Maths › Further 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.
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 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.
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.
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.