Question

Select all the correct answers.Which vectors are unit vectors?1 一32l -一》《完,表》口 u={1, 1}1-(美)

Answer 1

The unit vector has a magnitude = 1

So, for the given vectors, we will find the magnitude of every vector

$\begin{gathered} u=<\frac{\sqrt[]{3}}{2},-\frac{1}{2}> \\ |u|=\sqrt[]{(\frac{\sqrt[]{3}}{2})^2+(-\frac{1}{2})^2}=\sqrt[]{\frac{3}{4}+\frac{1}{4}}=\sqrt[]{\frac{4}{4}}=\sqrt[]{1}=1 \end{gathered}$

So, it is a unit vector

$\begin{gathered} u=<-\frac{2}{\sqrt[]{5}},\frac{1}{\sqrt[]{5}}> \\ |u|=\sqrt[]{(-\frac{2}{\sqrt[]{5}})^2+(\frac{1}{\sqrt[]{5}})^2}=\sqrt[]{\frac{4}{5}+\frac{1}{5}}=\sqrt[]{\frac{5}{5}}=1 \end{gathered}$

So, it is a unit vector

$\begin{gathered} u=<1,1> \\ |u|=\sqrt[]{1^2+1^2}=\sqrt[]{1+1}=\sqrt[]{2}=1.414 \end{gathered}$

So, it is not a unit vector

$\begin{gathered} u=<-\frac{5}{\sqrt[]{6}},\frac{1}{\sqrt[]{6}}> \\ |u|=\sqrt[]{(-\frac{5}{\sqrt[]{6}})^2+(\frac{1}{\sqrt[]{6}})^2}=\sqrt[]{\frac{25}{6}+\frac{1}{6}}=\sqrt[]{\frac{26}{6}}=2.08 \end{gathered}$

So, it is not a unit vector

So, the correct options are: 1 and 2