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 factors.
- 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.
- Find distances in three dimensions using the vector and triple products.
COMMON MISCONCEPTION
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
The booklet prints it under Vectors, in components, as a 3 by 3 determinant and in the |a||b| sin θ form, so the sign pattern is not yours to drill. It is a genuine vector, perpendicular to both factors. Its magnitude is |a||b| sin θ, the area of the parallelogram the two vectors span, and half of that is the triangle's area. Order matters, since b × a = −(a × b).
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 and coplanarity
Dotting a third vector in gives the scalar triple product a·(b × c), whose modulus is the volume of the parallelepiped built on the three vectors. The tetrahedron on the same three edges holds a sixth of that. A zero triple product means the three vectors are coplanar, a test that is worth its own mark and takes one line.
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, so the parallelepiped has volume 6.
The tetrahedron takes a sixth of it, volume 1. That sixth is the constant students forget, and the parallelepiped never needs one.
Lines, distances and direction cosines
A line through a with direction b can be written as (r − a) × b = 0, since a displacement parallel to b has a vanishing product with it. The components of any direction vector are its direction ratios, and dividing by the length gives the direction cosines (l, m, n), which satisfy l² + m² + n² = 1.
Both products measure distances. The perpendicular distance from a point P to that line is |(p − a) × b|/|b|, because the numerator is the area of a parallelogram whose base is |b|. For two skew lines the shortest distance is |(a₂ − a₁)·(b₁ × b₂)|/|b₁ × b₂|, which is the volume of a parallelepiped divided by the area of its base. Both formulae are worth quoting with the geometry attached, since that is what stops you inverting them under pressure.
GUIDED PRACTICE
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, which the cosines of the angles with the three axes must always satisfy.
ASSESSMENT FOCUS
- After every cross product, dot the answer with both factors; two zeros cost seconds and catch sign slips.
- Areas come from a magnitude and 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, so quote it instead of building a plane.
- Take the modulus of a triple product before calling it a volume, since the sign only records orientation.
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.
Show the answer
sin θ = 0, so the vectors are parallel, and indeed v = 2u. A zero vector product tests for parallel vectors in the same way that a zero scalar product tests for perpendicular ones.
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, and 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.
Or read them with their worked answers on the the vector product and the scalar triple product questions page.
CHECK YOUR PROGRESS
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- Compute a × b from components and check it is perpendicular to both factors.
- 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.
- Find distances in three dimensions using the vector and triple products.
Open the full revision checklist to see every objective in the course in one place.