incorrect
Option (2) \(\sqrt{21}\) square units — Incorrect. It treats the diagonals as adjacent sides; \(|\vec{d}_1\times\vec{d}_2|\) is twice the area.
Option (3) \(\displaystyle\frac{\sqrt{21}}{4}\) square units — Incorrect. It halves twice; the factor is \(\frac{1}{2}\) only.
Option (4) \(2\sqrt{21}\) square units — Incorrect. It doubles \(|\vec{d}_1\times\vec{d}_2|\) instead of halving it.