# Find the unit vector that has the same direction as \vec{v}= 2i - 13j.

## Question:

Find the unit vector that has the same direction as {eq}\vec{v}= 2i - 13j. {/eq}

## Angle Between Two Vectors:

A quite interesting and important characteristic derived from the scalar product is related to the cosine of the angle of the two vectors: the dot product of two vectors {eq}\vec u{\text{ and }}\vec v {/eq} is equivalent to the modulus of vector {eq}{\vec u} {/eq} by the modulus of vector {eq}{\vec v} {/eq} by the cosine of the angle formed between them: {eq}\vec u \cdot \vec v = \left\| {\vec u} \right\|\left\| {\vec v} \right\|\cos \theta {/eq}.

{eq}\eqalign{ & {\text{If we have a vector }}\,\vec v = {v_1}\hat i + {v_2}\hat j = \left\langle {{v_1},{v_2}} \right\rangle {\text{, then the unit vector }}\,\hat v{\text{ in the direction of vector }} \cr & \vec v\,{\text{ is given by: }} \cr & \,\,\,\,\,\,\,\hat v = \frac{{\vec v}}{{\left\| {\vec v} \right\|}} = \frac{{\left\langle {{v_1},{v_2}} \right\rangle }}{{\sqrt {{v_1}^2 + {v_2}^2} }} \cr & {\text{In this particular case we have the vector }}\,\vec v = 2\hat i - 13\hat j = \left\langle {2, - 13} \right\rangle {\text{, thus:}} \cr & \,\,\,\,\,\,\,\hat v = \frac{{\vec v}}{{\left\| {\vec v} \right\|}} = \frac{{\left\langle {2, - 13} \right\rangle }}{{\sqrt {{2^2} + {{\left( { - 13} \right)}^2}} }} = \frac{{\left\langle {2, - 13} \right\rangle }}{{\sqrt {173} }} = \left\langle {\frac{2}{{\sqrt {173} }}, - \frac{{13}}{{\sqrt {173} }}} \right\rangle \cr & {\text{Therefore}}{\text{, a unit vector that points in the direction of the vector }}\,\left\langle {2, - 13} \right\rangle \,{\text{ is }}\,\boxed{\hat v = \left\langle {\frac{2}{{\sqrt {173} }}, - \frac{{13}}{{\sqrt {173} }}} \right\rangle } \cr} {/eq} 