Part 4: Volume of a Parallelepiped

If u,v,and w share a common initial point, then the set of terminal points of the vectors su+tv+rw for 0 £ r,s,t £ 1 is called the parallelepiped spanned by u,v, and w:

Maple/javaview image

In order to find the volume of the parallelepiped, we first notice that the parallelepiped can be "sliced" into parts that can be rearranged to form a new parallelepiped spanned by u, v, and proju×v(w) .  Click the slideshow arrows below the image to see this slicing and rearranging in action.
 
Let's calculate the volume of a parallelepiped spanned by
vectors u, v, and w.
 
The new parallelepiped and the old parallelepiped have the same volume, and because proju×v( w) is perpendicular to the base spanned by u and v, the volume of the new parallelepiped is
Volume
=
( Area of base) ( height)
=
||u×v|| ||proju×v( w) ||
=
||u×v|| ||  w·( u×v)
( u×v) ·( u×v)
( u×v) ||
=
||u×v||   | w·( u×v) |
||u×v||2
 ||u×v||
=
| w·( u×v)|
Thus, the volume of the parallelepiped spanned by u, v, and w is
Equivalently, Volume = | ( u×v) · w |      

EXAMPLE 7    Find the volume of the parallelpiped spanned by u = á 2,0,0 ñ,  v = á1,3,0 ñ, and w = á 1,0,3 ñ . The figure below is drawn as if all vectors have their initial points at the origin.

Maple/javaview image

Solution: The cross product of u and v  is
u×v = 

0
0
3
0
 ,  
0
2
0
1
 ,  
2
0
1
3

= á 0,0,6 ñ
The dot product with w yields the volume:
Volume = | ( u×v) ·w| = | á 1,0,3 ñ · á0,0,6 ñ | = 18