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

Jul 11, 2017

The vector is =〈3,6,-3〉

Explanation:

The (cross product) is calculated with the determinant

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

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

Here, we have veca=〈-3,1,-1〉 and vecb=〈0,1,2〉

Therefore,

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

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

$= \vec{i} \left(1 \cdot 2 + 1 \cdot 1\right) - \vec{j} \left(- 3 \cdot 2 + 0 \cdot 1\right) + \vec{k} \left(- 3 \cdot 1 - 0 \cdot 1\right)$

=〈3,6,-3〉=vecc

Verification by doing 2 dot products

〈3,6,-3〉.〈-3,1,-1〉=-3*3+6*1+3*1=0

〈3,6,-3〉.〈0,1,2〉=3*0+6*1-3*2=0

So,

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