# What is the cross product of <-1, 2 ,7 > and <-3 ,1 ,-1 >?

Jul 12, 2016

$\left(- 1 , 2 , 7\right) \times \left(- 3 , 1 , - 1\right) = \left(- 9 , - 22 , 5\right)$

#### Explanation:

Cross products can be evaluated using the aptly named determinant rule. If you are trying to determine $\vec{A} \times \vec{B}$ with components:
$\vec{A} = \left({a}_{1} , {a}_{2} , {a}_{3}\right)$
$\vec{B} = \left({b}_{1} , {b}_{2} , {b}_{3}\right)$

Then you can construct a matrix of the form:
$M = \left(\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ {a}_{1} & {a}_{2} & {a}_{3} \\ {b}_{1} & {b}_{2} & {b}_{3}\end{matrix}\right)$

Where $\hat{i}$, $\hat{j}$, and $\hat{k}$ are your orthonormal basis vectors. (This will work in higher dimensions, but quickly becomes cumbersome). Then you can write the cross product as:
$\vec{A} \times \vec{B} = \det \left(M\right) = | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left({a}_{1} , {a}_{2} , {a}_{3}\right) , \left({b}_{1} , {b}_{2} , {b}_{3}\right) |$
$= \hat{i} | \left({a}_{2} , {a}_{3}\right) , \left({b}_{2} , {b}_{3}\right) | - \hat{j} | \left({a}_{1} , {a}_{3}\right) , \left({b}_{1} , {b}_{3}\right) | + \hat{k} | \left({a}_{1} , {a}_{2}\right) , \left({b}_{1} , {b}_{2}\right) |$
$= \left({a}_{2} {b}_{3} - {a}_{3} {b}_{2}\right) \hat{i} - \left({a}_{1} {b}_{3} - {a}_{3} {b}_{1}\right) \hat{j} + \left({a}_{1} {b}_{2} - {a}_{2} {b}_{1}\right) \hat{k}$

You can see that the result is going to be a vector. Plugging in the initial conditions you gave us, we end up with:
$\left(- 1 , 2 , 7\right) \times \left(- 3 , 1 , - 1\right) = | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 1 , 2 , 7\right) , \left(- 3 , 1 , - 1\right) |$
$= \hat{i} | \left(2 , 7\right) , \left(1 , - 1\right) | - \hat{j} | \left(- 1 , 7\right) , \left(- 3 , - 1\right) | + \hat{k} | \left(- 1 , 2\right) , \left(- 3 , 1\right) |$
$= \left(- 2 - 7\right) \hat{i} - \left(1 + 21\right) \hat{j} + \left(- 1 + 6\right) \hat{k}$
$= - 9 \hat{i} - 22 \hat{j} + 5 \hat{k}$
$= \left(- 9 , - 22 , 5\right)$