Skip to main content
Chemistry LibreTexts

14.3: The Vector Product

  • Page ID
    106892
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    The vector product of two vectors is a vector defined as

    \[\mathbf{u}\times \mathbf{v}=|\mathbf{u}| |\mathbf{v}| \mathbf{n} \sin\theta \nonumber\]

    where \(\theta\) is again the angle between the two vectors, and \(\mathbf{n}\) is the unit vector perpendicular to the plane formed by \(\mathbf{u}\) and \(\mathbf{v}\). The direction of the vector \(\mathbf{n}\) is given by the right-hand rule. Extend your right hand and point your index finger in the direction of \(\mathbf{u}\) (the vector on the left side of the \(\times\) symbol) and your forefinger in the direction of \(\mathbf{v}\). The direction of \(\mathbf{n}\), which determines the direction of \(\mathbf{u}\times \mathbf{v}\), is the direction of your thumb. If you want to revert the multiplication, and perform \(\mathbf{v}\times \mathbf{u}\), you need to point your index finger in the direction of \(\mathbf{v}\) and your forefinger in the direction of \(\mathbf{u}\) (still using the right hand!). The resulting vector will point in the opposite direction (Figure \(\PageIndex{1}\)).

    The magnitude of \(\mathbf{u}\times \mathbf{v}\) is the product of the magnitudes of the individual vectors times \(\sin \theta\). This magnitude has an interesting geometrical interpretation: it is the area of the parallelogram formed by the two vectors (Figure \(\PageIndex{1}\)).

    crossproduct.jpg
    Figure \(\PageIndex{1}\): The vector product (CC BY-NC-SA; Marcia Levitus)

    The cross product can also be expressed as a determinant:

    \[\mathbf{u}\times \mathbf{v}= \begin{vmatrix} \hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\ u_x&u_y&u_z\\ v_x&v_y&v_z\\ \end{vmatrix} \nonumber\]

    Example \(\PageIndex{1}\):

    Given \(\mathbf{u}=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{v}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\), calculate \(\mathbf{w}=\mathbf{u}\times \mathbf{v}\) and verify that the result is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\).

    Solution

    \[ \begin{align*} \mathbf{u}\times \mathbf{v} &= \begin{vmatrix} \hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\ u_x&u_y&u_z\\ v_x&v_y&v_z\\ \end{vmatrix}=\begin{vmatrix} \hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\ -2&1&1\\ 3&-1&1\\ \end{vmatrix} \\[4pt] &=\hat{\mathbf{i}}(1+1)-\hat{\mathbf{j}}(-2-3)+\hat{\mathbf{k}}(2-3) \\[4pt] &=\displaystyle{\color{Maroon}2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-\hat{\mathbf{k}}} \end{align*} \nonumber\]

    To verify that two vectors are perpendicular we perform the dot product:

    \[\mathbf{u} \cdot \mathbf{w}=(-2)(2)+(1)(5)+(1)(-1)=0 \nonumber\]

    \[\mathbf{v} \cdot \mathbf{w}=(3)(2)+(-1)(5)+(1)(-1)=0 \nonumber\]

    An important application of the cross product involves the definition of the angular momentum. If a particle with mass \(m\) moves a velocity \(\mathbf{v}\) (a vector), its (linear) momentum is \(\mathbf{p}=m\mathbf{v}\). Let \(\mathbf{r}\) be the position of the particle (another vector), then the angular momentum of the particle is defined as

    \[\mathbf{l}=\mathbf{r}\times\mathbf{p} \nonumber\]

    The angular momentum is therefore a vector perpendicular to both \(\mathbf{r}\) and \(\mathbf{p}\). Because the position of the particle needs to be defined with respect to a particular origin, this origin needs to be specified when defining the angular momentum.

    angularmomentum.jpg
    Figure \(\PageIndex{2}\): The angular momentum of a particle of position \(\mathbf{r}\) from the origin and momentum \(\mathbf{p}=m\mathbf{v}\) (CC BY-NC-SA; Marcia Levitus)

    This page titled 14.3: The Vector Product is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Marcia Levitus via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.