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

Jan 7, 2017

$| | \vec{w} | | - 1 = \sqrt{5} - 1 = 1.24$ (2dp)

#### Explanation:

We have:

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

The question does seem a little bit confused as there is no further reference to $\vec{u}$ or $\vec{v}$!

$| | \vec{w} | | = \sqrt{\vec{w} \cdot \vec{w}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{\left(1\right)}^{2} + {\left(- 2\right)}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{1 + 4}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{5}$

So then

$| | \vec{w} | | - 1 = \sqrt{5} - 1$