If sum of two unit vectors is also a unit vector, then magnitude of their difference and angle between the two given unit vectors is:
A) \[\sqrt 3 ,60^\circ \]
B) $\sqrt 3 ,120^\circ $
C) \[\sqrt 2 ,60^\circ \]
D) $\sqrt 2 ,120^\circ $

Hint: For any vector to be a unit vector, the modulus of the vector or the scalar component of the vector has to be $1$ . Suppose the vector given in the question are $\overrightarrow a $ and $\overrightarrow b $ , then the question implies that;
$\left| {\overrightarrow a } \right| = 1,\left| {\overrightarrow b } \right| = 1$ and $\left| {\overrightarrow a + \overrightarrow b } \right| = 1$
Also according to Parallelogram law of vector addition;
$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {{{\left| {\overrightarrow a } \right|}^2} + {{\left| {\overrightarrow b } \right|}^2} + 2\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta } $
Where $\overrightarrow a $ and $\overrightarrow b $ are the unit vectors and $\theta $ is the angle between the vectors.

Formulae used:
Parallelogram law of vector addition;
$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {{{\left| {\overrightarrow a } \right|}^2} + {{\left| {\overrightarrow b } \right|}^2} + 2\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta } $
Parallelogram law of vector subtraction;
\[\left| {\overrightarrow a - \overrightarrow b } \right| = \sqrt {{{\left| {\overrightarrow a } \right|}^2} + {{\left| {\overrightarrow b } \right|}^2} - 2\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta } \]
Where $\overrightarrow a $ and $\overrightarrow b $ are the unit vectors and $\theta $ is the angle between the vectors.

Complete step by step solution:
Given that;
$\left| {\overrightarrow a } \right| = 1,\left| {\overrightarrow b } \right| = 1$ and $\left| {\overrightarrow a + \overrightarrow b } \right| = 1$
Also according to vector addition property;
$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {{{\left| {\overrightarrow a } \right|}^2} + {{\left| {\overrightarrow b } \right|}^2} + 2\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta } $
Where $\overrightarrow a$ and $\overrightarrow b$ are the unit vectors and $\theta $ is the angle between the vectors.
For the first part of the question we have to find the value of $\theta $ such that the addition of the two unit vectors also gives rise to a vector whose modulus or scalar component is $1$. To do this we equation the formula of addition of vectors with the value $1$ .
$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {{{\left| {\overrightarrow a } \right|}^2} + {{\left| {\overrightarrow b } \right|}^2} + 2\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta } $ $...\left( 1 \right)$
$\left| {\overrightarrow a + \overrightarrow b } \right| = 1$ $...\left( 2 \right)$
Equating $\left( 1 \right)$ and $\left( 2 \right)$
$ \Rightarrow \sqrt {{{\left| {\overrightarrow a } \right|}^2} + {{\left| {\overrightarrow b } \right|}^2} + 2\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta } = 1$
$ \Rightarrow \sqrt {{{(1)}^2} + {{(1)}^2} + 2(1)(1)\cos \theta } = 1$ (Squaring both sides)
$ \Rightarrow {(\sqrt {2 + 2\cos \theta } )^2} = {1^2}$
$ \Rightarrow 2(1 + \cos \theta ) = 1$
$ \Rightarrow \cos \theta = - \dfrac{1}{2}$

To find the angle between the two vectors $\overrightarrow a $ and $\overrightarrow b $, we find the principal value of $\theta $ for which $\cos \theta = - \dfrac{1}{2}$ .
$ \Rightarrow \theta = {\cos ^{ - 1}}( - \dfrac{1}{2})$
$ \Rightarrow \theta = 120^\circ $
Therefore the vectors $\overrightarrow a $ and $\overrightarrow b $ have an angle of $120^\circ $ between them.
For the second part of the question, we have to find the magnitude of their difference and for that we use the formula for subtraction of vectors;
\[\left| {\overrightarrow a - \overrightarrow b } \right| = \sqrt {{{\left| {\overrightarrow a } \right|}^2} + {{\left| {\overrightarrow b } \right|}^2} - 2\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta } \]
\[ \Rightarrow \left| {\overrightarrow a - \overrightarrow b } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} - 2\left( 1 \right)\left( 1 \right)\cos 120^\circ } \]
$ \Rightarrow \left| {\overrightarrow a - \overrightarrow b } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} - 2\left( 1 \right)\left( 1 \right)\left( { - \dfrac{1}{2}} \right)} $
$ \Rightarrow \left| {\overrightarrow a - \overrightarrow b } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} $
$ \Rightarrow \left| {\overrightarrow a - \overrightarrow b } \right| = \sqrt 3 $
Hence, the magnitude of the difference of the vectors is $\sqrt 3$.

Therefore the option that matches the solution is (B) $\sqrt 3 ,120^\circ.$

Note: During addition of subtraction of vectors, there are two approaches that can be used: Parallelogram law of vector addition/subtraction or triangle law of vector addition/subtraction. The approach we choose depends on our level of comfort and the approach that best matches the data given in the question.