# The Volume of a Parallelepiped in 3-Space

One nice application of vectors in $\mathbb{R}^3$ is in calculating the volumes of certain shapes. One such shape that we can calculate the volume of with vectors are parallelepipeds.

Theorem 1: If $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, then the volume of the parallelepiped formed between these three vectors can be calculated with the following formula: $\mathrm{Volume} = \mathrm{abs} ( \vec{u} \cdot (\vec{v} \times \vec{w}) ) = \mathrm{abs} \begin{vmatrix}u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3 \end{vmatrix}$. |

**Proof:**Recall that the formula of a parallelepiped is defined by the formula $V = (\mathrm{Area \: of \: base})(\mathrm{height})$. The area of the base of the parallelepiped will be the area of the parallelogram defined by the vectors $\vec{u}$ and $\vec{v}$, which we already calculated to be $A = \| \vec{u} \times \vec{v} \|$. We will now need to calculate the height of the parallelepiped.

- First, let’s consult the following image:

- We note that the height of the parallelepiped is simply the norm of projection of the cross product $\vec{u} \times \vec{v}$ onto $\vec{w}$, that is $h = \| \mathrm{proj}_{\vec{u} \times \vec{v}} \vec{w} \|$. S

(1)

- Substituting this back into our formula for the volume of a parallelepiped we get that:

(2)

- We note that this formula gives up the absolute value of the scalar triple product between the vectors $\vec{u}, \vec{v}, \vec{w}$, that is:

(3)

- Of course the interchanging of rows does in this determinant does not affect the determinant when we absolute value the result, and so our proof is complete. $\blacksquare$

Theorem 2: If $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ and the scalar triple product (or parallelepiped volume) is equal to zero, that is $\begin{vmatrix}u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3 \end{vmatrix} = 0$. Then $\vec{u}, \vec{v}, \vec{w}$ lie on the same plane. |

**Proof:**If $\mathrm{abs} ( \vec{u} \cdot (\vec{v} \times \vec{w})) = 0$, then the volume formed from the parallelepiped in $\mathbb{R}^3$ is zero. Since every vector in $\{ \vec{u}, \vec{v}, \vec{w} \}$ must already lie on the same plane as another from the set, then since the volume is zero, there must exist a vector in this set that lies on the same plane as the other two. Without loss of generality, suppose that $\vec{u}$ lies on the same plane as $\vec{v}$ and $\vec{w}$. Then $\vec{v}$ must also lie on the same plane as $\vec{u}$ and $\vec{w}$, and the same goes for $\vec{w}$. Therefore all three vectors lie on the same plane. $\blacksquare$

## Example 1

**Given that $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ and $\vec{u} = (1, 0, 1)$, $\vec{v} = (1, 1, 0)$, and $\vec{w} = (w_1, 0, 1)$, find a value of $w_1$ that makes all three vectors lie on the same plane.**

As we just learned, three vectors lie on the same plane if their scalar triple product is zero, and thus we must evaluate the following determinant to equal zero:

(4)

Let’s evaluate this determinant along the third row to get $w_1 \begin{vmatrix}0 & 1\\ 1 & 0\end{vmatrix} + \begin{vmatrix} 1 & 0\\ 1 & 1 \end{vmatrix} = 0$, which when simplified is $-w_1 + 1 = 0$. Therefore if $w_1 = 1$, then all three vectors lie on the same plane.