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Dot Product and Cross Product in 3D

Key Terms

Dot product (algebraic)
a · b = a₁b₁ + a₂b₂ + a₃b₃. The result is a scalar.
Dot product (geometric)
a · b = |a||b|cosθ, where θ is the angle between the vectors (0 ≤ θ ≤ π).
Perpendicularity test
a ⊥ b ⇔ a · b = 0 (provided neither is the zero vector).
Projection
The scalar projection of a onto b is (a · b)/|b|. The vector projection is [(a · b)/|b|²]b.
Cross product
a × b is a vector computed via the 3×3 determinant with i, j, k in the first row:
   a × b = (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k
Cross product magnitude
|a × b| = |a||b|sinθ ≥ 0.
Perpendicularity
a × b is perpendicular to both a and b.
Area of parallelogram
with sides a and b: Area = |a × b|.
Area of triangle
with sides a and b: Area = ½|a × b|.
Anti-commutativity
a × b = −(b × a).
Key Formulas:
  • Dot product: a · b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b|cosθ
  • Angle: cosθ = (a · b) / (|a||b|)
  • Cross product: a × b = (a₂b₃−a₃b₂, a₃b₁−a₁b₃, a₁b₂−a₂b₁)
  • |a × b| = |a||b|sinθ
  • Area of parallelogram = |a × b|; Area of triangle = ½|a × b|
Hot Tip Remember the sign pattern for the cross product: i gets a +, j gets a −, k gets a +. A reliable method is to expand along the first row of the 3×3 determinant, using cofactors. For the angle between two vectors, always use the dot product formula — the cross product gives |sinθ|, which loses the obtuse/acute distinction.

Worked Example 1 — Angle between two vectors

Find the angle between a = (1, 2, 2) and b = (0, 3, 4).

a · b = (1)(0) + (2)(3) + (2)(4) = 0 + 6 + 8 = 14

|a| = √(1 + 4 + 4) = √9 = 3

|b| = √(0 + 9 + 16) = √25 = 5

cosθ = 14 / (3 × 5) = 14/15

θ = cos−1(14/15) ≈ 21.1°

Worked Example 2 — Cross product

Find a × b where a = (1, 2, 3) and b = (4, 5, 6).

a × b = ( (2)(6) − (3)(5) )i − ( (1)(6) − (3)(4) )j + ( (1)(5) − (2)(4) )k

= (12 − 15)i − (6 − 12)j + (5 − 8)k

= −3i + 6j − 3k = (−3, 6, −3)

Verify perpendicularity: a · (a × b) = (1)(−3) + (2)(6) + (3)(−3) = −3 + 12 − 9 = 0 ✓

The Dot Product: Measuring How “Aligned” Two Vectors Are

The dot product of two vectors a and b quantifies how much they point in the same direction. Algebraically, it is computed component-wise: a · b = a₁b₁ + a₂b₂ + a₃b₃. The result is a scalar — there is no direction attached to it. Geometrically, a · b = |a||b|cosθ, where θ is the angle between the vectors when placed tail-to-tail.

When θ = 90°, cosθ = 0, so a · b = 0 regardless of the magnitudes. This gives the elegant perpendicularity test: two non-zero vectors are perpendicular if and only if their dot product is zero. This is one of the most-used results in all of 3D geometry.

Geometric Interpretation: Projection

The scalar projection of a onto b is the signed length of the shadow that a casts in the direction of b:

  • Scalar projection: compₛa = (a · b) / |b|
  • Vector projection: projₛa = [(a · b) / |b|²]b

Projections appear in physics (component of force along a direction), computer graphics (shading), and geometry (finding the closest point on a line to a given point).

The Cross Product: A Vector Perpendicular to Both

Unlike the dot product, the cross product a × b produces a vector. It is defined only in three dimensions and is computed as the determinant:

  |i  j  k |
  |a₁ a₂ a₃|
  |b₁ b₂ b₃|

Expanding: a × b = (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k.

The resulting vector is perpendicular to both a and b. Its direction is given by the right-hand rule: point your fingers from a towards b and your thumb points in the direction of a × b.

The Right-Hand Rule

To find the direction of a × b: align your right hand with a, curl your fingers toward b, and your thumb points in the direction of a × b. Reversing the order gives b × a = −(a × b), pointing in the opposite direction.

Magnitude and Area

The magnitude |a × b| = |a||b|sinθ equals the area of the parallelogram formed by a and b. This is because the area of a parallelogram is base × height = |b| × |a|sinθ. The area of the triangle with sides a and b is half of this.

This geometric interpretation makes the cross product essential in: finding normal vectors to planes, computing areas of triangular surfaces, and determining if three points are collinear (cross product of two edges = 0).

Normal Vectors to a Plane

If two vectors u and v lie in a plane, then n = u × v is a normal to that plane. This is used to find the equation of a plane through three points: compute two edge vectors, take their cross product to get n, then use the point-normal form.

Exam Tip: To verify your cross product, always check that it is perpendicular to both original vectors by computing two dot products. Each should equal zero.
Exam Tip: When finding the angle between two vectors, the dot product formula always gives the angle in [0°, 180°]. If you need an acute angle specifically, and you get an obtuse answer, that is still correct — it is the angle between the vectors as positioned.

Mastery Practice

See Answers ➔

  1. Compute the dot product a · b for each pair. Fluency

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

  2. Find the angle θ between each pair of vectors. Give your answer in degrees, correct to one decimal place. Fluency

    1. (a) a = (1, 0, 1), b = (0, 1, 1)
    2. (b) a = (2, 1, −2), b = (1, 2, 2)

  3. Determine whether each pair of vectors is perpendicular. Fluency

    1. (a) u = (3, −2, 1), v = (1, 2, 1)
    2. (b) u = (1, 4, −2), v = (2, −1, 1)
    3. (c) u = (2, 0, −3), v = (3, 1, 2)

  4. Find the cross product a × b for each pair. Fluency

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

  5. Find the area of the parallelogram with adjacent sides a = (1, 3, −2) and b = (2, 1, 4). Understanding

  6. Find the area of the triangle with vertices A = (1, 0, 0), B = (0, 2, 0), C = (0, 0, 3). Understanding

  7. Find a unit vector that is perpendicular to both a = (1, 1, 0) and b = (0, 1, 1). Understanding

  8. Find the value of k such that a = (k, 2, −1) and b = (3, k, 4) are perpendicular. Understanding

  9. Show that the vectors a = (3, −1, 2), b = (1, 4, 1) and c = (5, −6, 3) satisfy a × b = c up to sign. Hence find the area of the parallelogram formed by a and b. Problem Solving

  10. Two forces F₁ = (2, −3, 1) N and F₂ = (1, 2, −4) N act at a point. Find: (a) the angle between the forces, (b) a vector perpendicular to both forces. Problem Solving