# What is the cross product of (- 5 i + 4 j - 5 k) and (4 i + 4 j + 2 k)?

Jan 14, 2016

If we call the first vector $\vec{a}$ and the second $\vec{b}$, the cross product, $\vec{a} \times \vec{b}$ is $\left(28 \vec{i} - 10 \vec{j} - 36 \vec{k}\right)$.

#### Explanation:

Sal Khan of Khan academy does a nice job of calculating a cross product in this video: https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/linear-algebra-cross-product-introduction

It's something that's easier to do visually, but I'll try to do it justice here:

$\vec{a} = \left(- 5 \vec{i} + 4 \vec{j} - 5 \vec{k}\right)$
$\vec{b} = \left(4 \vec{i} + 4 \vec{j} + 2 \vec{k}\right)$

We can refer to the coefficient of $i$ in $\vec{a}$ as ${a}_{i}$, the coefficient of $j$ in $\vec{b}$ as ${b}_{j}$ and so on.

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

Sal's video above and the Wikipedia article on the cross product will do a better job of explaining why the next step is as follows than I can here:

$\vec{a} \times \vec{b} = \left({a}_{j} {b}_{k} - {a}_{k} {b}_{j}\right) \vec{i} + \left({a}_{k} {b}_{i} - {a}_{i} {b}_{k}\right) \vec{j} + \left({a}_{i} {b}_{j} - {a}_{j} {b}_{i}\right) \vec{k}$

$= \left(4 \cdot 2 - \left(- 5\right) \cdot 4\right) \vec{i} + \left(\left(- 5\right) \cdot 4 - \left(- 5\right) \cdot 2\right) \vec{j} + \left(\left(- 5\right) \cdot 4 - 4 \cdot 4\right) \vec{k} = 28 \vec{i} - 10 \vec{j} - 36 \vec{k}$