# What is the cross product of [-1, -1, 2] and [2, 5, 4] ?

Mar 17, 2016

$\left[- 14 , 8 , - 3\right]$

#### 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.

$\left[- 1 , - 1 , 2\right] \times \left[2 , 5 , 4\right]$

$= \left(- \hat{i} - \hat{j} + 2 \hat{k}\right) \times \left(2 \hat{i} + 5 \hat{j} + 4 \hat{k}\right)$

$= \textcolor{w h i t e}{\begin{matrix}\textcolor{b l a c k}{- \hat{i} \times 2 \hat{i} - \hat{i} \times 5 \hat{j} - \hat{i} \times 4 \hat{k}} \\ \textcolor{b l a c k}{- \hat{j} \times 2 \hat{i} - \hat{j} \times 5 \hat{j} - \hat{j} \times 4 \hat{k}} \\ \textcolor{b l a c k}{+ 2 \hat{k} \times 2 \hat{i} + 2 \hat{k} \times 5 \hat{j} + 2 \hat{k} \times 4 \hat{k}}\end{matrix}}$

$= \textcolor{w h i t e}{\begin{matrix}\textcolor{b l a c k}{- 2 \left(\vec{0}\right) - 5 \hat{k} + 4 \hat{j}} \\ \textcolor{b l a c k}{+ 2 \hat{k} \quad - 5 \left(\vec{0}\right) - 4 \hat{i}} \\ \textcolor{b l a c k}{\quad + 4 \hat{j} \quad - 10 \hat{i} + 8 \left(\vec{0}\right)}\end{matrix}}$

$= - 14 \hat{i} + 8 \hat{j} + 8 \hat{k}$

$= \left[- 14 , 8 , - 3\right]$