# How do you find the magnitude of <-15,-12> and write it as a sum of the unit vectors?

May 3, 2017

See below

#### Explanation:

The way to find the magnitude is to take the square root of the sum of the squares of the elements in the vectors.

If a vector is $< a , b >$, then the magnitude $= \sqrt{{a}^{2} + {b}^{2}}$

This is true no matter how many dimensions the vector is (means that the vector can go on forever: <a,b,c,d....> and you are going to do the same)

So for this vector, you can apply that formula:

$\sqrt{{\left(- 15\right)}^{2} + {12}^{2}} = \sqrt{225 + 144} = \sqrt{369}$ $= 3 \sqrt{41}$

Unit vector means that you want to have a vector with a magnitude of 1 in the same direction, and you can get that by dividing each element of the vector by the magnitude.

So we have: $< \frac{- 45 \sqrt{41}}{369} , \frac{36 \sqrt{41}}{369} >$

Simplifying made it $\frac{- 5 \sqrt{41}}{41} , \frac{4 \sqrt{41}}{41}$

Hope that helps!