Let ABCD be a parallelogram with vertices \( A(2,2,a),\ B(1,0,0),\ C(1,3,a) \) and D, where \( a\gt 0 \). If the area of parallelogram is 5 square units, then the value of \( a \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2\sqrt{2} \)
In parallelogram ABCD, AB and BC are adjacent sides, so the area is \( |\vec{AB}\times\vec{BC}| \).
\( \displaystyle \vec{AB}=(1-2)\hat{\imath}+(0-2)\hat{\jmath}+(0-a)\hat{k}=-\hat{\imath}-2\hat{\jmath}-a\hat{k} \)
\( \displaystyle \vec{BC}=(1-1)\hat{\imath}+(3-0)\hat{\jmath}+(a-0)\hat{k}=3\hat{\jmath}+a\hat{k} \)
\( \displaystyle \vec{AB}\times\vec{BC}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\-1 & -2 & -a\\0 & 3 & a\end{vmatrix} \)
\( \displaystyle =(-2a+3a)\hat{\imath}-(-a-0)\hat{\jmath}+(-3-0)\hat{k} \)
\( \displaystyle =a\hat{\imath}+a\hat{\jmath}-3\hat{k} \)
\( \displaystyle |\vec{AB}\times\vec{BC}|=\sqrt{a^{2}+a^{2}+9}=\sqrt{2a^{2}+9} \)
Area = 5:
\( \displaystyle 2a^{2}+9=25\ \Rightarrow\ a^{2}=8 \)
\( \displaystyle a=2\sqrt{2}\ \ (a\gt 0) \)