How does one define cross product Cross product has a wide application in many field like in physics.Torque and circular motion are its application. But how does one define cross product and why they defined in that way??
 A: In $3$-dimensional space, we are blessed with the following. Any two nonparallel vectors define a plane through the origin. And that plane has a unique line through the origin that is perpendicular to the plane. Maybe we can define $\vec{u}\times\vec{v}$ in a well-defined way to be some vector on this unique line.
We define the cross product of $\vec{u}$ and $\vec{v}$ to be a third vector $\vec{w}$ such that


*

*$\vec{w}$ lies on that perpendicular line

*the magnitude of $\vec{w}$ is proportional to the product of the magnitudes of $\vec{u}$ and $\vec{v}$


This is not enough to uniquely define $\vec{w}$. To proceed, consider when $\vec{u}$ and $\vec{v}$ are parallel. We have no plane to consider. We have infinitely many directions that are simultaneously perpendicular to $\vec{u}$ and $\vec{v}$. Not finding any good geometric reason to place $\vec{w}$ along any one of these directions, we decide that in this case $\vec{w}$ should be the zero vector. We can make this consistent with our earlier needs using:


*

*the magnitude of $\vec{w}$ is proportional to the sine of the angle between $\vec{u}$ and $\vec{v}$ (in using sine, we don't care if you measure the angle between parallel vectors as $0$, $\pi$, $2\pi$,...)


At this point, we declare the magnitude of $\vec{w}$ to be $\|\vec{u}\|\|\vec{v}\|\sin(\theta)$ where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$. We recognize  that there are still two possibilities for the vector $\vec{w}$ - two vectors that point in exactly opposite directions. We humans have arbitrarily adopted the convention to define $\vec{w}$ using the right-hand rule: if your right hand fingers point parallel to  $\vec{u}$ and curl towards $\vec{v}$'s direction, then your thumb should point parallel with $\vec{w}$.
This is the geometric definition of $\vec{u}\times\vec{v}$.

The above turns out to be equivalent to an algebraic definition of $\vec{u}\times\vec{v}$ given by 
$$\vec{u}\times\vec{v}=(u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1)$$
Proving the above equivalence is easier if you have some linear algebra and trigonometry under your belt. 

I recommend using the geometric definition for better understanding of problems involving the cross-product, and the algebraic formula for doing computations with it.
A: The cheeky answer is that the reason the cross product is defined that way is because it has a wide application in many fields like physics. :)
While one can try and trace the history of the operation, the reason it has stuck around and we keep using it for things is precisely because of its wide application and simplicity.
The history, incidentally, is that it comes from multiplication of quaternions. I believe using quaternions for this purpose was an attempt to generalize the fact that complex number arithmetic is so very useful for two-dimensional geometry.
