The Volume of a Parallelepiped in 3-Space

# 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)
\begin{align} h = \| \mathrm{proj}_{\vec{u} \times \vec{v}} \vec{w} \| = \frac{ \mid \vec{w} \cdot (\vec{u} \times \vec{v}) \mid}{\| \vec{u} \times \vec{v} \|} \end{align}
• Substituting this back into our formula for the volume of a parallelepiped we get that:
(2)
\begin{align} V = \| \vec{u} \times \vec{v} \| \frac{ \mid \vec{w} \cdot (\vec{u} \times \vec{v}) \mid}{\| \vec{u} \times \vec{v} \|} \\ V = \mid \vec{w} \cdot (\vec{u} \times \vec{v}) \mid \end{align}
• 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)
\begin{align} V = \mathrm{abs} \begin{vmatrix} w_1 & w_2 & w_3 \\ v_1 & v_2 & v_3\\ u_1 & u_2 & u_3 \end{vmatrix} \end{align}
• 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)
\begin{align} \begin{vmatrix} 1 & 0 & 1\\ 1 & 1 & 0\\ w_1 & 0 & 1 \end{vmatrix} = 0 \end{align}

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.