The Areas of Parallelograms and Triangles in 3-Space

# The Area of a Parallelogram in 3-Space

Given two vectors $\vec{u} = (u_1, u_2, u_3)$ and $\vec{v} = (v_1, v_2, v_3)$, if we place $\vec{u}$ and $\vec{v}$ so that their initial points coincide, then a parallelogram is formed as illustrated:

Calculating the area of this parallelogram in 3-space can be done with the formula $A= \| \vec{u} \| \| \vec{v} \| \sin \theta$. We will now begin to prove this.

 Theorem 1: If $\vec{u}, \vec{v} \in \mathbb{R}^3$, then the area of the parallelogram formed by $\vec{u}$ and $\vec{v}$ can be computed as $\mathrm{Area} = \| \vec{u} \| \| \vec{v} \| \sin \theta$.
• Proof: First construct some vectors $\vec{u}$ and $\vec{v}$ in 3-space such that their initial points coincide and let theta be the angle between these two vectors. Geometrically, we know that the area for a parallelogram is $A = bh$ where $b$ is the base of the parallelogram and $h$ is the height.
• Making appropriate substitutions, we see that the base of the parallelogram is the length of $\vec{v}$ or rather the its norm $\| \vec{v} \|$. Furthermore, we can calculate the height of this parallelogram using right-triangle properties from the following illustration:
• We know that $\sin \theta = \frac{opposite}{hypotenuse}$, and thus it follows that we need to solve for the opposite side of this constructed triangle (our height). It thus follows that $\sin \theta = \frac{height}{\| \vec{u} \| }$ or more appropriately, $h = \sin \theta \| \vec{u} \|$. Since we now know the base and height of the parallelogram, we can substitute this back into the formula for the area of a parallelogram to get:
(1)
\begin{align} A = \| \vec{u} \| \| \vec{v} \| \sin \theta \\ \blacksquare \end{align}

# The Relationship of the Area of a Parallelogram to the Cross Product

As we will soon see, the area of a parallelogram formed from two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$ can be seen as a geometric representation of the cross product $\vec{u} \times \vec{v}$. First, recall Lagrange's Identity:

(2)
\begin{align} \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}\|^2 \|\vec{v} \|^2 - (\vec{u} \cdot \vec{v})^2 \end{align}

We can instantly make a substitution into Lagrange's formula as we have a convenient substitution for the dot product, that is $\vec{u} \cdot \vec{v} = \| \vec{u} \| \| \vec{v} \| \cos \theta$. Making this substitution and the substitution that $\cos ^ \theta = 1 - \sin^2 \theta$ we get that:

(3)
\begin{align} \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}\|^2 \|\vec{v} \|^2 - (\| \vec{u} \| \| \vec{v} \| \cos\theta)^2 \\ \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}\|^2 \|\vec{v} \|^2 - \| \vec{u} \|^2 \| \vec{v} \|^2 \cos^2\theta \\ \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}\|^2 \|\vec{v} \|^2 (1 - \cos^2\theta) \\ \| \vec{u} \times \vec{v} \|^2 = \|\vec{u}||^2 \|\vec{v} \|^2 \sin^2\theta \end{align}

The last step is to square root both sides of this equation. Since the length/norm of a vector will always be positive and that $\sin \theta > 0$ for $0 ≤ \theta < \pi$, it follows that all parts under the square root are positive, therefore:

(4)
\begin{align} \| \vec{u} \times \vec{v} \| = \|\vec{u}\| \|\vec{v}\| \sin \theta \end{align}

Note that this is the same formula as the area of a parallelogram in 3-space, and thus it follows that $A = \| \vec{u} \times \vec{v} \| = \| \vec{u} \| \| \vec{v} \| \sin \theta$.

# The Area of a Triangle in 3-Space

We note that the area of a triangle defined by two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. Thus we can give the area of a triangle with the following formula:

(5)
\begin{align} \: A = \frac{1}{2} \| \vec{u} \times \vec{v} \| = \frac{1}{2} \|\vec{u}\| \|\vec{v}\| \sin \theta \end{align}
 Corollary 1: If $\vec{u}, \vec{v} \in \mathbb{R}^3$, then the area of the triangle formed by $\vec{u}$ and $\vec{v}$ is $\mathrm{Area} = \frac{1}{2} \| \vec{u} \| \| \vec{v} \| \sin \theta$.