# Thread: parallelogram area with 3D vectors

1. ## parallelogram area with 3D vectors

How do I find the area of a parallelogram with the given vertices
K (1, 3, 2) L (1, 4, 4) M (4, 9, 4) N (4, 8, 2)

I have to use the cross product, but I'm pretty lost.

2. Originally Posted by acg716
How do I find the area of a parallelogram with the given vertices
K (1, 3, 2) L (1, 4, 4) M (4, 9, 4) N (4, 8, 2)

I have to use the cross product, but I'm pretty lost.
For any $\vec{u},\vec{v}\in\mathbb{R}^3$, $\|\vec{u}\times \vec{v}\|=\|\vec{u}\| \|\vec{v}\| |\sin(\vec{u},\vec{v})|$ equals the area of the parallelogram defined by the points $O,O+\vec{u},O+\vec{u}+\vec{v},O+\vec{v}$.
Thanks to the above, the area of $KLMN$ is $\|\overrightarrow{KL}\times\overrightarrow{KM}\|$: compute the vectors, the cross-product, and then the norm.
Then (like I wrote earlier) the area is given by the norm of this vector: $\|(x,y,z)\|=\sqrt{x^2+y^2+z^2}$, where $(x,y,z)$ is to be replaced by the cross product.