Question

The sum of two unit vectors is also a vector of unit magnitude, then the magnitude of the difference of the two unit vectors is
(A) 1 unit
(B) 2 units
(C) $\sqrt 3$ units
(D) zero

Answer 1

Hint
From the formula for the sum of the two unit vectors, we get the value of the dot product of the two unit vectors. By substituting this value in the formula for the difference of the two vectors we get the magnitude of the difference.
In this solution, we will be using the following formula,
$\Rightarrow {\left| {\vec a + \vec b} \right|^2} = {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} + 2\vec a \cdot \vec b$
and ${\left| {\vec a - \vec b} \right|^2} = {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} - 2\vec a \cdot \vec b$
where $\vec a$ and $\vec b$ are two vectors.

Complete step by step answer
Let us consider 2 unit vectors $\vec a$ and $\vec b$. Since they are unit vectors, the magnitude of these will be $\left| {\vec a} \right| = \left| {\vec b} \right| = 1$
We are given the magnitude of the sum of these two unit vectors is also 1. Therefore we can write, $\left| {\vec a + \vec b} \right| = 1$. Now from the formula of the square of the 2 vectors we can write,
$\Rightarrow {\left| {\vec a + \vec b} \right|^2} = {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} + 2\vec a \cdot \vec b$
Now substituting the values in this equation we get,
$\Rightarrow {\left( 1 \right)^2} = {\left( 1 \right)^2} + {\left( 1 \right)^2} + 2\vec a \cdot \vec b$
This gives us,
$\Rightarrow 1 = 2 + 2\vec a \cdot \vec b$
So from here we can find the dot product of the 2 vectors $\vec a$ and $\vec b$ as,
$\Rightarrow - 1 = 2\vec a \cdot \vec b$
Therefore we get,
$\Rightarrow \vec a \cdot \vec b = - \dfrac{1}{2}$
Now we need to find the difference between $\vec a$ and $\vec b$. So we use the formula for the square of the difference of the 2 vectors given by,
$\Rightarrow {\left| {\vec a - \vec b} \right|^2} = {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} - 2\vec a \cdot \vec b$
Therefore by substituting the values we get,
$\Rightarrow {\left| {\vec a - \vec b} \right|^2} = {\left( 1 \right)^2} + {\left( 1 \right)^2} - 2\left( { - \dfrac{1}{2}} \right)$
Therefore we get from here,
$\Rightarrow {\left| {\vec a - \vec b} \right|^2} = 1 + 1 + 1 = 3$
Hence by taking square root on both the sides we get the magnitude of the difference of 2 unit vectors as,
$\Rightarrow \left| {\vec a - \vec b} \right| = \sqrt 3$ units
Therefore the difference will be $\sqrt 3$ units.
So the correct answer will be option (C).

Note
The unit vectors are vectors quantities that have direction but a magnitude of 1. They are very useful in specifying directions. The unit vector of a certain vector can be given by dividing that vector with its own magnitude. That is,
$\Rightarrow \hat A = \dfrac{{\vec A}}{{\left| {\vec A} \right|}}$