Question

The summation of two unit vectors is a third unit vector, then the modulus of the difference of the unit vectors is?

• A. $\sqrt{3}$
• B. $1-\sqrt{3}$
• C. $1+\sqrt{3}$
• D. $-\sqrt{3}$

Answer 1

The correct option is A

$\sqrt{3}$

Explanation for the correct option:

Let $a,b$ be two unit vectors and the summation of these two unit vectors be equal to $c$ which is also a unit vector. Thus,

$a+b=c$

by squaring both sides,

$\begin{array}{cc}\Rightarrow & {\left|a+b\right|}^2={\left|c\right|}^2\\ \Rightarrow & {\left|a\right|}^2+{\left|b\right|}^2+a·b+a·b={\left|c\right|}^2\end{array}$

Since, $\left|a\right|=\left|b\right|=\left|c\right|=1$

$$\begin{gathered} \Rightarrow 1+1+2\left(a·b\right)=1 \\ \Rightarrow 2\left(a·b\right)=-1 \end{gathered}$$

$\begin{array}{cc}\Rightarrow & {\left|a-b\right|}^2={\left|a\right|}^2+{\left|b\right|}^2-2ab\\ \Rightarrow & {\left|a-b\right|}^2=1+1-(-1)\\ \Rightarrow & {\left|a-b\right|}^2=3\\ \Rightarrow & a-b=\sqrt{3}\end{array}$

Hence, option (A) is correct.