# How do you find the component form and magnitude of the vector v given initial point (1,11) and terminal point (9,3)?

Aug 18, 2017

$\boldsymbol{\underline{v}} = \left(\begin{matrix}8 \\ - 8\end{matrix}\right) = \left\langle8 , - 8\right\rangle = 8 \boldsymbol{\underline{\hat{i}}} - 8 \boldsymbol{\underline{\hat{j}}}$

$| | \boldsymbol{\underline{v}} | | = 8 \sqrt{2}$

#### Explanation:

Denote the given coordinates by:

$O = \left(0 , 0\right)$
$A = \left(1 , 11\right)$
$B = \left(9 , 3\right)$

Then the vector $\boldsymbol{\underline{v}}$ is given by:

$\boldsymbol{\underline{v}} = \boldsymbol{\vec{O B}} - \boldsymbol{\vec{O A}}$
$\setminus \setminus \setminus = \left(\begin{matrix}9 \\ 3\end{matrix}\right) - \left(\begin{matrix}1 \\ 11\end{matrix}\right)$
$\setminus \setminus \setminus = \left(\begin{matrix}8 \\ - 8\end{matrix}\right)$

Alternatively, depending upon the desired notation we can also write:

$\boldsymbol{\underline{v}} = \left\langle8 , - 8\right\rangle = 8 \boldsymbol{\underline{\hat{i}}} - 8 \boldsymbol{\underline{\hat{j}}}$

And we can calculate the magnitude, using the metric norm:

$| | \boldsymbol{\underline{v}} | | = \sqrt{{\left(8\right)}^{2} + {\left(- 8\right)}^{2}}$
$\text{ } = \sqrt{64 + 64}$
$\text{ } = \sqrt{128}$
$\text{ } = 8 \sqrt{2}$