Learning Objectives
Use a dot product to test perpendicularity and find an angle.
Project a vector onto a nonzero direction.
Use a cross product to find a perpendicular vector and an area.
Prerequisites: vector components, magnitudes, and trigonometric ratios. Dot and cross products answer different geometric questions, so check the type of output you need before calculating.
1. The Dot Product Produces a Scalar
For u = ⟨u₁, u₂, u₃⟩ and v = ⟨v₁, v₂, v₃⟩, the dot product is u · v = u₁v₁ + u₂v₂ + u₃v₃. Multiply matching components and add. The result is a number, not a vector.
For nonzero vectors, u · v = |u||v| cos θ, where 0 ≤ θ ≤ π is the angle between them. Thus cos θ = (u · v)/(|u||v|). A zero dot product means nonzero vectors are perpendicular. A positive dot product indicates an acute angle, and a negative one indicates an obtuse angle.
Worked Example: An Angle
Let u = ⟨1, 1, 0⟩ and v = ⟨1, 0, 1⟩. Their dot product is 1, and both magnitudes are √2.
Therefore cos θ = 1/(√2√2) = 1/2, so θ = π/3 radians, or 60°.
The positive dot product agrees with the acute angle. This sign check helps catch arithmetic errors.
2. Projection Is the Part Along a Direction
The vector projection of u onto a nonzero vector v is projᵥ u = [(u · v)/(v · v)]v. The scalar in brackets tells you how much of v to use. The result is parallel to v and represents the component of u along that line.
Worked Example: Project onto an Axis
Project u = ⟨3, 4, 0⟩ onto v = ⟨2, 0, 0⟩. We have u · v = 6 and v · v = 4.
The projection is (6/4)⟨2, 0, 0⟩ = ⟨3, 0, 0⟩.
The remainder u − projᵥ u = ⟨0, 4, 0⟩ is perpendicular to v; their dot product is zero.
3. The Cross Product Produces a Vector
In three dimensions, u × v = ⟨u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁⟩. For nonparallel nonzero vectors, the result is perpendicular to both, with direction set by the right-hand rule. Reversing the order reverses the result: v × u = −(u × v).
Its magnitude is |u × v| = |u||v| sin θ. This is the area of the parallelogram spanned by u and v. The triangle with those two side vectors has half that area. Parallel vectors have cross product zero.
Worked Example: Perpendicular Direction and Area
Let u = ⟨1, 2, 0⟩ and v = ⟨0, 1, 3⟩. Their cross product is ⟨6, −3, 1⟩.
Check perpendicularity: ⟨6, −3, 1⟩ · u = 6 − 6 = 0, and its dot product with v is −3 + 3 = 0.
The parallelogram area is √(36 + 9 + 1) = √46. The corresponding triangle area is √46/2.
Common Mistakes
A dot product is a scalar; a cross product is a vector. Do not swap the operations.
Projection onto v requires v ≠ 0. Angle formulas also require both vectors to be nonzero.
Cross product order matters. Verify the middle component carefully and check perpendicularity afterward.
4. Practice
1. Are ⟨2, −1, 3⟩ and ⟨1, 5, 1⟩ perpendicular?
2. Project ⟨2, 2, 1⟩ onto ⟨1, 1, 0⟩.
3. Find ⟨2, 0, 0⟩ × ⟨0, 3, 0⟩ and the area of the triangle these vectors span.
Worked Solutions
1. Their dot product is 2 − 5 + 3 = 0. Both are nonzero, so they are perpendicular.
2. The projection coefficient is (2 + 2)/(1 + 1) = 2. The projection is ⟨2, 2, 0⟩. The remainder ⟨0, 0, 1⟩ is perpendicular to the target direction.
3. The cross product is ⟨0, 0, 6⟩. The parallelogram area is 6 and the triangle area is 3 square units.
5. Summary
- Use the dot product for angles, perpendicularity, and projection.
- Use the cross product for a normal direction or a spanned area in three dimensions.
- Check the output type and any nonzero-vector requirements before applying a formula.