The Area of a Parallelogram: Angle Not Required

The area of a parallelogram can be found using two vectors.

Let the two vectors be:

A and B

If A is the base of the parallelogram, then the height is the perpendicular part of B.

From the diagram:

height = |B|sin(θ)

Therefore:

Area = base × height

Area = |A||B|sin(θ)

This is the standard formula for the area of a parallelogram formed by two vectors.

Removing the Angle

The formula

Area = |A||B|sin(θ)

uses the angle θ between the two vectors. But the angle is not always given.

To remove the angle, start with the dot product identity:

A · B = |A||B|cos(θ)

Now square both sides:

(A · B)² = |A|²|B|²cos²(θ)

Using the trigonometric identity:

cos²(θ) = 1 − sin²(θ)

we get:

(A · B)² = |A|²|B|²(1 − sin²(θ))

Expand the right-hand side:

(A · B)² = |A|²|B|² − |A|²|B|²sin²(θ)

Now rearrange:

|A|²|B|²sin²(θ) = |A|²|B|² − (A · B)²

The Area Identity

Since

Area = |A||B|sin(θ)

squaring both sides gives:

Area² = |A|²|B|²sin²(θ)

But we already showed that:

|A|²|B|²sin²(θ) = |A|²|B|² − (A · B)²

Therefore:

Area² = |A|²|B|² − (A · B)²

Taking the square root:

Area = √(|A|²|B|² − (A · B)²)

Using Coordinates

If

A = (a₁, a₂, a₃)

and

B = (b₁, b₂, b₃)

then:

|A|² = a₁² + a₂² + a₃²

|B|² = b₁² + b₂² + b₃²

A · B = a₁b₁ + a₂b₂ + a₃b₃

So the area can be found directly from the coordinates:

Area = √((a₁² + a₂² + a₃²)(b₁² + b₂² + b₃²) − (a₁b₁ + a₂b₂ + a₃b₃)²)

Main Point

The area of the parallelogram can be found without knowing the angle.

The identity

Area = √(|A|²|B|² − (A · B)²)

allows the area to be calculated directly from vector lengths and the dot product.

Angle not required. Just coordinates.