# Given the vectors u=<2,2>, v=<-3,4>, and w=<1,-2>, how do you find (v*u)-(w*v)?

Nov 1, 2016

$\left(\vec{v} \cdot \vec{u}\right) - \left(\vec{w} \cdot \vec{v}\right) = 13$

#### Explanation:

Inner Product Definition
If $\vec{u} = \left\langle\left({u}_{1} , {u}_{2}\right)\right\rangle$, and $\vec{v} = \left\langle\left({v}_{1} , {v}_{2}\right)\right\rangle$, then the inner product (or dot product), a scaler quantity, is given by:
$\vec{u} \cdot \vec{v} = {u}_{1} {v}_{1} + {u}_{2} {v}_{2}$
=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+

So with, $\vec{u} = \left\langle2 , 2\right\rangle$, and $\vec{v} = \left\langle- 3 , 4\right\rangle$, and $\vec{w} = \left\langle1 , - 2\right\rangle$ we get:

$\left(\vec{v} \cdot \vec{u}\right) - \left(\vec{w} \cdot \vec{v}\right) = \left\{\left(- 3\right) \left(2\right) + \left(4\right) \left(2\right)\right\} - \left\{\left(1\right) \left(- 3\right) + \left(- 2\right) \left(4\right)\right\}$
$\therefore \left(\vec{v} \cdot \vec{u}\right) - \left(\vec{w} \cdot \vec{v}\right) = \left\{- 6 + 8\right\} - \left\{- 3 - 8\right\}$
$\therefore \left(\vec{v} \cdot \vec{u}\right) - \left(\vec{w} \cdot \vec{v}\right) = 2 - \left\{- 11\right\}$
$\therefore \left(\vec{v} \cdot \vec{u}\right) - \left(\vec{w} \cdot \vec{v}\right) = 13$