the volume of a parallelepiped

Take a cardboard box and push it sideways so its right angles lean over; it becomes a slanted box, a parallelepiped, with six parallelogram faces. How much space does it still enclose? Surprisingly, the leaning makes no difference to the rule for volume, and three edge vectors meeting at one corner tell you the answer instantly through a single number.

Let three edges from one vertex be the vectors u, v, w. The volume is V = |u . (v x w)|, the absolute value of the scalar triple product. The reasoning is clean: v x w is a vector perpendicular to the base parallelogram, and its magnitude |v x w| equals that base's area. Dotting with u picks out the component of u perpendicular to the base — exactly the height. Since volume = base area times height, and the dot product folds the height-times-area together, |u . (v x w)| is the volume. The absolute value matters because the triple product can come out negative (a left-handed corner), and a volume must be non-negative. Equivalently, the volume is the absolute value of the 3-by-3 determinant whose rows are the components of u, v, w.

This single formula does heavy lifting: it gives the volume of any parallelepiped, and one sixth of it gives the volume of the tetrahedron on the same three edges. When the triple product is zero the box is flat — the three edges are coplanar and enclose no volume — which is also the coplanarity test. A subtle point about the box-shape itself: pushing it over (a shear) changes the shape and the face areas but, remarkably, NOT the volume, as long as the base and height are unchanged. This is Cavalieri's principle showing up in vector form, and it is why the simple base-times-height rule survives the slant.