Two vectors $\vec a$ and $\vec b$ are collinear if and only if there exist scalar m and n, at least one of them is non-zero such that:
(A) $m\vec a - n\vec b = 0$
(B) $m\vec a + n\vec b = 0$
(C) $\dfrac{{m\vec a}}{{n\vec b}} = 0$
(D) None of these

Hint:
When two vectors $\vec a$ and $\vec b$ are collinear then there exists a constant $\lambda $ which satisfies the equation $\vec a = \lambda \cdot \vec b$ . According to the given options, assume the values of lambda $'\lambda '$ as $\lambda = \dfrac{n}{m}{\text{ and }}\lambda = - \dfrac{n}{m}$ to obtain the equations and match the answers from options.

Complete Step by Step Solution:
Here in this problem, we are given two collinear vectors $\vec a$ and $\vec b$ . Also, there exist scalars $m$ and $n$ where at least one of them is non-zero. With this information, we need to find which of the given four options is correct.
Before starting with solutions we must understand a few terms related to this problem. A vector is an object that has both a magnitude and a direction. Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. The direction of the vector is from its tail to its head.
Vectors that lie along the same line or parallel lines are known to be collinear vectors. They are also known as parallel vectors. Two vectors $\vec a$ and $\vec b$ are collinear if there exists a number $\lambda $ such that:
$ \Rightarrow \vec a = \lambda \cdot \vec b$
Now as per the given information, let us assume that $\lambda = \dfrac{n}{m}$
Therefore, on substituting this assumption in the above equation, we will get:
$ \Rightarrow \vec a = \lambda \cdot \vec b \Rightarrow \vec a = \left( {\dfrac{n}{m}} \right) \cdot \vec b$
On transforming this further by transposing ‘m’ from the denominator to the LHS, we get:
$ \Rightarrow \vec a = \left( {\dfrac{n}{m}} \right) \cdot \vec b \Rightarrow m \cdot \vec a = n \cdot \vec b$
On taking RHS to LHS, we get the equation:
$ \Rightarrow m \cdot \vec a = n \cdot \vec b \Rightarrow m \cdot \vec a - n \cdot \vec b = 0$
Similarly, let’s now assume that $\lambda = - \left( {\dfrac{n}{m}} \right)$
Now again substituting this value in the equation $\vec a = \lambda \cdot \vec b$ we get:
$ \Rightarrow \vec a = \lambda \cdot \vec b \Rightarrow \vec a = - \left( {\dfrac{n}{m}} \right) \cdot \vec b$
This can be further simplified as:
$ \Rightarrow \vec a = - \left( {\dfrac{n}{m}} \right) \cdot \vec b \Rightarrow m \cdot \vec a = - n \cdot \vec b$
On taking RHS to LHS, we get:
$ \Rightarrow m \cdot \vec a = - n \cdot \vec b \Rightarrow m \cdot \vec a + n \cdot \vec b = 0$
Therefore, we get that equations $m \cdot \vec a - n \cdot \vec b = 0$ and $m \cdot \vec a + n \cdot \vec b = 0$

Hence, the option (A) and (B) are the correct answer.

Note:
In this question, the equation of condition of collinearity played a crucial role in the solution of this problem. An alternative approach to solve this problem can be taken by analyzing the options given. Using the equation given in the option $m \cdot \vec a - n \cdot \vec b = 0$ can be transformed such that $m \cdot \vec a - n \cdot \vec b = 0 \Rightarrow \vec a = \dfrac{n}{m} \cdot \vec b$ . This can be then compared with the condition of collinearity.