# 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:

## 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:

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

Now when we expand across the first row to obtain:

(3)Or simply:

(4)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:

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

(6)Therefore we obtain:

(7)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:

# 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:

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

Note that the volume of a parallelepiped is defined as:

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

(11)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:

The diagram below illustrates this:

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

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

(14)Which when simplified we obtain:

(15)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: