The summation of two unit vectors is a third unit vector, then the modulus of the difference of the unit vectors is?
$1.\sqrt 3 $
$ 2.1 - \sqrt 3 $
$ 3.1 + \sqrt 3 $
$ 4. - \sqrt 3 $

Hint: We are supposed to find the modulus of difference of two unit vectors. So, we can start by considering these vectors as two scalars first. It is also given that their summation would also give a unit vector. Using this, we can find a generalised equation and get the value of the difference.
Formulas used for this question are:
$\left( {a + b} \right)^2 = {a^2} + {b^2} + 2ab $
 $\left( {a - b} \right)^2 = {a^2} + {b^2} - 2ab $

Complete step-by-step solution:
Let us consider the given two unit vectors as $a$and $b$.
It is given that their summation also gives a unit vector. We can call it $c$.
According to the question,
$a + b = c$…..(1)
The only way we can find out the difference is using the above given formulas.
Therefore, ${\left| {a + b} \right|^2} = {\left| a \right|^2} + {\left| b \right|^2} + 2\left( {ab} \right)$
Substituting $1$for $\left| a \right|$and $\left| b \right|$since they are unit vectors.
$\eqalign{
  & \Rightarrow 1 = 1 + 1 + 2\left( {ab} \right) \cr
  & \Rightarrow 1 = 2 + 2\left( {ab} \right) \cr
  & \Rightarrow ab = \dfrac{{ - 1}}{2} \cr} $
Now, we can substitute the above value in ${\left( {a - b} \right)^2}$
${\left| {a - b} \right|^2} = {\left| a \right|^2} + {\left| b \right|^2} - 2\left( {ab} \right)$
Substituting for the above equation, we get,
$\eqalign{
  & \Rightarrow {\left| {a - b} \right|^2} = 1 + 1 - \left( {2 \times \left( {\dfrac{{ - 1}}{2}} \right)} \right) \cr
  & \Rightarrow {\left| {a - b} \right|^2} = 2 + 1 \cr
  & \Rightarrow {\left| {a - b} \right|^2} = 3 \cr
  & \Rightarrow \left| {a - b} \right| = \sqrt 3 \cr} $
The final answer is $\sqrt 3 $
Hence, option (1) is the right answer.
Additional Information:
A vector is a quantity that has both magnitude and direction. A unit vector is the one which has a magnitude $1$. It is also known as Direction Vector.

Note: The given question has mentioned sum and differences. It also includes modulus. Find the best relationship that includes both the mentioned criteria. Substitute $1$ for only unit vectors that is $\left| a \right|,\left| b \right|$ and not for $a,b$. When we square on both sides, a normal integer gets the root and an already squared number’s root will cancel.