# Question #afdb4

##### 1 Answer
Aug 18, 2016

We add the negative of the second vector to the positive of the first to get the magnitude of the resultant vector of $1$ unit.

#### Explanation:

This question asks us to take the difference between two unit vectors where we are only given the angle between them (which we will assume is given in degrees). Let's call our two vectors $\vec{a}$ and $\vec{b}$. We are looking for the difference between them which we can write:

$\vec{d} = \vec{a} - \vec{b}$

Subtracting two vectors graphically is not obvious, so what we can do is replace the vector $\vec{b}$ with it's negative where we define

$\vec{c} = - \vec{b}$

so our first equation becomes:

$\vec{d} = \vec{a} + \left(- \vec{b}\right) = \vec{a} + \vec{c}$

plotting these vectors we get the following (where we have arbitrarily chosen to put vector $\vec{a}$ on the $x$-axis - everything else falls into place from there): Note that vector $\vec{c}$ is the same length as $\vec{b}$ but it points in exactly the opposite direction, so that if you add the two together you get zero, which is what the negative of a quantity should do.

Now we just add $\vec{a}$ and $\vec{c}$ in the normal way. Finding the coordinates of their endpoints we get:

$\vec{a} \implies \left(1 , 0\right)$
$\vec{c} \implies \left(- \frac{1}{2} , - \frac{\sqrt{3}}{2}\right)$

now we add the $x$ and $y$ components of each vector to get our resultant:

$\vec{d} \implies \left(1 - \frac{1}{2} , 0 - \frac{\sqrt{3}}{2}\right) = \left(\frac{1}{2} , - \frac{\sqrt{3}}{2}\right)$

The length and angle of $\vec{d}$ can be found as

$| \vec{d} | = \sqrt{{x}_{d}^{2} + {y}_{d}^{2}} = \sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} = 1$

$\angle \vec{d} = {\tan}^{-} 1 \left({y}_{d} / {x}_{d}\right) = {\tan}^{-} 1 \left(\frac{- \frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = - {60}^{\text{o}}$

so in polar form, the resultant is

$\vec{d} \implies 1 \angle - {60}^{\text{o}}$

The question asks us for the value of the resulting vector, which we can assume is the magnitude, being $1$ unit.