PLZ HELP ME UNDERSTAND: WHY the direction of cross products of 2 unit vectors is perpendicular to both.?

Question

why the direction is kept only "perpendicular" to both.Is that only a convention assigned or it has some practical significance.

a person below quoted
"it is a property of cross product to generate a new vector perpendicular to the two(unit or not) vectors"
exactly that i am trying to know that how this property comes.it must have some logic or reason or explaination or anyway u can justify this line.

________________________________________________________________
hey vidigod.thanks a lot fr this GR8 answer!!!!!!
I hav worked out exact answer fr this question based on information provided by u.and whole credit fr this goes to u.
hats off for u !!!!!!

Answer 1

I'm pretty tired, but here's what I suppose. There is a practical purpose. You could define the cross product to give you some third non-perpendicular vector so long as it's not in the plane of the two unit vectors you've taken the cross-product of.

What is necessary is that a third dimension is utilized (according to the source below, this is why cross-product was invented/discovered). If you want to work in three dimensions, you need to have a set of basis vectors to create the three dimensional space. The easiest set of basis vectors to work with and understand are the unit vectors that are all perpendicular to one another. In three dimensions, they are:

i = [1 0 0]
j = [0 1 0]
k = [0 0 1]

Instead of using k = [0 0 1], you might try

k = [1 2 3]

This would not be perpendicular to either i or j, but I think it would suffice in defining a cross product. However, I think results would be less intuitive and more difficult to apply to 'real world' ideas like torque (see source again). And using this k to calculate vector magnitude or in defining other three dimensional vectors is going to be a pain. Even to define:

[0 0 1] with k=[1 2 3] as a basis vector (and i and j as before) would be:

[0 0 1] = -(1/3)i -(2/3)j + (1/3)k

We know the magnitude of [0 0 1] is one, but imagine trying to calculate that with -(1/3)i..etc.

Like I mentioned though, I'm very tired, and I could be completely off here. I don't want to misinform. Could somebody else confirm any of this? Or am I way off the mark?

Hope this helps.

Answer 2

Let c be the vector (x, y, z). The dot product of perpendicular vectors is equal to zero. Find a dot c and b dot c and set them equal to zero. The magnitude of c needs to be one so √(x^2+y^2+z^2) = 1. Or more simply x^2 + y^2 + z^2 = 1. This will give you a system of three equations with three unknowns. Solve the system to get your answer. P.S. It would be a lot easier to use the cross product.

Answer 3

It is a property of cross product to generate a new vector perpendicular to the two(unit or not) vectors. Significance of perpendicularity...who will know better than a physicist!

Answer 4

it is by definition, the cross product was "invented" or "discovered" to be the vector that has that property. Not only that, it also has the property that u,v uxv is positively oriented (which is not emphized that much).
the amazing thing is that, dimension 3 is the only dimension in which such a cross product exists!!!!

Answer 5

no reason, it's definition of cross product to be ppdicular (Lord what a weird word!) that's all!

Answer 6

because they are thge two sides of the vectors