To find the unit vector in the same direction as a vector, we divide it by its magnitude. 

The magnitude of \(\displaystyle \vec{v}\) is \(\displaystyle \left | \vec{v} \right |= \sqrt{3^2+6^2+1^2}=\sqrt{46}\).

We divide vector \(\displaystyle \vec{v}\) by its magnitude to get the unit vector \(\displaystyle \vec{u}_v\):

\(\displaystyle \vec{u}_v= \frac{\vec{v}}{\left | \vec{v} \right |}=\frac{1}{\sqrt{46}}\cdot[3,6,1]\) 

or 

\(\displaystyle \vec{u}_v= \left [ {\frac{3}{\sqrt{46}}},\frac{6}{\sqrt{46}} , \frac{1}{\sqrt{46}} \right ]\)

All unit vectors have a magnitude of \(\displaystyle 1\), so to verify we are correct:

\(\displaystyle \left | \vec{u}_v \right | = \sqrt{ (\frac{3}{\sqrt{46}})^2 + (\frac{6}{\sqrt{46}})^2 + (\frac{9}{\sqrt{46}})^2 } = \sqrt{\frac{9}{46} + \frac{36}{46} + \frac{1}{46} } = 1\)

First, you must find the length of the vector. This is given by the equation:

\(\displaystyle \sqrt{3^2+6^2+44^2+(-5)^2+12^2}=\left \| v\right \|\)

Then, dividing the vector by its length gives the unit vector in the same direction. 

\(\displaystyle \frac{v}{\left \| v\right \|}=\frac{< 3,6,44,-5,12>}{\sqrt{2150}}\\ \\ \approx< 0.065,0.129,0.949,-0.108,0.259>\)

To get the unit vector that is in the same direction as the original vector \(\displaystyle \small v\), we divide the vector by the magnitude of the vector.

For \(\displaystyle \small (1,3)\), the magnitude is:

\(\displaystyle \sqrt{x^2+y^2}\)

 \(\displaystyle \small \sqrt{1^2+3^2}=\sqrt{10}\).

 This means the unit vector in the same direction of \(\displaystyle \small (1,3)\) is, 

\(\displaystyle \small \frac{1}{\sqrt{10}}(1,3)=\left(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right)\). 

In order to find the unit vector u of a given vector v, we follow the formula

\(\displaystyle u=\frac{v}{\left | v\right |}\)

Let

\(\displaystyle v=\left< a,b\right>\)

The magnitude of v follows the formula

\(\displaystyle \left | v\right |=\sqrt{a^{2}+b^{2}}\).

For this vector in the problem

\(\displaystyle \left | v\right |=\sqrt{3^{2}+4^{2}}\)

\(\displaystyle =\sqrt{9+16}\)

\(\displaystyle =\sqrt{25}\)

\(\displaystyle \left | v\right |=5\).

Following the unit vector formula and substituting for the vector and magnitude

\(\displaystyle u=\frac{\left< 3,4\right>}{5}\).

As such,

\(\displaystyle u\,=\,\left< \frac{3}{5},\, \frac{4}{5}\right>\).

In order to find the unit vector u of a given vector v, we follow the formula

\(\displaystyle u=\frac{v}{\left | v\right |}\)

Let

\(\displaystyle v=\left< a,b\right>\)

The magnitude of v follows the formula

\(\displaystyle \left | v\right |=\sqrt{a^{2}+b^{2}}\)

For this vector in the problem

\(\displaystyle \left | v\right |=\sqrt{5^{2}+12^{2}}\)

\(\displaystyle =\sqrt{25+144}\)

\(\displaystyle =\sqrt{169}\)

\(\displaystyle \left | v\right |=13\)

Following the unit vector formula and substituting for the vector and magnitude

\(\displaystyle u=\frac{\left< 5,12\right>}{13}\)

As such,

\(\displaystyle u\,=\,\left< \frac{5}{13},\, \frac{12}{13}\right>\)