MathsFurther Pure 1 › The vector product and the scalar triple product

The vector product and the scalar triple product

Multiply two vectors and get a third, perpendicular to both, whose length is an area; dot it with a third vector and the result is a volume. Geometry becomes arithmetic.

Year FMEDEXCEL 9FM0 FP1

Builds on Lines and planes in three dimensions and Determinants and inverses.

IN THIS TOPIC

  • Compute a × b from components and check it is perpendicular to both.
  • Use |a × b| as an area and a·b × c as a volume, with the tetrahedron's sixth.
  • Write a line as (r − a) × b = 0 and read off direction ratios and cosines.

WHAT YOU PROBABLY THINK

Multiplying two vectors can only sensibly produce a number, as the scalar product does.

A product that is a vector

The vector product a × b is the determinant-style expansion

a × b = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

a genuine vector, perpendicular to both factors, so the opening claim undersells what multiplication can do. Its magnitude is |a||b| sin θ, the area of the parallelogram the two vectors span; half of it is the triangle's area.

a × b stands perpendicular to the parallelogram spanned by a and b, and its length is that parallelogram's areaaba × barea |a × b|perpendicular to both, by construction
FIG. 1The parallelogram spanned by a and b with a × b standing perpendicular to both: the product's length is the shaded area.

WORKED EXAMPLE

A product, checked twice

Find a × b for a = (1, 2, 0) and b = (3, 1, 2), and the area of the triangle the vectors span.

a × b = (2×2 − 0×1, 0×3 − 1×2, 1×1 − 2×3) = (4, −2, −5).

Perpendicularity check: (4, −2, −5)·(1, 2, 0) = 0 and (4, −2, −5)·(3, 1, 2) = 0. Both pass.

|a × b| = √(16 + 4 + 25) = 3√5, so the triangle has area 3√5/2.

Volumes, lines and cosines

Dotting a third vector in gives the scalar triple product a·b × c: the volume of the parallelepiped on the three vectors, with the tetrahedron on the same edges holding a sixth of it. A zero triple product means the three vectors are coplanar, a test worth its own mark.

WORKED EXAMPLE

A volume from three edges

Find the volume of the parallelepiped with edges a = (1, 0, 0), b = (1, 2, 0), c = (1, 1, 3), and of the tetrahedron on the same edges.

b × c = (2×3 − 0×1, 0×1 − 1×3, 1×1 − 2×1) = (6, −3, −1).

a·(b × c) = 6: parallelepiped volume 6.

Tetrahedron: a sixth of it, volume 1. The sixth is the constant students forget; the parallelepiped never needs one.

The parallelepiped on a, b and c: volume a·b × c = 6, and the tetrahedron on the same edges takes a sixthabcvolume = a·b × ctetrahedron: a sixth
FIG. 2The parallelepiped on a, b and c: its volume is the scalar triple product, and the tetrahedron on the same edges takes one sixth.

A line through a with direction b is also (r − a) × b = 0: displacement parallel to b means their product vanishes. The components of any direction vector are its direction ratios; dividing by the length gives direction cosines (l, m, n) with l² + m² + n² = 1.

YOUR TURN

Cosines of a direction

Find the direction cosines of the vector (2, −1, 2), and verify their defining property.

Show the working

The length is √(4 + 1 + 4) = 3.

Cosines: l = 2/3, m = −1/3, n = 2/3.

l² + m² + n² = (4 + 1 + 4)/9 = 1, as the cosines of the angles with the three axes must always satisfy.

THE EXAM BIT

  • After every cross product, dot the answer with both factors; two zeros cost seconds and catch sign slips.
  • Areas come from a magnitude, so they carry a square root; volumes come from a triple product and do not.
  • The tetrahedron takes one sixth of the parallelepiped, not one third.
  • a·b × c = 0 is the coplanarity test; quote it rather than building planes.

CHECK YOURSELF

Vectors u = (2, 0, 1) and v = (4, 0, 2) satisfy u × v = 0. What does this say about them?

Show a hint

The magnitude of the product is |u||v| sin θ.

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a × b is perpendicular to both factors with |a × b| = |a||b| sin θ, the parallelogram's area.

a·b × c is the parallelepiped's volume; the tetrahedron takes a sixth; zero means coplanar.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

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  • Compute a × b from components and check it is perpendicular to both.
  • Use |a × b| as an area and a·b × c as a volume, with the tetrahedron's sixth.
  • Write a line as (r − a) × b = 0 and read off direction ratios and cosines.

Open the full revision checklist to see every objective in the course in one place.

No animated video for this topic yet; these notes stand alone.