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

Jun 24, 2017

#### Answer:

The vector is =〈-42,-33,-37〉

#### Explanation:

The cross product of $2$ vectors is a vector perpendicular to the $2$ vectors. It is calculated with the determinant

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(d , e , f\right) , \left(g , h , i\right) |$

where 〈d,e,f〉 and 〈g,h,i〉 are the 2 vectors

Here, we have veca=〈4,5,-9〉 and vecb=〈5,-3,-3〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(4 , 5 , - 9\right) , \left(5 , - 3 , - 3\right) |$

$= \vec{i} | \left(5 , - 9\right) , \left(- 3 , - 3\right) | - \vec{j} | \left(4 , - 9\right) , \left(5 , - 3\right) | + \vec{k} | \left(4 , 5\right) , \left(5 , - 3\right) |$

$= \vec{i} \left(5 \cdot - 3 - - 3 \cdot - 9\right) - \vec{j} \left(4 \cdot - 3 - 5 \cdot - 9\right) + \vec{k} \left(4 \cdot - 3 - 5 \cdot 5\right)$

=〈-42,-33,-37〉=vecc

Verification by doing 2 dot products

〈-42,-33,-37〉.〈4,5,-9〉=-42*4-33*5+37*9=0

〈-42,-33,-37〉.〈5,-3,-3〉=-42*5+33*3+37*3=0

So,

$\vec{c}$ is perpendicular to $\vec{a}$ and $\vec{b}$