Area Of A Parallelogram In 2 Space And Volume Of A Parallelepiped

# Area of a Parallelogram in 2-Space

Remember that the area of a parallelogram defined by vectors u and v that exist in 3-Space required the use of the cross product. However, the cross product applies only to vectors in 3-Space. Thus, we will derive the following equation for the area of a parallelogram defined by two vectors that exist in 2-Space:

(1)
\begin{align} || \vec{u} x \vec{v} || = | \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} | = abs( \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} ) \end{align}

## Proof of the Area of a Parallelogram in 2-Space

Hence, we will write a general vector u from 2-Space to 3-Space to form the following determinant:

(2)
\begin{align} \vec{u} x \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ u_1 & u_2 & 0\\ v_1 & v_ 2& 0 \end{vmatrix} \end{align}

Where u = (u1, u2), and v = (v1, v2).

Now when we expand across the first row to obtain:

(3)
\begin{align} \vec{u} x \vec{v} = \vec{i} \begin{vmatrix} u_2 & 0 \\ v_2 & 0 \end{vmatrix} - \vec{j} \begin{vmatrix} u_1 & 0 \\ v_1 & 0 \end{vmatrix} + \vec{k} \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} \end{align}

Or simply:

(4)
\begin{align} \vec{u} x \vec{v} = \vec{k} \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} \end{align}

But remember that the area of a parallelogram is defined by the norm of the cross product of vector u and v. Thus, we obtain:

(5)
\begin{align} || \vec{u} x \vec{v} || = || \vec{k} \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} || \end{align}

But remember, the determinant will result in a scalar. Hence, we can take it out via this property:

(6)
\begin{align} ||k\vec{u} || = | k | || \vec{u} || \end{align}

Therefore we obtain:

(7)
\begin{align} || \vec{u} x \vec{v} || = | \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} | || \vec{k} || \end{align}

But remember that k is a standard unit vector with a norm of 1. Thus, we obtain the final equation for the area of a parallelogram in 2-Space:

(8)
\begin{align} || \vec{u} x \vec{v} || = | \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} | = abs( \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} ) \end{align}

# Volume of a Parallelepiped in 3-Space

The volume of a parallelepiped is defined to be the absolute value of the scalar triple product of vectors u, v, and w in 3-Space, or the following:

(9)
\begin{align} V = abs \begin{bmatrix} \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3 \end{vmatrix} \end{bmatrix} \end{align}

## Proving the Volume of a Parallelepiped in 3-Space

Note that the volume of a parallelepiped is defined as:

(10)
$$V = (area-of-base)(height)$$

Now we know that the area of the base is as follows:

(11)
\begin{align} Area-of-base = || \vec{u} x \vec{v} || \end{align}

We must now determine the height of the parallelogram. Simply, it will be equal to the norm of the projection of vector w onto the cross product u x v. Thus we obtain:

(12)
\begin{align} h = || proj_{\vec{u} x \vec{v}} \vec{w} || \end{align}

The diagram below illustrates this:

But according to the formula for the norm of a projection, we obtain:

(13)
\begin{align} h = || proj_{\vec{u} x \vec{v}} \vec{w} || = \frac{| \vec{w} \cdot (\vec{u} x \vec{v}) |}{|| \vec{u} x \vec{v} ||} \end{align}

We can thus substitute this into the general volume of a parallelepiped equation to obtain:

(14)
\begin{align} V = || \vec{u} x \vec{v} || \frac{| \vec{w} \cdot (\vec{u} x \vec{v}) |}{|| \vec{u} x \vec{v} ||} \end{align}

Which when simplified we obtain:

(15)
\begin{align} V = | \vec{w} \cdot (\vec{u} x \vec{v}) | \end{align}

But the portion of the equation inside the absolute value is the Scalar Triple Product of vectors w, u, and v, thus, we can write it in determinant form to obtain:

(16)
\begin{align} V = | \begin{vmatrix} w_1 & w_ 2& w_3 \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} | \end{align}