Question

If $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are mutually perpendicular unit vectors, then $\left|\overrightarrow{a}+\overrightarrow{b}\right|=\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

Given:
$\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are unit vectors
$\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are mutually perpendicular vectors


$$\begin{gathered} {\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2=\left(\overrightarrow{a}+\overrightarrow{b}\right)·\left(\overrightarrow{a}+\overrightarrow{b}\right) \\ =\overrightarrow{a}·\overrightarrow{a}+\overrightarrow{b}·\overrightarrow{b}+2\overrightarrow{a}·\overrightarrow{b} \\ ={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}·\overrightarrow{b} \\ =1^2+1^2+2\left(0\right)\left(\because \left|\overrightarrow{a}\right|=1,\left|\overrightarrow{b}\right|=1,\overrightarrow{a}·\overrightarrow{b}=0\right) \\ =1+1 \\ =2 \\ \Rightarrow \left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{2} \end{gathered}$$

Hence, $\left|\overrightarrow{a}+\overrightarrow{b}\right|=\underline{\sqrt{2}}$.