What is the cross product of `[3,-1,2]` and `[5,1,-3]`?

We know that `vecA xx vecB = ||vecA|| * ||vecB|| * sin(theta) hatn`, where `hatn` is a unit vector given by the right hand rule.

So for of the unit vectors `hati`, `hatj` and `hatk` in the direction of `x`, `y` and `z` respectively, we can arrive at the following results.

`color(white)( (color(black){hati xx hati = vec0}, color(black){qquad hati xx hatj = hatk}, color(black){qquad hati xx hatk = -hatj}), (color(black){hatj xx hati = -hatk}, color(black){qquad hatj xx hatj = vec0}, color(black){qquad hatj xx hatk = hati}), (color(black){hatk xx hati = hatj}, color(black){qquad hatk xx hatj = -hati}, color(black){qquad hatk xx hatk = vec0}))`

Another thing that you should know is that cross product is distributive, which means

`vecA xx (vecB + vecC) = vecA xx vecB + vecA xx vecC`.

We are going to need all of these results for this question.

`[3,-1,2] xx [5,1,-3]`

`= (3hati - hatj + 2hatk) xx (5hati + hatj - 3hatk)`

`= color(white)( (color(black){qquad 3hati xx 5hati + 3hati xx hatj + 3hati xx (-3hatk)}), (color(black){-hatj xx 5hati - hatj xx hatj - hatj xx (-3hatk)}), (color(black){+2hatk xx 5hati + 2hatk xx hatj + 2hatk xx (-3hatk)}) )`

`= color(white)( (color(black){15(vec0) + 3hatk + 9hatj}), (color(black){+5hatk qquad - vec0 quad + 3hati}), (color(black){quad +10hatj quad - 2hati - 6(vec0)}) )`

`= hati + 19hatj + 8hatk`

`= [1,19,8]`