Online, I have not found materials about it, and, still I am not convinced myself about the properties of the cross product of unit vectors. It is pretty clear that : $$\hat i \times \hat i = i \cdot i \cdot \sin \phi = 1 \cdot 1 \cdot \sin (0°) = 0 = \hat j \times \hat j = \hat k \times \hat k$$
where $\phi$ is the angle between the unit vectors.
but I don't fully understand why for example we have:
$$ \begin{array}{lcl}
1) \quad \hat i \times \hat j & = & \hat k \\
2) \quad \hat i \times \hat k & = & - \hat j \end{array}$$
I have tried to prove it using the right-hand screw rule :
This is my figure:
Using that rule:
for 1):
as in fig.2: looking from the top, we can see the xy plane and the z-axis has direction perpendicular to the plane, and its way towards the top, i.e. towards our eyes. Of the z-axis, or of $\hat k$, I can see the tip.
I consider a screw on the intersection of the unit vectors $\hat i, \hat j$, and with a counterclockwise rotation, I go to unscrew it, or better, the screw moves towards our eyes, the same direction and way of the z-axis: so the sign is positive.
for 2):
starting from fig.2, I rotate of 90° keeping firm the x-axis, therefore I obtain the fig.3, where the y-axis has gone behind, if we look from the top we can see the zx-plane and the y-axis has the direction perpendicular to the plane, but, its way it is towards the bottom, i.e. inside the plane (or inside our paper, or inside our monitor). Of y-axis, or $\hat j$, I can see the tail.
So, I consider a screw on the intersection of the unit vectors $\hat i, \hat k$, and with a counterclockwise rotation, I go to unscrew it, or better, the screw moves towards our eyes, the same direction of y-axis, but, the opposite way of the y-axis: so the sign is negative.
$\blacksquare$
What do you think, is it a good way to proceed, to prove that using the right-hand screw rule? Thanks!
Cross product between two vectors
$$\vec A=(x_1,x_2,x_3)\;,\;\;\;\vec B=(y_1,y_2,y_3)$$
is defined as
$$\vec C=\vec A\times \vec B:=\begin{vmatrix}e_1&e_2&e_3\\x_1&x_2&x_3\\y_1&y_2&y_3\end{vmatrix}=(x_2y_3-x_3y_2\,,\,x_3y_1-x_1y_3\,,\,x_1y_2-x_2y_1)$$
and it turns out to be a vector $\vec C$ orthogonal to the plane spanned by $\vec A$ and $\vec B$.
The direction of $\vec C$ is given by the right hand rule.
From the defintiion we have that
$\hat i \times \hat j=\hat k, \quad \hat j \times \hat i=-\hat k$
$\hat j \times \hat k=\hat i, \quad \hat k \times \hat j=-\hat i$
$\hat k \times \hat i=\hat j, \quad \hat i \times \hat k=-\hat j$