# What is the cross product of <-1, 2 ,27 > and <-3 ,1 ,-1 >?

Dec 29, 2016

The answer is =〈-29,-82,5〉

#### 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 〈-1,2,27〉 and 〈-3,1,-1〉

And the cross product is

$| \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 1 , 2 , 27\right) , \left(- 3 , 1 , - 1\right) |$

$= \hat{i} | \left(2 , 27\right) , \left(1 , - 1\right) | - \hat{j} | \left(- 1 , 27\right) , \left(- 3 , - 1\right) | + \hat{k} | \left(- 1 , 2\right) , \left(- 3 , 1\right) |$

$= \hat{i} \left(- 2 - 27\right) - \hat{i} \left(1 + 81\right) + \hat{k} \left(- 1 + 6\right)$

=〈-29,-82,5〉

Verification, by doing the dot product

〈-29,-82,5〉.〈-1,2,27〉=29-164+135=0

〈-29,-82,5〉.〈-3,1,-1〉=87-82-5=0

Therefore, the vector is perpendicular to the other 2 vectors