Question

If $a$and $b$are unit vectors$\left|a+b\right|=1$, then $\left|a-b\right|$is equal to?

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

Answer 1

The correct option is D

$\sqrt{3}$

Explanation for correct option:

Step 1: Taking the dot product of a vector$(\mathit{a}\mathbf{+}\mathit{b})$ with itself.

We have given$a=1$ and $b=1$

$\left|a+b\right|=1$

Now, taking the squaring on both sides of $\left|a+b\right|=1$

$$\begin{gathered} {\left|a+b\right|}^2=1^2 \\ \Rightarrow (\mathit{a}\mathbf{+}\mathit{b}).(\mathit{a}\mathbf{+}\mathit{b})=1 \\ \Rightarrow a^2+b^2+2ab\cos \left(\theta \right)=1\because a=1,b=1 \\ \Rightarrow 2+2\cos \left(\theta \right)=1 \\ \Rightarrow c\mathrm{os}\left(\theta \right)=\frac{-1}{2} \end{gathered}$$

Step 2: Now, take the dot-product of $(\mathit{a}-\mathit{b}\mathbf{)}$with itself.

$\begin{array}{rcl}\mathbf{(}\mathit{a}-\mathit{b}).(\mathit{a}-\mathit{b})& =& a^2+b^2-2ab\cos \left(\theta \right)\\ & \Rightarrow & {\left|a-b\right|}^2=1+1-2\left(\frac{-1}{2}\right)\\ & \Rightarrow & {\left|a-b\right|}^2=3\\ & \Rightarrow & \left|a-b\right|=\sqrt{3}\end{array}$

Hence, the option $\mathbf{\left(}\mathit{D}\mathbf{\right)}$ is the correct option.