The diagonals of a parallelogram are represented by \( 4\hat{\imath}-\hat{\jmath}-3\hat{k} \) and \( \hat{\imath}+2\hat{\jmath}-2\hat{k} \).
The area of parallelogram is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\sqrt{170}}{2} \) Sq. units
When the diagonals \( \vec{d}_1 \) and \( \vec{d}_2 \) are given, the area of the parallelogram is \( \frac{1}{2}|\vec{d}_1\times\vec{d}_2| \).
\( \displaystyle \vec{d}_1\times\vec{d}_2=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\4 & -1 & -3\\1 & 2 & -2\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(2+6)-\hat{\jmath}(-8+3)+\hat{k}(8+1) \)
\( \displaystyle \vec{d}_1\times\vec{d}_2=8\hat{\imath}+5\hat{\jmath}+9\hat{k} \)
\( \displaystyle |\vec{d}_1\times\vec{d}_2|=\sqrt{64+25+81}=\sqrt{170} \)
\( \displaystyle \text{Area}=\frac{1}{2}\sqrt{170}\ \text{sq. units} \)
Tip: Sides \( \vec{a},\vec{b} \): area \( |\vec{a}\times\vec{b}| \). Diagonals \( \vec{d}_1,\vec{d}_2 \): area \( \frac{1}{2}|\vec{d}_1\times\vec{d}_2| \), because \( |\vec{d}_1\times\vec{d}_2|=|(\vec{a}+\vec{b})\times(\vec{b}-\vec{a})|=2|\vec{a}\times\vec{b}| \). Forgetting the \( \frac{1}{2} \) gives \( \sqrt{170} \).