The cross product defined.
The cross product of two vectors is a third vector, perpendicular to both operands, with a magnitude equal to the product of the magnitudes of the operands and the sine of the angle between the operand directions. The cross product between parallel vectors is zero, and is the full value of the product of the operands between perpendicular vectors.
The direction of the cross product is found by rotating the direction of the first operand towards the direction of the second operand (in the shortest direction) with the fingers of your right hand  your right thumb points in the direction of the cross product.
What good is it?
The cross product is an extremely useful mathematical operation in physics. For example, if angular velocity is represented by a vector, Omega, with a magnitude of radians per unit time and a direction along the axis of rotation (using the right hand rule), then the time derivative of any other vector being subjected to this rotation rate is simply the cross product of Omega and the vector.
Consider an object on a circular trajectory with a radius of curvature, R:
Tangential velocity is the time derivative of the radius vector, R.
Note that a counterclockwise rotation rate, Omega, points towards you, perpendicular to your computer screen. (The direction of your right thumb if you curl your fingers counterclockwise.)
If the radius vector points horizontally to your right, rotating Omega into R with the fingers of your right hand, your thumb points vertically in the plane of the screen  this is the direction of the tangential velocity vector.
Centripetal acceleration is the time derivative of the tangential velocity vector.
Continuing the above example, rotating Omega into the vertical velocity vector, your right thumb points horizontally towards the left, i.e., in the minus R direction.
For a simple circular trajectory, Omega is perpendicular to R, v is perpendicular to both Omega and R, and the above cross products all evaluate to the products of the operands. The following section explains how cross products are calculated for the more complicated general case.
Right hand coordinates
The cross product is used to define the positive directions in a right hand coordinate system. For this discussion, a local coordinate system called ENU will be used. The initials stand for east, north and up. The directions east and north are tangent to the surface of the earth and up is perpendicular to the surface. For simplicity, the earth will be considered to be a perfect sphere so that the radius vector from the center of the earth is in the up direction with no east or north components.
The positive direction of each unit vector is defined by taking the cross product of the other two unit vectors in the following order: Note that reversing the order of the above cross products would reverse the sign of the unit vector on the rhs.
There is a similar coordinate system, NED, (for north, east, down) which is often used in flight dynamics. The NED system is slightly more awkward to visualize  rotating a north vector into an east vector so your thumb points down requires you to twist your right arm. Besides, If you understand ENU, you wont have any trouble figuring out NED.
The cross product is distributable across addition.
This is an important concept which allows us to evaluate a cross product by crossing all the components of the first operand with all the components of the second operand.
While far from a general proof, the following diagram demonstrates the above equation for the special case in which all three of the vectors are in the same plane. Since the cross product is equal to the area of the parallelogram defined by the two operands, it is apparent that the sum of the two parallelograms, B X A, and C X A is equal to the parallelogram defined by (B + C) and A. The two triangles with sides B, C, and B+C are equal. The triangle on the left is included in the sum of the two small parallelograms and outside the large parallelogram. The triangle on the right is inside the large parallelogram and outside the sum of the small parallelograms. (We could obviously cut off the left triangle and lay it on top of the right triangle to fill the entire large parallelogram.) This page is a work in progress.

