From area to volume
The dot product gives lengths and angles. The cross product gives areas. Put them together and you get volume.
Three vectors a, b and c starting at one point determine a slanted box called a parallelepiped (par-uh-lel-uh-PIE-ped): a solid whose six faces are parallelograms, with opposite faces parallel. A shoebox is a special case where every angle is 90°. A stack of paper pushed sideways, or a crystal of calcite, has this shape with slanted sides.
The volume of any prism is base area × height. For a parallelepiped:
- The base is the parallelogram on b and c, so its area is |b × c| (lesson 8).
- b × c is perpendicular to the base. The height h is the length of the part of a in that perpendicular direction: the absolute value of the scalar projection of a on b × c, which is |a · (b × c)| / |b × c| (lesson 5).
Multiply: V = |b × c| × |a · (b × c)| / |b × c|, so
V = |a · (b × c)|
The number a · (b × c) is called the triple scalar product (or scalar triple product). Notice the order of operations: the cross product first (it must be, since a · b is a number and you cannot cross a number with c).
Worked exampleExample 1: A volume you can check
Find the volume of the parallelepiped determined by a = [2, 0, 0], b = [1, 3, 0] and c = [1, 1, 4]. Check it with base × height.
- b × c = [(3)(4) − (0)(1), (0)(1) − (1)(4), (1)(1) − (3)(1)] = [12, −4, −2].
- a · (b × c) = 2(12) + 0(−4) + 0(−2) = 24.
- V = |24| = 24.
- Check: a and b lie in the xy-plane, so they form a base of area |a × b| = |[0, 0, 6]| = 6. The third edge c rises 4 units (its z-component), so the height is 4. Volume 6 × 4 = 24 ✓.
Worked exampleExample 2: A negative triple product
Find the volume of the parallelepiped determined by a = [1, 2, −1], b = [3, 0, 2] and c = [−1, 4, 1].
- b × c = [(0)(1) − (2)(4), (2)(−1) − (3)(1), (3)(4) − (0)(−1)] = [−8, −5, 12].
- a · (b × c) = −8 − 10 − 12 = −30.
- V = |−30| = 30.
V = 30 cubic units. The negative sign only means a is on the opposite side of the base from b × c (the three vectors, in this order, form a left-handed set). Volume is always the absolute value.
Coplanar vectors
Three vectors are coplanar if they lie in one plane when placed tail to tail. Then the parallelepiped is flat and has no volume. So:
a · (b × c) = 0 if and only if a, b and c are coplanar.
This also tests whether four points lie in one plane: build three vectors from one of the points to the other three, and compute their triple scalar product.
Worked exampleExample 5: Volume from vertices
A parallelepiped has a vertex P(1, 1, 1), and the three vertices joined to P by edges are Q(3, 1, 2), R(1, 4, 1) and S(2, 2, 5). Find its volume.
- Edges from P: a = PQ = [2, 0, 1], b = PR = [0, 3, 0], c = PS = [1, 1, 4].
- b × c = [(3)(4) − (0)(1), (0)(1) − (0)(4), (0)(1) − (3)(1)] = [12, 0, −3].
- a · (b × c) = 24 + 0 − 3 = 21.
Hands-on activityThe sheared deck
You need: a deck of playing cards (or a thick pad of sticky notes, or a stack of 50 to 100 sheets of paper), a ruler and a calculator.
- Stack the cards neatly. Measure the length ℓ, the width w and the thickness t of the stack in centimetres. The volume is ℓwt.
- Push the top of the stack forward and sideways so it leans in two directions at once. Now it is a parallelepiped.
- Set up axes: x along the card length, y along the card width, z up, with the origin at a bottom corner. The bottom edges are a = [ℓ, 0, 0] and b = [0, w, 0]. Measure how far the top card has moved forward (p) and sideways (q) compared with the bottom card. The slanted edge is c = [p, q, t].
- Compute |c · (a × b)| and compare with ℓwt.
What to look for: a × b = [0, 0, ℓw], so c · (a × b) = ℓwt no matter what p and q are. The cards are the same cards, so the volume cannot change, and the formula agrees. For an 8.9 cm × 6.4 cm × 1.6 cm deck, the volume stays about 91 cm³.