Scalar (dot) product

For non-zero a and b with angle θ between them (0 ≤ θ ≤ π), a·b = |a||b| cos θ. If either vector is 0, a·b = 0. The result is a real number, not a vector.

For non-zero vectors, a·b = 0 exactly when a ⊥ b. At θ = 0, a·b = |a||b|; at θ = π, a·b = −|a||b|. In particular a·a = |a|², so |a| = √(a·a).

The unit vectors give î·î = ĵ·ĵ = k̂·k̂ = 1 and î·ĵ = ĵ·k̂ = k̂·î = 0.

The angle between non-zero vectors: cos θ = (a·b)/(|a||b|), so θ = cos⁻¹[(a·b)/(|a||b|)].

The dot product is commutative (a·b = b·a) and distributive (a·(b + c) = a·b + a·c), and (λa)·b = λ(a·b) = a·(λb). So in components a·b = a₁b₁ + a₂b₂ + a₃b₃.

Worked: if |a| = 1, |b| = 2 and a·b = 1, then cos θ = 1/2 and θ = π/3. For î + ĵ − k̂ and î − ĵ + k̂, the dot product is 1 − 1 − 1 = −1 and cos θ = −1/3.

Worked: for a = 5î − ĵ − 3k̂ and b = î + 3ĵ − 5k̂, (a + b)·(a − b) = (6)(4) + (2)(−4) + (−8)(2) = 24 − 8 − 16 = 0, so a + b and a − b are perpendicular.