# What is the cross product of [-1, -1, 2] and [-1, 2, 2] ?

Feb 1, 2017

$\left[- 1 , - 1 , 2\right] \times \left[- 1 , 2 , 2\right] = \left[- 6 , 0 , - 3\right]$

#### Explanation:

The cross product between two vectors $\vec{A}$ and $\vec{B}$ is defined to be

$\vec{A} \times \vec{B} = | | \vec{A} | | \cdot | | \vec{B} | | \cdot \sin \left(\theta\right) \cdot \hat{n}$,

where $\hat{n}$ is a unit vector given by the right hand rule, and $\theta$ is the angle between $\vec{A}$ and $\vec{B}$ and must satisfy $0 \le \theta \le \pi$.

For of the unit vectors $\hat{i}$, $\hat{j}$ and $\hat{k}$ in the direction of $x$, $y$ and $z$ respectively, using the above definition of cross product gives the following set of results.

$\textcolor{w h i t e}{\begin{matrix}\textcolor{b l a c k}{\hat{i} \times \hat{i} = \vec{0}} & \textcolor{b l a c k}{q \quad \hat{i} \times \hat{j} = \hat{k}} & \textcolor{b l a c k}{q \quad \hat{i} \times \hat{k} = - \hat{j}} \\ \textcolor{b l a c k}{\hat{j} \times \hat{i} = - \hat{k}} & \textcolor{b l a c k}{q \quad \hat{j} \times \hat{j} = \vec{0}} & \textcolor{b l a c k}{q \quad \hat{j} \times \hat{k} = \hat{i}} \\ \textcolor{b l a c k}{\hat{k} \times \hat{i} = \hat{j}} & \textcolor{b l a c k}{q \quad \hat{k} \times \hat{j} = - \hat{i}} & \textcolor{b l a c k}{q \quad \hat{k} \times \hat{k} = \vec{0}}\end{matrix}}$

Also, note that cross product is distributive.

$\vec{A} \times \left(\vec{B} + \vec{C}\right) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}$.

So for this question.

$\left[- 1 , - 1 , 2\right] \times \left[- 1 , 2 , 2\right]$

$= \left(- \hat{i} - \hat{j} + 2 \hat{k}\right) \times \left(- \hat{i} + 2 \hat{j} + 2 \hat{k}\right)$

$= \textcolor{w h i t e}{\begin{matrix}\textcolor{b l a c k}{- \hat{i} \times \left(- \hat{i}\right) - \hat{i} \times 2 \hat{j} - \hat{i} \times 2 \hat{k}} \\ \textcolor{b l a c k}{- \hat{j} \times \left(- \hat{i}\right) - \hat{j} \times 2 \hat{j} - \hat{j} \times 2 \hat{k}} \\ \textcolor{b l a c k}{+ 2 \hat{k} \times \left(- \hat{i}\right) + 2 \hat{k} \times 2 \hat{j} + 2 \hat{k} \times 2 \hat{k}}\end{matrix}}$

$= \textcolor{w h i t e}{\begin{matrix}\textcolor{b l a c k}{\vec{0} - 2 \hat{k} \quad q \quad + 2 \hat{j}} \\ \textcolor{b l a c k}{- \hat{k} - 2 \left(\vec{0}\right) - 2 \hat{i}} \\ \textcolor{b l a c k}{- 2 \hat{j} - 4 \hat{i} \quad - 4 \left(\vec{0}\right)}\end{matrix}}$

$= - 6 \hat{i} - 3 \hat{k}$

$= \left[- 6 , 0 , - 3\right]$