# Area of a Parallelogram

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• September 6th 2008, 03:14 PM
JonathanEyoon
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 =/
• September 6th 2008, 05:32 PM
skeeter
$\vec{KL} = \langle 0, 1, 3 \rangle$
$\vec{KN} = \langle 2, 5, 0 \rangle$

$\vec{KL} \times \vec{KN} = \langle -15, 6, -2 \rangle$

$|\vec{KL} \times \vec{KN}| = \sqrt{(-15)^2 + 6^2 + (-2)^2} \approx 16.28$