How to find unit vector

Jan 2010
205
0
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?
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
Essentially, we want to find the k such that:

\(\displaystyle k\cdot\|\vec{v}\|=1\)

\(\displaystyle k=\frac{1}{\|\vec{v}\|}\)
 
Jan 2010
205
0
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?
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
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:

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