If \( \vec{a}=2\hat{\imath}-3\hat{\jmath}+\hat{k},\vec{b}=-\hat{\imath}+\hat{k},\vec{c}=2\hat{\jmath}-\hat{k} \) are three vectors, then area of the parallelogram having diagonals \( (\vec{a}+\vec{b}) \) and \( (\vec{b}+\vec{c}) \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{2}\sqrt{21} \) sq. units
The diagonals are
\( \displaystyle \vec{d}_{1}=\vec{a}+\vec{b}=\hat{\imath}-3\hat{\jmath}+2\hat{k} \)
\( \displaystyle \vec{d}_{2}=\vec{b}+\vec{c}=-\hat{\imath}+2\hat{\jmath} \)
For a parallelogram with diagonals \( \vec{d}_{1},\vec{d}_{2} \), the area is \( \frac{1}{2}|\vec{d}_{1}\times\vec{d}_{2}| \).
\( \displaystyle \vec{d}_{1}\times\vec{d}_{2}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 1 & -3 & 2\\ -1 & 2 & 0\end{vmatrix} \)
\( \displaystyle =(0-4)\hat{\imath}-(0+2)\hat{\jmath}+(2-3)\hat{k} \)
\( \displaystyle =-4\hat{\imath}-2\hat{\jmath}-\hat{k} \)
\( \displaystyle |\vec{d}_{1}\times\vec{d}_{2}|=\sqrt{16+4+1} \)
\( \displaystyle =\sqrt{21} \)
\( \displaystyle \text{Area}=\frac{1}{2}\sqrt{21}\text{ sq. units} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{2}\sqrt{61} \) sq. units — Incorrect. \( |\vec{d}_{1}\times\vec{d}_{2}|^{2}=21 \), not 61.
Option (2) \( \displaystyle\frac{1}{3}\sqrt{21} \) sq. units — Incorrect. The factor for diagonals is \( \frac{1}{2} \), not \( \frac{1}{3} \).
Option (4) \( \sqrt{21} \) sq. units — Incorrect. \( \sqrt{21} \) is \( |\vec{d}_{1}\times\vec{d}_{2}| \), which would be the area only if \( \vec{d}_{1},\vec{d}_{2} \) were adjacent sides; for diagonals it must be halved.