Question

Can we divide two vectors? If this is possible then how?

Answer 1

The definition of a Vector space allows us to add two vectors, subtract two vectors, and multiply a vector by a scalar. When we say "division," we really mean the inverse operation of multiplication, so that $ab=c$ just means that $c$ is the only number with the property $bc=a$. None of the vector products mentioned above can have division defined because of the uniqueness issue.

Complete step by step solution:
We cannot divide two vectors. The definition of a Vector space allows us to add two vectors, subtract two vectors, and multiply a vector by a scalar.
In addition, in some vector spaces, we can have different types of multiplication of vectors. For instance, vector spaces over the real numbers can have a Dot product, which multiplies two vectors to get a real number. Also, some other vector spaces, like ${{\mathbb{R}}^{3}}$ can have a Cross product, which multiplies two vectors and produces another vector. Other vector spaces can have other sorts of multiplication like the Exterior product and other wacky things.
However, none of these sorts of multiplication let you divide. When we say "division," we really mean the inverse operation of multiplication, so that $ab=c$ just means that $c$ is the only number with the property $bc=a$. None of the vector products mentioned above can have division defined because of the uniqueness issue.
For example, if we try to use the cross product to define vector division, we run into the problem that
$\left( 1,0,0 \right)\times \left( 0,1,0 \right)=\left( 0,0,1 \right)\left( 1,0,0 \right)\times \left( 0,1,0 \right)=\left( 0,0,1 \right)$ and
$\left( 1,0,0 \right)\times \left( 1,1,0 \right)=\left( 0,0,1 \right)\left( 1,0,0 \right)\times \left( 1,1,0 \right)=\left( 0,0,1 \right)$
Because of this, we can't uniquely define the quantity $\left( 0,0,1 \right)\left( 1,0,0 \right)$.
Hence, we cannot divide two vectors.

Note:
Some students can go wrong by interpreting the division of two vectors as the division of coefficients of their directions which is completely wrong; there's no such division defined for vectors.