Topic 3 / 17basic

Dot Product

Compute dot products component-wise and geometrically; find angles, test orthogonality, project, and compute work done by a force.

The dot product takes two vectors and returns a scalar. It encodes the angle between them: parallel vectors give large positive values, perpendicular vectors give zero, anti-parallel vectors give large negatives. From the dot product we get angles, projections, and the formula for work.

Two formulas for the dot product

Component form:

u \cdot v = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}

. Geometric form:

u \cdot v = ∣ u ∣∣ v ∣ cos θ

. Setting them equal lets us solve for

θ

Key takeaways

Component form is fastest for computation.
Geometric form is best for interpretation (angles, orthogonality).

Orthogonality test

u \cdot v = 0 ⟺ u ⊥ v

(assuming neither is the zero vector). This is the workhorse: any time you need to check perpendicularity, dot.

Angle between two vectors

cos θ = \frac{u \cdot v}{∣ u ∣∣ v ∣}

. The angle

θ \in [0, π]

. Negative dot product → obtuse angle; positive → acute.

Orthogonal projection

The projection of

u

onto

v

proj_{v} u = \frac{u \cdot v}{∣ v ∣ ^{2}} v

. Be careful: the **denominator is

∣ v ∣^{2}

**, not

∣ v ∣

. The scalar projection (length, signed) is

u \cdot v /∣ v ∣

Strategy tips

Always include the factor $v$ at the end so the answer is a vector.
If you only want the length, divide $u \cdot v$ by $∣ v ∣$ (not $∣ v ∣^{2}$ ).

Work done by a constant force

If a constant force

F

moves an object through displacement

d

, the work done is

W = F \cdot d

. Only the component of force along the displacement contributes.

Algebraic properties

Commutative:

u \cdot v = v \cdot u

. Distributive:

u \cdot (v + w) = u \cdot v + u \cdot w

v \cdot v = ∣ v ∣^{2}

u \cdot 0 = 0

Worked examples

Example 1

Find the angle between

u = ⟨ 1, 2, 3 ⟩

and

v = ⟨ - 2, 1, 4 ⟩

1
Dot product.
$u \cdot v = - 2 + 2 + 12 = 12$
2
Magnitudes.
$∣ u ∣ = 14, ∣ v ∣ = 21$
3
Cosine of the angle.
$cos θ = \frac{12}{14 \cdot 21} = \frac{12}{294}$
4
Take inverse cosine.
$θ = cos^{- 1} (\frac{12}{294}) \approx 45.6°$

Answer

θ = cos^{- 1} (12/ 294) \approx 45.6°

Example 2

Find

proj_{v} u

where

u = ⟨ 4, 1 ⟩

and

v = ⟨ 2, 3 ⟩

1
Dot product.
$u \cdot v = 8 + 3 = 11$
2
Magnitude squared.
$∣ v ∣^{2} = 4 + 9 = 13$
3
Projection formula.
$proj_{v} u = \frac{11}{13} ⟨ 2, 3 ⟩ = ⟨ 22/13, 33/13 ⟩$

Answer

⟨ 22/13, 33/13 ⟩

Example 3

A force

F = 4 i + 5 j + 10 k

moves an object from

(1, 0, 2)

(5, 3, 8)

. Find the work done.

1
Displacement vector.
$d = ⟨ 5 - 1, 3 - 0, 8 - 2 ⟩ = ⟨ 4, 3, 6 ⟩$
2
Work is dot product.
$W = F \cdot d = 16 + 15 + 60 = 91$

Answer

W = 91

(units of force × distance).

Interactive visualizations

Loading visualization…

u_x3.00

u_y1.00

v_x2.00

v_y3.00

u \cdot v = 3.0 \cdot 2.0 + 1.0 \cdot 3.0 = 9.00

cos θ = \frac{9.00}{3.16 \cdot 3.61} = 0.789 \Rightarrow θ \approx 37. 9^{\circ}

proj_{v} u has length \frac{u \cdot v}{∣ v ∣} = 2.496

Adjust the components of

u

and

v

. Watch the dot product, angle, and projection update live.

Formulas in this topic

Dot product (components)

u \cdot v = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}

Compute via components.

Dot product (geometric)

u \cdot v = ∣ u ∣∣ v ∣ cos θ

Use to extract the angle between vectors.

Angle between vectors

cos θ = \frac{u \cdot v}{∣ u ∣∣ v ∣}

Solve for θ ∈ [0, π].

Vector projection

proj_{v} u = \frac{u \cdot v}{∣ v ∣ ^{2}} v

Component of u along v as a vector.

Scalar projection

comp_{v} u = \frac{u \cdot v}{∣ v ∣}

Signed length of the projection.

Work (constant force)

W = F \cdot d

Constant force times displacement, dotted.

Two formulas for the dot product

Orthogonality test

Angle between two vectors

Orthogonal projection

Work done by a constant force

Algebraic properties

Worked examples

Interactive visualizations

Formulas in this topic

Related topics