# Area of a Parallelogram with Vertices..

• Jan 11th 2009, 10:00 PM
sfgiants13
Area of a Parallelogram with Vertices..
[1,2,3]
[1,3,6]
[3,8,6]
[3,7,3]

I know I have to find 2 (I think) vectors that determine the area and then I take the cross product of the 2. That's about all I know I don't know which 2 determine the area.
• Jan 11th 2009, 10:14 PM
earboth
Quote:

Originally Posted by sfgiants13
[1,2,3]
[1,3,6]
[3,8,6]
[3,7,3]

I know I have to find 2 (I think) vectors that determine the area and then I take the cross product of the 2. That's about all I know I don't know which 2 determine the area.

The given vectors are the stationary vectors of the vertices of the parallelogram. Calculate the vectors describing the sides of the parallelogram:

$\overrightarrow{AD} = \vec d - \vec a = [2,5,0]$

$\overrightarrow{BC} = \vec c - \vec b = [2,5,0]$
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

$\overrightarrow{AB} = \vec b - \vec a = [0,1,3]$

$\overrightarrow{DC} = \vec c - \vec d = [0,1,3]$
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

You now know the pairs of parallels. As you have written, the area is

$a_{parallelogram} = |\overrightarrow{AD} \times \overrightarrow{AB}| =| [2,5,0] \times[0,1,3] |= | [15,-6,2] | = \sqrt{265} \approx 16.2788$