The vector product and the scalar triple product

A product that is a vector

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

Check the current formula booklet for your awarding organisation; whether or not the components, the 3 × 3 determinant form and the |a||b| sin θ form are supplied, you must know when each applies. The result is a 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).

Perpendicular to both still leaves two opposite directions, and the geometrical definition settles which one. Writing n for the unit vector along the product, a × b = |a||b| sin θ n, where a, b and n in that order form a right-handed triple, the same handedness as i, j and k. Curl the fingers of your right hand from a round to b and your thumb lies along n. The components say the same thing: i × j = k, j × k = i and k × i = j, while each reversed product turns the sign over.

Describe the drawing

A cyan arrow a and an amber arrow b spanning a shaded parallelogram, with the coral arrow a × b standing straight up out of it from their common tail. A curved arrow sweeps round that upright, from a towards b, in the direction the fingers of a right hand take when the thumb points along the product. Beside it the unit vectors i, j and k are drawn with the same handedness, so that i × j = k reads off the picture. A grey arrow runs downwards from the same tail for b × a, which is the same length pointing the opposite way.

The algebra needs one more rule. The vector product distributes over addition, a × (b + c) = a × b + a × c, so a product of two sums expands as brackets usually do, as long as each term keeps its two factors in the original order: (a + b) × (c + d) = a × c + a × d + b × c + b × d. Anti-commutativity then simplifies the result: (a + b) × (a − b) = −2(a × b), because a × a and b × b vanish while the two middle terms −(a × b) and b × a are equal rather than cancelling.

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 has a sixth of that volume. A zero triple product means the three vectors are coplanar.

Describe the drawing

A slanted box drawn on three edge arrows a, b and c from one corner. Its volume is the modulus of the scalar triple product, written either way round as a dot b cross c or a cross b dot c, six for the worked example's vectors, and the tetrahedron sharing those three edges occupies exactly one sixth of it.

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.

Describe the drawing

A coral arrow for the direction vector two comma one comma two rising through a dashed component box in a three-dimensional sketch, with three arcs marking its angles alpha, beta and gamma to the x, y and z axes. Dividing each component by the length three gives the direction cosines two thirds, one third and two thirds, and their squares add to exactly one, the identity every set of direction cosines must satisfy.

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 lines with non-parallel directions b₁ and b₂, the shortest distance is the absolute projection of a joining vector onto b₁ × b₂, that is |(a₂ − a₁)·(b₁ × b₂)|/|b₁ × b₂|, which is the volume of a parallelepiped divided by the area of its base. If the directions are parallel, b₁ × b₂ = 0 and the formula does not apply: treat that case as a point-to-line distance instead.