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

Nov 1, 2016

$\left(\vec{u} \cdot 2 \vec{v}\right) \vec{w} = \left\langle4 , - 8\right\rangle$

#### Explanation:

Inner Product Definition
If $\vec{A} = \left\langle\left({a}_{1} , {a}_{2} , {a}_{3}\right)\right\rangle$, and $\vec{B} = \left\langle\left({b}_{1} , {b}_{2} , {b}_{3}\right)\right\rangle$, then the inner product (or dot product), a scaler quantity, is given by:
$\vec{A} \cdot \vec{B} = {a}_{1} {b}_{1} + {a}_{2} {b}_{2} + {a}_{3} {b}_{3}$

Inner Product = 0 $\Leftrightarrow$ vectors are perpendicular

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

Then;
$\left(\vec{u} \cdot 2 \vec{v}\right) \vec{w} = 2 \left(\vec{u} \cdot \vec{v}\right) \vec{w}$
$\therefore \left(\vec{u} \cdot 2 \vec{v}\right) \vec{w} = 2 \left\{\left(2\right) \left(- 3\right) + \left(2\right) \left(4\right)\right\} \vec{w}$
$\therefore \left(\vec{u} \cdot 2 \vec{v}\right) \vec{w} = 2 \left(- 6 + 8\right) \vec{w}$
$\therefore \left(\vec{u} \cdot 2 \vec{v}\right) \vec{w} = 2 \left(2\right) \vec{w}$
$\therefore \left(\vec{u} \cdot 2 \vec{v}\right) \vec{w} = 4 \vec{w}$
$\therefore \left(\vec{u} \cdot 2 \vec{v}\right) \vec{w} = 4 \left\langle1 , - 2\right\rangle$
$\therefore \left(\vec{u} \cdot 2 \vec{v}\right) \vec{w} = \left\langle4 , - 8\right\rangle$