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

Apr 9, 2017

The vector is =〈45,-15,-15〉

#### Explanation:

The cross product of 2 vectors is

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

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

Here, we have veca=〈-3,-6,-3〉 and vecb=〈-2,1,-7〉

Therefore,

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

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

$= \vec{i} \left(- 6 \cdot - 7 + 3 \cdot 1\right) - \vec{j} \left(- 3 \cdot - 7 - 3 \cdot 2\right) + \vec{k} \left(- 3 \cdot 1 - 6 \cdot 2\right)$

=〈45,-15,-15〉=vecc

Verification by doing 2 dot products

〈45,-15,-15〉.〈-3,-6,-3〉=-45*3+6*15+3*15=0

〈45,-15,-15〉.〈-2,1,-7〉=-45*2-15*1+15*7=0

So,

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