# How do you find the magnitude of YZ given Y(5,0) and Z(7,6)?

Sep 15, 2017

$\boldsymbol{\vec{Y Z}} = \left(\begin{matrix}2 \\ 6\end{matrix}\right) \setminus \setminus$ and $\setminus \setminus \left\mid \boldsymbol{\vec{Y Z}} \right\mid = 2 \sqrt{10}$

#### Explanation:

We have $Y$ and $Z$ with coordinates $\left(5 , 0\right)$ and $\left(7 , 6\right)$ respectively.

So in vector notation we can write:

$\boldsymbol{\vec{O Y}} = \left(\begin{matrix}5 \\ 0\end{matrix}\right) \setminus \setminus$ and $\setminus \setminus \boldsymbol{\vec{O Z}} = \left(\begin{matrix}7 \\ 6\end{matrix}\right)$

We can calculate $\left\mid \boldsymbol{\vec{Y Z}} \right\mid$ in several ways:

Method 1:

Using the coordinates along, we can apply pythagoras theorem:

$Y {Z}^{2} = {\left(7 - 5\right)}^{2} + {\left(6 - 0\right)}^{2}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {2}^{2} + {6}^{2}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 4 + 36$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 40$

And so $Y Z = \sqrt{40} = 2 \sqrt{10}$

Method 2:

Using vector notation we can calculate the vector $\boldsymbol{\vec{Y Z}}$ and then calculate its magnitude.

We have:

$\boldsymbol{\vec{Y Z}} = \boldsymbol{\vec{O Z}} - \boldsymbol{\vec{O Y}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}7 \\ 6\end{matrix}\right) - \left(\begin{matrix}5 \\ 0\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}7 - 5 \\ 6 - 0\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}2 \\ 6\end{matrix}\right)$

And so:

$\left\mid \boldsymbol{\vec{Y Z}} \right\mid = \sqrt{{2}^{2} + {6}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 \sqrt{10}$, as before.