# TrigonometryHow to find unit vector

#### daigo

##### Member
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?

#### CaptainBlack

##### Well-known member
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?
The unit vector in the same direction as $${\bf{v}}$$ is a scalar multiple of $${\bf{v}}$$, call it $$\lambda {\bf{v}}$$, then as it is a unit vector $\lambda {\bf{v}}.\lambda {\bf{v}}=\lambda^2 {\bf{v}}. {\bf{v}}=\lambda^2 ||v||^2=1$

etc

CB

#### Jameson

Staff member
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?
This is the same thing CB wrote essentially, just added an example to see the steps.

Looking at the vector you provided we can derive the scalar with which we must divide to get a unit vector. If v = <3,4> then |v|=5 as you already posted. Now let's look at the vector $$\displaystyle v_1=\frac{1}{S}<3,4>$$. When you multiply a vector by a scalar the magnitude changes but the direction does not so this is perfect for what we are trying to find.

So $$\displaystyle |v_1|=\sqrt{ \left( \frac{3}{S} \right)^2+ \left( \frac{4}{S} \right)^2}$$. Now set the magnitude equal to one and solve for S.

$$\displaystyle 1^2 = \left( \frac{3}{S} \right)^2+ \left( \frac{4}{S} \right)^2$$

$$\displaystyle 1^2 = \left( \frac{9}{S^2} \right)+ \left( \frac{16}{S^2} \right)=\frac{9+16}{S^2}=\frac{25}{S^2}$$

$$\displaystyle S^2=25$$ thus $$\displaystyle S= 5$$. Even though -5 is another choice if we divide by -5 we change the direction of the vector so it can't be the unit vector, thus 5 is our only option.

If you go back and change the vector <3,4> for a generalized vector <a,b> the same steps will lead to $$\displaystyle S^2=a^2+b^2$$ and when we take the square root of both sides we get that S is the same as the magnitude.

I don't believe I've committed any major errors that aren't allowed with scalars and vectors as the conclusion is correct so I hope this helps show where that formula comes from.

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