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Wize University Linear Algebra Textbook > Products of Vectors

Volume of a Parallelepiped (Triple Product)

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Volume of a Parallelepiped

Volume=∣w⃗ ⋅ (u⃗×v⃗)∣\boxed{\quad \text{Volume} =\lvert \vec{w}\ \cdot\ (\vec{u}\times\vec{v})\rvert \quad}
Note: The dot product of a vector with the cross product of two other vectors is called the triple product (or box product).

Steps
Given three vectors u⃗, v⃗, w⃗∈R3\vec{u},\ \vec{v},\ \vec{w} \in \reals^3 that define a parallelepiped:

1. Find the cross product of two of the vectors

Wize Tip
Pick the two "simplest" vectors when taking the cross product (look for 0s and 1s)

2. Find the dot product of the resulting vector and the remaining vector

3. Take the absolute value of the answer
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Example: Volume of a Parallelepiped

Find the volume of the parallelepiped defined by u⃗=⟨−2,3,0⟩\vec{u}= \lang −2,3,0 \rang, v⃗=⟨−1,−1,4⟩ \vec{v} = \lang −1, −1,4 \rang and w⃗=⟨0,1,5⟩\vec{w}=\lang 0,1,5 \rang.
Each axis is measured in centimetres; don't forget units!

We choose to start with u⃗\vec u and w⃗\vec w since they contain 0s.
u⃗×w⃗=[−230]×[015]=[(3)(5) − (0)(1)(0)(0) − (−2)(5)(−2)(1) − (3)(0)]=[1510−2]\begin{aligned} \vec{u} \times \vec{w} &= \begin{bmatrix} -2\\ 3\\ 0\\ \end{bmatrix} \times \begin{bmatrix} 0\\ 1\\ 5\\ \end{bmatrix}\\[2em] &= \begin{bmatrix} (3)(5)\ -\ (0)(1)\\ (0)(0)\ -\ (-2)(5)\\ (-2)(1)\ -\ (3)(0)\\ \end{bmatrix}\\[2em] &= \begin{bmatrix} 15\\ 10\\ -2\\ \end{bmatrix}\\[2em] \end{aligned}


∣v⃗⋅(u⃗×w⃗)∣=∣[−1−14]⋅[1510−2]∣=∣(−1)(15) + (−1)(10) + (4)(−2)∣=33 cm3\begin{aligned} \lvert \vec v \cdot (\vec{u} \times \vec{w}) \rvert &= \left\lvert \begin{bmatrix} -1\\ -1\\ 4\\ \end{bmatrix} \cdot \begin{bmatrix} 15\\ 10\\ -2\\ \end{bmatrix} \right\rvert\\[1.5em] &= \lvert (-1)(15) \ +\ (-1)(10) \ +\ (4)(-2) \rvert\\[0.4em] &= \boxed{33 \text{ cm}^3} \end{aligned}

Practice: Volume of a Parallelepiped

A parallelepiped is defined by the vectors u⃗=⟨0,3,1⟩\vec{u}=\lang 0,3,1\rang , v⃗=⟨1,2,0⟩\vec{v}=\lang1,2,0\rang and w⃗=⟨1,1,5⟩\vec{w}=\lang 1,1,5\rang.
Each axis is measured in feet.

Extra Practice
Practice Question: Cross Product and Volume
Practice Question: Application of Cross Product
Find the volume of the parallelepiped defined by the vectors u⃗=(1,2,0), v⃗=(0,3,1)\vec{u}=\left(1,2,0\right),\ \vec{v}=\left(0,3,1\right) and w⃗=(1,1,5)\vec{w}=\left(1,1,5\right).
133 - FML 3 - 18.1W e.g. 39.2
Consider the points P1=(1,4,1)\bcb{P_1 = (1,4,1)}, P2=(3,4,−1)\bcb{P_2 = (3,4,-1)}, P3=(2,4,−2)\bcb{P_3 = (2,4,-2)}, P4=(0,2,0)\bcb{P_4 = (0,2,0)}.
Find the volume of the parallelepiped defined by the vectors P1P2⃗\bcb{\vec{P_1P_2}}, P1P3⃗\bcb{\vec{P_1P_3}} and P1P4⃗\bcb{\vec{P_1P_4}}.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 39.2_$\tkcth{Mock F}\tkct{?}$_$\tkct{no soln / vid}$
Consider the points P1=(1,4,1)\bcb{P_1 = (1,4,1)}, P2=(3,4,−1)\bcb{P_2 = (3,4,-1)}, P3=(2,4,−2)\bcb{P_3 = (2,4,-2)}, P4=(0,2,0)\bcb{P_4 = (0,2,0)}.
Find the volume of the parallelepiped defined by the vectors P1P2⃗\bcb{\vec{P_1P_2}}, P1P3⃗\bcb{\vec{P_1P_3}} and P1P4⃗\bcb{\vec{P_1P_4}}.
Practice Question: Cross Product and Volume
Practice Question: Application of Cross Product
Find the volume of the parallelepiped defined by the vectors u⃗=(1,2,0), v⃗=(0,3,1)\vec{u}=\left(1,2,0\right),\ \vec{v}=\left(0,3,1\right) and w⃗=(1,1,5)\vec{w}=\left(1,1,5\right).
Find the volume of the parallelepiped determined by 𝑢⃗=(1,3,3),𝑣⃗=(1,0,2),𝑢⃗ = (1,3,3), \vec𝑣 = (1,0,2), and 𝑤⃗=(0,2,3). 𝑤⃗= (0,2,3).
Additional practice problems--vector products
Additional Practice Problems--Vector Products
1. If u⃗=(−2,3,−1),  v⃗=(−1,−2,3)\vec{u}=\left(-2,3,-1\right),\ \ \vec{v}=\left(-1,-2,3\right), find
a) u⃗ ∙ v⃗\vec{u}\ \bullet\ \vec{v}
Find the volume of the parallelepiped defined by the vectors u→=(1,0,1)\overrightarrow{u}=\left(1,0,1\right), v→=(2,−1,0)\overrightarrow{v}=\left(2,-1,0\right)and w→=(3,−3,4)\overrightarrow{w}=\left(3,-3,4\right).