Question

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

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

Answer 1

Answer: (B)

$\sqrt{3}$

Answer 2

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

${\overset{\rightarrow}{n}}_{s} = {\widehat{n}}_{1} + {\widehat{n}}_{2}$ or $n_{s}^{2} = n_{1}^{2} + n_{2}^{2} + 2n_{1}n_{2}\cos\theta = 1 + 1 + 2\cos\theta$

Since it is given that $n_{s}$ is also a unit vector, therefore

$1 = 1 + 1 + 2\cos\theta$⇒$\cos\theta = - \frac{1}{2}$ ∴$\theta = 120^{0}$

Now the difference vector is ${\widehat{n}}_{d} = {\widehat{n}}_{1} - {\widehat{n}}_{2}$ or

$n_{d}^{2} = n_{1}^{2} + n_{2}^{2} - 2n_{1}n_{2}\cos\theta = 1 + 1 - 2\cos(120^{0})$

∴ $n_{d}^{2} = 2 - 2( - 1/2) = 2 + 1 = 3 \Rightarrow n_{d} = \sqrt{3}$