# How do you compute the dot product to find the magnitude of u=12i-16j?

Jul 6, 2016

$\left\mid \vec{u} \right\mid = 20$

#### Explanation:

you should dot it against itself

by definition $\vec{u} \cdot \vec{v} = \left\mid \vec{u} \right\mid \left\mid \vec{v} \right\mid \cos \psi$

if $\vec{v} = \vec{u}$ then $\psi = 0$ and so $\vec{u} \cdot \vec{u} = \left\mid \vec{u} \right\mid \left\mid \vec{u} \right\mid \cos 0 = {\left(\left\mid \vec{u} \right\mid\right)}^{\setminus} \textcolor{red}{2}$

so $< 12 , - 16 > \cdot < 12 , - 16 > = \left(12\right) \left(12\right) + \left(- 16\right) \left(- 16\right) = 400 = {\left(\left\mid \vec{u} \right\mid\right)}^{\setminus} \textcolor{red}{2}$

so $\left\mid \vec{u} \right\mid = \sqrt{400} = 20$