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Vectors and the Volume of Parallelepipeds

Date: 03/08/2003 at 21:52:49
From: William
Subject: Vectors and the volume of Parallelepipeds

Could you explain for me how the formula V = |a.(b x c)| (the volume
of a parallelepiped is equal to the magnitude of the scalar triple
product of the vectors that determine the parallelepiped; where a, b,
and c are those vectors) is derived?

I have a suspicion that it is quite similar to the derivation of a
similar formula for the volume of the tetrahedron found at

   Volume of a Tetrahedron

but could you explain it to me in terms of a parallelepiped?  Your 
response will be greatly appreciated.

Date: 03/12/2003 at 08:56:31
From: Doctor Rick
Subject: Re: Vectors and the volume of Parallelepipeds

Hi, William.

The elements we need are indeed found in that derivation. 

First, as stated there, b x c is a vector perpendicular to the plane 
containing b and c, and its magnitude is the area of the parallelogram 
with sides b and c. This is easy to see, because the vector product 
has magnitude |b| |c| sin(theta), where theta is the angle between the 
vectors, and |c| sin(theta) is the altitude of the parallelogram:

        /|                 /
       / |                /
     c/  |               /
     /   |h             /
    /    |             /
   /th   |            /

You see that h = |c|*sin(theta). The area of the parallelogram is the 
base times the altitude, or |b| * h = |b| * |c| * sin(theta).

Next, the dot product a . d is the length of d times the component of 
a parallel to d. Since d = b x c is perpendicular to the base 
parallelogram and its length is the area of the base parallelogram, 
a . (b x c) is the area of the base times the component of a 
perpendicular to the base. What is the component of a perpendicular 
to the base? It is the altitude of the parallelepiped. Therefore 
a . (b x c) is the area of the base times the altitude. And what is 
the volume of a parallelepiped? The area of the base times the 

The absolute value is put in there just because, depending on the 
order in which you put the vectors, you might get the vector product 
on the opposite side of the base parallelogram from the third side, 
in which case you'll get a negative number. Taking the absolute value 
gives you the correct volume.

The dot product is the same as the scalar product. It is equal to the 
product of the magnitudes of the vectors with the cosine of the angle 
between them:

  a . b = |a| |b| cos(theta)

Consider the triangle here:

         / |
       b/  |
       /   |
      /th  |

You can see that |b| cos(theta) is the length of the bottom leg of 
the triangle. Are you familiar with resolving a vector into two 
perpendicular components? If vector a lies parallel to the x axis, 
then the component we are looking at is the x component of the 
vector, and the vertical line in the figure is the y component. If a 
lies in a different direction, we can still speak of the component 
parallel to a and the component perpendicular to a, but they aren't 
simply the x and y components; they are the magnitude of b times the 
cosine and sine respectively of the angle between a and b.

Thus |b| cos(theta) is the component of b parallel to vector a, and 
therefore the dot product a . b = |a| |b| cos(theta) is the magnitude 
of a times the component of b parallel to a. (It's also the magnitude 
of b times the component of a parallel to b; you can see it either 
way. The dot product is commutative: a . b = b . a)

Does this make things clear?

- Doctor Rick, The Math Forum 

Date: 03/13/2003 at 08:49:56
From: William
Subject: Thank you (Vectors and the volume of Parallelepipeds)

Yes, it makes it perfectly clear. Thank you for your help. It is 
greatly appreciated.
Associated Topics:
College Higher-Dimensional Geometry
College Linear Algebra

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