Question 10.4.4: Finding the Area of a Parallelogram Using the Cross Product ...

First notice that

a \times b =\left|\begin{array}{ccc} i & j & k \\1 & 2 & 3 \\4 & 5 & 6\end{array}\right|= i \left|\begin{array}{ll}2 & 3 \\5 & 6\end{array}\right|- j \left|\begin{array}{ll}1 & 3 \\4 & 6\end{array}\right|+ k \left|\begin{array}{ll}1 & 2 \\4 & 5\end{array}\right|=\langle-3,6,-3\rangle.

From (4.5), the area of the parallelogram is given by

=\| b \|\| a \| \sin \theta=\| a \times b \|, (4.5)

\| a \times b \|=\|\langle-3,6,-3\rangle\|=\sqrt{54} \approx 7.348.