The area (in sq. units) of the parallelogram whose diagonals are represented by the vectors \( \vec{a}=2\hat{\imath}-\hat{\jmath}+\hat{k} \) and \( \vec{b}=3\hat{\imath}+4\hat{\jmath}-\hat{k} \), is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{2}\sqrt{155} \)
If \( \vec{a} \) and \( \vec{b} \) are the diagonals of a parallelogram, its area is \( \frac{1}{2}|\vec{a}\times\vec{b}| \).
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & -1 & 1\\3 & 4 & -1\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(1-4)-\hat{\jmath}(-2-3)+\hat{k}(8+3) \)
\( \displaystyle =-3\hat{\imath}+5\hat{\jmath}+11\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{9+25+121}=\sqrt{155} \)
\( \displaystyle \text{Area}=\frac{1}{2}\sqrt{155} \)
Why the other options are incorrect
Option (1) \( \sqrt{155} \) — Incorrect. \( \sqrt{155}=|\vec{a}\times\vec{b}| \) is the area when \( \vec{a} \) and \( \vec{b} \) are adjacent sides, not diagonals.
Option (2) \( \displaystyle\frac{1}{3}\sqrt{155} \) — Incorrect. The factor is \( \frac{1}{2} \), not \( \frac{1}{3} \).
Option (3) \( \displaystyle\frac{1}{4}\sqrt{155} \) — Incorrect. The factor is \( \frac{1}{2} \), not \( \frac{1}{4} \).
Tip: Adjacent sides: area = \( |\vec{a}\times\vec{b}| \). Diagonals: area = \( \frac{1}{2}|\vec{a}\times\vec{b}| \). Triangle with sides \( \vec{a} \), \( \vec{b} \): area = \( \frac{1}{2}|\vec{a}\times\vec{b}| \).