
Area of a Parallelogram
I'm given 4 points and asked to find the area of the parallelogram.
K(1,2,3) , L(1,3,6), M(3,8,6), and N(3,7,3)
I want to use the cross product to do this. Don't I just have to take the cross product of KN and KL? I did this and got an answer of 11 cubic units which is not the answer. It's 16.28 and I'm not seeing what I did wrong =/

$\displaystyle \vec{KL} = \langle 0, 1, 3 \rangle$
$\displaystyle \vec{KN} = \langle 2, 5, 0 \rangle$
$\displaystyle \vec{KL} \times \vec{KN} = \langle 15, 6, 2 \rangle$
$\displaystyle \vec{KL} \times \vec{KN} = \sqrt{(15)^2 + 6^2 + (2)^2} \approx 16.28$