# What is the cross product of [-1,0,1] and [3, 1, -1] ?

Mar 16, 2016

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

#### Explanation:

We know that $\vec{A} \times \vec{B} = | | \vec{A} | | \cdot | | \vec{B} | | \cdot \sin \left(\theta\right) \hat{n}$, where $\hat{n}$ is a unit vector given by the right hand rule.

So for of the unit vectors $\hat{i}$, $\hat{j}$ and $\hat{k}$ in the direction of $x$, $y$ and $z$ respectively, we can arrive at the following results.

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

Another thing that you should know is that cross product is distributive, which means

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

We are going to need all of these results for this question.

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

$= \left(- \hat{i} + \hat{k}\right) \times \left(3 \hat{i} + \hat{j} - \hat{k}\right)$

$= \textcolor{w h i t e}{\begin{matrix}\textcolor{b l a c k}{- \hat{i} \times 3 \hat{i} - \hat{i} \times \hat{j} - \hat{i} \times \left(- \hat{k}\right)} \\ \textcolor{b l a c k}{+ \hat{k} \times 3 \hat{i} + \hat{k} \times \hat{j} + \hat{k} \times \left(- \hat{k}\right)}\end{matrix}}$

$= \textcolor{w h i t e}{\begin{matrix}\textcolor{b l a c k}{- 3 \left(\vec{0}\right) - \hat{k} - \hat{j}} \\ \textcolor{b l a c k}{+ 3 \hat{j} q \quad - \hat{i} - \vec{0}}\end{matrix}}$

$= - \hat{i} + 2 \hat{j} + - 1 \hat{k}$

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