# If your given the plane -x+6y+5z=21. How do you find the normal vector for that plane and a point on that plane?

Aug 31, 2016

$\vec{n} = \left(- 1 , 6 , 5\right)$
${p}_{0} = \left(0 , 0 , \frac{21}{5}\right)$

#### Explanation:

A plane can be represented as

$\Pi \to \left\langlep - {p}_{0} , \vec{n}\right\rangle = 0$

where

$p = \left(x , y , z\right) \in \Pi$ is a generic point
${p}_{0} = \left({x}_{0} , {y}_{0} , {z}_{0}\right) \in \Pi$ is a given point
and $\vec{n} = \left({n}_{x} , {n}_{y} , {n}_{z}\right)$ is a vector normal to $\Pi$

Developping

${n}_{x} \left(x - {x}_{0}\right) + {n}_{y} \left(y - {y}_{0}\right) + {n}_{z} \left(z - {z}_{0}\right) =$
$x {n}_{x} + y {n}_{y} + z {n}_{z} = - \left({x}_{0} {n}_{x} + {y}_{0} {n}_{y} + {z}_{0} {n}_{z}\right)$

by comparisson we have

${n}_{x} = - 1$
${n}_{y} = 6$
${n}_{z} = 5$

and

$- \left(- {x}_{0} + 6 {y}_{0} + 5 {z}_{0}\right) = 21$

choosing ${x}_{0} = {y}_{0} = 0$ we obtain

${z}_{0} = \frac{21}{5}$

Finally

$\vec{n} = \left(- 1 , 6 , 5\right)$
${p}_{0} = \left(0 , 0 , \frac{21}{5}\right)$