Question

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

• A. √2
• B. √3
• C. 1/√2
• D. √5

Answer 1

The correct option is B √3

Let ${\hat{n}}_1$ and ${\hat{n}}_2$ are the two unit vectors, then the sum is
${\overrightarrow{n}}_s$ = ${\hat{n}}_1$ + ${\hat{n}}_2$ or ${n_s}^2$ = ${n_1}^2$ + ${n_2}^2$ + 2$n_1{n}_{2}cos\theta$
= 1 + 1 + 2cos$\theta$
Since it is given that $n_s$ is also a unit vector, therefore 1 = 1+ 1 + 2 cos $\theta$ = $\Rightarrow$ cos $\theta$ = - $\frac{1}{2}$ $\therefore$ $\theta$ = ${120}^{\circ }$
Now the difference vector is ${\hat{n}}_d$ = ${\hat{n}}_1$ - ${\hat{n}}_2$ or ${n_1}^2$ + ${n_2}^2$ - 2$n_1{n}_{2}cos\theta$ = 1 + 1 - 2 cos(${120}^{\circ }$)
$\therefore$ ${n_d}^2$ = 2 - 2(- $\frac{1}{2}$) = 2 + 1 = 3 $\Rightarrow$ $n_d$ = $\sqrt{3}$