# How to find unit vector

#### daigo

Say for example v = <3,4>

I was taught to divide each component by the magnitude in order to get the unit vector, i.e.

3^2 + 4^2 = ||v||^2
5 = ||v||

So the unit vector of that vector is <3/5,4/5> or 1/5<3,4>

But if I forgot that I had to divide the components by the magnitude, I would not know how to get the unit vector. So I guess I am asking why you divide the components by the magnitude in order to get the unit vector?

#### MarkFL

Essentially, we want to find the k such that:

$$\displaystyle k\cdot\|\vec{v}\|=1$$

$$\displaystyle k=\frac{1}{\|\vec{v}\|}$$

#### daigo

So is it safe to say:

$$\displaystyle \frac{1}{5}\sqrt{3^{2}+4^{2}}$$

$$\displaystyle \sqrt{(\frac{1}{25}) 3^{2} + (\frac{1}{25}) 4^{2}}$$

$$\displaystyle \sqrt{(\frac{3^{2}}{25}) + (\frac{4^{2}}{25})}$$

$$\displaystyle \sqrt{(\frac{3^{2}}{5^{2}}) + (\frac{4^{2}}{5^{2}})}$$

$$\displaystyle \sqrt{(\frac{3}{5})^{2} + (\frac{4}{5})^{2}}$$

Which means 3/5 and 4/5 are the components of the unit vector?

#### MarkFL

Yes. We want:

$$\displaystyle \sqrt{\(k\cdot v_x$$^2+$$k\cdot v_y$$^2}=1\)

$$\displaystyle \sqrt{k^2\(v_x^2+v_y^2$$}=1\)

$$\displaystyle k\|\vec{v}\|=1$$

$$\displaystyle k=\frac{1}{\|\vec{v}\|}$$

Hence, the unit vector is:

$$\displaystyle k\cdot\vec{v}=\frac{\vec{v}}{\|\vec{v}\|}=\left\langle \frac{v_x}{\|\vec{v}\|},\frac{v_y}{\|\vec{v}\|} \right\rangle$$