# What is the cross product of <4 , 5 ,-9 > and <5 ,1 ,-3 >?

Aug 2, 2017

The cross product is $< - 6 , - 33 , - 21 >$

#### Explanation:

Given vectors $u$ and $v$, the cross product of these two vectors, $u \times v$ is given by: Where

$u \times v = \left({u}_{2} {v}_{3} - {u}_{3} {v}_{2}\right) \hat{i} - \left({u}_{1} {v}_{3} - {u}_{3} {v}_{1}\right) \hat{j} + \left({u}_{1} {v}_{2} - {u}_{2} {v}_{1}\right) \hat{k}$

This process may look rather complicated but in reality isn't so bad once you get the hang of it.

We have vectors $< 4 , 5 , - 9 >$ and $< 5 , 1 , - 3 >$

This gives a $3 \times 3$ matrix:

$\left\mid \begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & - 9 \\ 5 & 1 & - 3\end{matrix} \right\mid$

To find the cross product, first imagine covering up the $i$ column (or actually do so if possible), and take the product of the $j$ and $k$ columns, similar to as you would using cross multiplication with proportions. Starting with the number at the top left, multiply the first number by its diagonal, then subtract the product of the second number and its diagonal. This is your new $i$ component.

$\left(5 \cdot - 3\right) - \left(- 9 \cdot 1\right) = - 15 - \left(- 9\right) = - 6$

$\implies - 6 \hat{i}$

Now imagine covering up the $j$ column. Similarly to above, take the cross product of the $i$ and $k$ columns. However, this time, whatever your answer is, you will multiply it by $- 1$.

$- \left[\left(4 \cdot - 3\right) - \left(- 9 \cdot 5\right)\right] = - \left[- 12 - \left(- 45\right)\right] = - 33$

$\implies - 33 \hat{j}$

Finally, imagine covering up the $k$ column. Now, take the cross product of the $i$ and $j$ columns.

$\left(4 \cdot 1\right) - \left(5 \cdot 5\right) = 4 - 25 = - 21$

$\implies - 21 \hat{k}$

Therefore, the cross product is $< - 6 , - 33 , - 21 >$ or, equivalently, $\left(- 6 i - 33 j - 21 k\right)$.