Scalar Triple Product (Parallelepiped Volume)

V=a(b×c)V = \left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|

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Cross two vectors to get a third whose length is the base area and whose direction is the base's normal, then dot that with the remaining vector to pick out the height — the product |a·(b × c)| is the volume of the box the three vectors lean out into. It is also, term for term, the determinant of the 3×3 matrix holding those vectors as rows, which is the cleanest way to see why a zero determinant means the rows are coplanar: the box is flat and holds nothing.

The three edge vectors (2, 0, 0), (0, 3, 0) and (0, 0, 4) give a rectangular box of 24 m³, which the formula returns as 24 000 L. Less obviously, (1, 2, 3), (0, 1, 4) and (5, 6, 0) span a badly skewed parallelepiped of exactly 1 m³. Lagrange used triple products in his 1773 work on the attraction of ellipsoids, and the operation is still the standard test in mesh software for whether a tetrahedron has been built inside-out — one sixth of the triple product is the tetrahedron's volume, and a negative sign means its vertices were listed in the wrong order.

Scalar Triple Product (Parallelepiped Volume)
V=a(b×c)V = \left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|
Where
  • VV= Parallelepiped volume
  • axa_x= x-component of a
  • aya_y= y-component of a
  • aza_z= z-component of a
  • bxb_x= x-component of b
  • byb_y= y-component of b
  • bzb_z= z-component of b
  • cxc_x= x-component of c
  • cyc_y= y-component of c
  • czc_z= z-component of c