# How do you find the cross product of <2,3,0> and 2<-1,2,4>?

Dec 16, 2016

The answer is =〈12,-8,7〉

#### Explanation:

The cross product of 2 vecors, 〈a,b,c〉 and 〈d,e,f〉 is given by the determinant

$| \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(a , b , c\right) , \left(d , e , f\right) |$

Here the 2 vectors are 〈2,3,0〉 and 〈-1,2,4〉

The cross product is

$| \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(2 , 3 , 0\right) , \left(- 1 , 2 , 4\right) |$

$= \hat{i} \left(12 - 0\right) - \hat{j} \left(8 - 0\right) + \hat{k} \left(4 + 3\right)$

=〈12,-8,7〉

Verification by doing the dot products

〈2,3,0〉.〈12,-8,7〉=24-24+0=0

〈-1,2,4〉.〈12,-8,7〉=-12-16+28=0

Since the dot products are $= 0$, the vector obtained is perpendicular to the original vectors