# How do you find the vector C that is perpendicular to A-> -3x+9y-z=0 and which vector C has a magnitude of 1?

Aug 18, 2016

$\frac{\left\{- 3 , 9 , - 1\right\}}{\sqrt{91}}$

#### Explanation:

One of the usual plane representations is

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

where

$p = \left\{x , y , z\right\}$ a generic point pertaining to $\Pi$
${p}_{0} = \left\{{x}_{0} , {y}_{0} , {z}_{0}\right\}$ a given point in $\Pi$
$\vec{n} = \left\{a . b . c\right\}$ a normal vector to $\Pi$

So

$\Pi \to \left(x - {x}_{0}\right) a + \left(y - {y}_{0}\right) b + \left(z - {z}_{0}\right) c = 0$

or

$a x + b y + c z = a {x}_{0} + b {y}_{0} + c {z}_{0}$

After that we can recognize

$\vec{n} = \left\{- 3 , 9 , - 1\right\}$
${p}_{0} = \left\{0 , 0 , 0\right\}$

Now, normalizing $\vec{n}$ we have

$\hat{n} = \frac{\vec{n}}{\left\lVert \vec{n} \right\rVert} = \frac{\left\{- 3 , 9 , - 1\right\}}{\sqrt{{3}^{2} + {9}^{2} + 1}} = \frac{\left\{- 3 , 9 , - 1\right\}}{\sqrt{91}}$