# What is the cross product of [1,-2,-1] and [-2,0,3] ?

Jan 6, 2017

The answer is =〈-6,-1,-4〉

#### Explanation:

The cross product of 2 vectors, 〈a,b,c〉 and d,e,f〉

is given by the determinant

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

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

and $| \left(a , b\right) , \left(c , d\right) | = a d - b c$

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

And the cross product is

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

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

$= \hat{i} \left(- 6 + 0\right) - \hat{i} \left(3 - 2\right) + \hat{k} \left(0 - 4\right)$

=〈-6,-1,-4〉

Verification, by doing the dot product

〈-6,-1,-4〉.〈1,-2,-1〉=-6+2+4=0

〈-6,-1,-4〉.〈-2,0,3〉=12+0-12=0

Therefore, the vector is perpendicular to the other 2 vectors