# What is the cross product of <3 ,1 ,-6 > and <7 ,3 ,2 >?

Dec 26, 2016

The cross product is =〈20,-48,2〉

#### 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 〈3,1,-6〉 and 〈7,3,2〉

And the cross product is

$| \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(3 , 1 , - 6\right) , \left(7 , 3 , 2\right) |$

$= \hat{i} | \left(1 , - 6\right) , \left(3 , 2\right) | - \hat{j} | \left(3 , - 6\right) , \left(7 , 2\right) | + \hat{k} | \left(3 , 1\right) , \left(7 , 3\right) |$

$= \hat{i} \left(2 + 18\right) - \hat{i} \left(6 + 42\right) + \hat{k} \left(9 - 7\right)$

=〈20,-48,2〉

Verification, by doing the dot product

〈20,-48,2〉.〈3,1,-6〉=60-48-12=0

〈20,-48,2〉.〈7,3,2〉=140-144+4=0

Therefore, the vector is perpendicular to the other 2 vectors