# surface area of parallelepiped

Can someone please explain to me how I would get the surface area of a parallelepiped if I'm given three vectors of the parallelepiped.

*parallelepiped is parallelogram only 3-dimensional.

Also what's the difference between geometric and algebraic vectors?

Edit: I know surface area = 2lw +2lh + 2wh but how would I solve this qusetion if I'm given 3 vectors. Please explain by giving me an example of random 3 vectors.

Last edited by cool_dude; Mar 1, 2007 at 02:12 PM.
 asterisk_man Posts: 476, Reputation: 32 Full Member #2 Mar 1, 2007, 05:03 PM
I believe you should be able to calculate the magnitude of each vector and randomly assign each one as l, w, and h and then apply that to the formula you provided.

sorry i didn't provide the example you wanted. I think my description should be simple enough that you don't need it.
I'm sure you know that the magnitude of a 3d vector is $\sqrt{x^2+y^2+z^2}$
 cool_dude Posts: 124, Reputation: 9 Junior Member #3 Mar 1, 2007, 05:09 PM
Thanks for the reply. Yes I did know that's how you get magnitude of the vector. What I didn't know was you can randomly assign the magnitude of each vector to l,w,h.

Edit: what's the difference between algebraic and geometric vectors? I think geometric has to do with 3-dimensional. Can't remember.
 asterisk_man Posts: 476, Reputation: 32 Full Member #4 Mar 2, 2007, 06:29 AM
I'm not sure what "algebraic" vs "geometric" vectors are either, do we need to figure this out or just curiosity? As far as randomly assigning to l,w,h. Imagine you're trying to calculate the surface area of a sheet of paper. If WxH is 8.5x11 or 11x8.5 it's still going to have the same surface area.
Now of course your 3 vectors must all share a common endpoint otherwise they won't really be describing the shape in question but I assume that much is a given.
Good luck
 Capuchin Posts: 5,255, Reputation: 655 Uber Member #5 Mar 2, 2007, 06:36 AM
How do you define length, width and height for a parallelepiped? If it's just along the edges then I don't think your surface area equation is right, that's for a cuboid.
 asterisk_man Posts: 476, Reputation: 32 Full Member #6 Mar 2, 2007, 07:05 AM
I think it's right. Area of a parallelogram is just w*h
 Capuchin Posts: 5,255, Reputation: 655 Uber Member #7 Mar 2, 2007, 07:11 AM
But the height isn't the length of a side, it's height perpendicular to the base...
 asterisk_man Posts: 476, Reputation: 32 Full Member #8 Mar 2, 2007, 07:30 AM
Oh capuchin, why'd you have to go and make it so difficult
 Capuchin Posts: 5,255, Reputation: 655 Uber Member #9 Mar 2, 2007, 07:31 AM
That's what i do
 asterisk_man Posts: 476, Reputation: 32 Full Member #10 Mar 2, 2007, 10:17 AM
sorry my original answer was wrong.

how's this sound:
I'll call the vectors x,y,z

first find the normalized vector for each vector.
$
\hat x = \frac x {\left|x\right|}\\
\hat y = \frac y {\left|y\right|}\\
\hat z = \frac z {\left|z\right|}
$

we have 3 unique sides, the xy, xz, yz sides. We'll find the area for those then multiply by 2.
Lets' work xy and the others will be obvious.
Pick one vector to be the base lets choose x. The length of this vector with be the width of the parallelogram. We need the height.
now we calculate the angle between x and y
$\theta_{xy}=\arccos \left(\hat x \cdot \hat y\right)$
(that's a dot product, not a multiply)

now the height of the parallelogram is
$height_{xy} = \left|y\right|\sin \theta_{xy}$

so
$area_{xy} = \left|x\right| * height_{xy}$

you can do the same for xz and yz sides

$totalSurfaceArea = 2\left(area_{xy} + area_{xz} + area_{yz}\right)$

how's that look?

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