Question

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

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{5}$

Answer 1

The correct option is B

$\sqrt{3}$

Explanation for the correct option.

Step 1: Find the angle between the two vectors.

Let, ${\hat{n}}_1$ and ${\hat{n}}_2$ be two vectors whose magnitude is $1$, the sum of these vectors is also a unit vector, so $\left|{\hat{n}}_1+{\hat{n}}_2\right|=1$. Now if $\theta$ is the angle between the two vectors then

$$\begin{gathered} {\left|{\hat{n}}_1+{\hat{n}}_2\right|}^2=1^2 \\ \Rightarrow {\left|{\hat{n}}_1\right|}^2+{\left|{\hat{n}}_2\right|}^2+2\left|{\hat{n}}_1\right|\left|{\hat{n}}_2\right|\cos \theta =1 \\ \Rightarrow 1+1+2\times 1\times 1\times \cos \theta =1\left[\left|{\hat{n}}_1\right|=\left|{\hat{n}}_2\right|=1\right] \\ \Rightarrow 2+2\cos \theta =1 \\ \Rightarrow 2\cos \theta =-1 \\ \Rightarrow \cos \theta =-\frac{1}{2} \\ \Rightarrow \theta =120° \end{gathered}$$

Step 2: Find the magnitude of the difference of vectors.

The magnitude of the difference of vectors is given as:

$\begin{array}{rcl}\left|{\hat{n}}_1-{\hat{n}}_2\right|& =& \sqrt{{\left|{\hat{n}}_1\right|}^2+{\left|{\hat{n}}_2\right|}^2-2\left|{\hat{n}}_1\right|\left|{\hat{n}}_2\right|\cos \theta }\\ & =& \sqrt{1+1-2\times 1\times 1\times \cos 120°}\left[\left|{\hat{n}}_1\right|=\left|{\hat{n}}_2\right|=1,\theta =120°\right]\\ & =& \sqrt{2-2\times \left(-\frac{1}{2}\right)}\left[\cos 120°=-\frac{1}{2}\right]\\ & =& \sqrt{2+1}\\ & =& \sqrt{3}\end{array}$

Hence, the correct option is B.