Dot (Scalar) Product

a \cdot b = ∣ a ∣∣ b ∣ cos θ = a_{x} b_{x} + a_{y} b_{y} + a_{z} b_{z}

The dot product is a scalar. It is zero for orthogonal vectors, |a|² for a vector with itself, and equals work W = F·d when force and displacement are vectors. It is commutative.

$a_{x}, a_{y}, a_{z}$: Components of a
$b_{x}, b_{y}, b_{z}$: Components of b
$θ$: Angle between a and b
$∣ a ∣$: Magnitude of a

Step-by-step solved example

a = ⟨3, 4, 0⟩, b = ⟨4, 3, 0⟩. Find a·b and cos θ.

Two vectors and the included angle for a·b = |a||b|cosθ.

1. Components
a·b = 3(4)+4(3)+0 = 24. |a| = 5, |b| = 5.
$a \cdot b = 24$
2. Angle
cos θ = 24/(5×5) = 0.96, θ ≈ 16.3°.
$cos θ = 0.96$

Answer: a·b = 24, cos θ = 0.96

10 practice questions

0/10 correct

1.⟨3, 4⟩ · ⟨3, 4⟩ equals

2.⟨1, 0⟩ · ⟨0, 1⟩ equals

3.Orthogonal vectors have a dot product of

4.|a| = 2, |b| = 3, θ = 60°. a·b =

5.⟨2, 3, 6⟩ · ⟨1, 1, 1⟩ equals

6.a·b = b·a means the dot product is

7.Work F·d with F = ⟨0, 10⟩ N and d = ⟨3, 4⟩ m is

8.If a·b = |a||b| then θ is

9.⟨4, 0, 3⟩ · ⟨0, 5, 0⟩ equals

10.|a|² equals