If the sum of two unit vectors is a unit vector, then magnitude of difference is

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$\sqrt{2}$

$\sqrt{3}$

$1 / \sqrt{2}$

$\sqrt{5}$

answer is B.

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Detailed Solution

Let ${\hat{n}}_{1}$ and ${\hat{n}}_{2}$ are the two unit vectors, then the sum is

$\begin{array}{l} {\vec{n}}_{s} = {\hat{n}}_{1} + {\hat{n}}_{2} \\ Magnitude of {\vec{n}}_{s} is given by \\ n_{s}^{2} = n_{1}^{2} + n_{2}^{2} + 2 n_{1} n_{2} c o s θ \\ = 1 + 1 + 2 c o s θ \end{array}$

Since it is given that n_s is also a unit vector, therefore

$\begin{array}{l} 1 = 1 + 1 + 2 \cos θ \\ \Rightarrow \cos θ = - \frac{1}{2} \\ ∴ θ = 120^{\circ} \end{array}$

Now the difference vector is

$\begin{array}{l} {\hat{n}}_{d} = {\hat{n}}_{1} - {\hat{n}}_{2} \\ Magnitude of {\hat{n}}_{d} is given by \\ n_{d}^{2} = n_{1}^{2} + n_{2}^{2} - 2 n_{1} n_{2} \cos θ = 1 + 1 - 2 \cos (120^{\circ}) \\ \Rightarrow n_{d}^{2} = 2 - 2 (- 1 / 2) = 2 + 1 = 3 \\ \Rightarrow n_{d} = \sqrt{3} \end{array}$