Answer
$\dfrac{5}{2}\sqrt {14}$
Work Step by Step
The cross product is defined as:
$u \times v=\begin{vmatrix}i&j&k\\m_1&m_2&m_3\\n_1&n_2&n_3\end{vmatrix}=\lt m_2n_3-m_3n_2, m_3n_1-m_1n_3, m_1n_2-m_2b_1 \gt$
The area of parallelogram $A$ is defined as the length of the cross product $u \times v$ .
Now,
$u \times v=\lt (-1)(-\dfrac{3}{2})-(4)(2),(4)(\dfrac{1}{2})-(2)(-\dfrac{3}{2}), (2)(2) -(-1)(\dfrac{1}{2}) \gt=\lt -\dfrac{13}{2},5,\dfrac{9}{2} \gt$
Then, we have
$A=|u \times v|=\sqrt {(-\dfrac{3}{2})^2+(5)^2+(\dfrac{9}{2})^2}$
or, $A=\dfrac{5}{2}\sqrt {14}$
