Answer
$2\sqrt {2}$
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,
$u \times v=\lt (-1)(-1)-(1)(1),(1)(1)-(1)(-1), (1)(1) -(-1)(1) \gt=\lt 0,2,2 \gt$
Then, we have
$A=|u \times v|=\sqrt {(0)^2+(2)^2+(2)^2}$
or, $A=\sqrt 8=2\sqrt {2}$
