# How do you find unit vector perpendicular to plane: 6x-2y+3z+8=0?

Oct 8, 2016

Unit vector perpendicular to plane $6 x - 2 y + 3 z + 8 = 0$ is $\left(\frac{6}{7} , - \frac{2}{7} , \frac{3}{7}\right)$

#### Explanation:

If we have a plane $a x + b y + c z = d$, the vector normal to the plane is given by $\left(a , b , c\right)$

Hence as given plane is $6 x - 2 y + 3 z + 8 = 0$ or $6 x - 2 y + 3 z = - 8$

and vector normal to this plane is $\left(6 , - 2 , 3\right)$

Now the magnitude of vector $\left(6 , - 2 , 3\right)$ is

$\sqrt{{6}^{2} + {\left(- 2\right)}^{2} + {3}^{2}} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$

Hence, unit vector perpendicular to plane $6 x - 2 y + 3 z + 8 = 0$ is $\left(\frac{6}{7} , - \frac{2}{7} , \frac{3}{7}\right)$

Oct 14, 2017

$\frac{6}{7} \vec{i} - \frac{2}{7} \vec{j} + \frac{3}{7} \vec{k}$

#### Explanation:

we can find a perpendicular vector by using the result

that $\phi \left(x , y , z\right) = 0 \implies \nabla \phi \left(x , y , z\right)$is a vector perpendicular to the plane

where $\nabla$ is the del operator and is defined as:

$\nabla \equiv \vec{i} \frac{\partial}{\mathrm{dx}} + \vec{j} \frac{\partial}{\partial y} + \vec{k} \frac{\partial}{\partial z}$

remembering that when we partially differentiate wrt a variable we treat the other variables as constant

$\phi \left(x , y , z\right) = 6 x - 2 y + 3 z + 8 = 0$

$\therefore \nabla \phi \left(x , y , z\right) =$

$\vec{i} \frac{\partial}{\partial x} \left(6 x - 2 y + 3 z + 8\right)$

$+ \vec{j} \frac{\partial}{\partial y} \left(6 x - 2 y + 3 z + 8\right)$

$+ \vec{k} \frac{\partial}{\partial z} \left(6 x - 2 y + 3 z + 8\right)$

$= 6 \vec{i} - 2 \vec{j} + 3 \vec{k}$

call this $\vec{n}$

then a unit vector is

$\hat{\vec{n}} = \frac{\vec{n}}{|} \vec{n} |$

$\therefore \hat{\vec{n}} = \frac{6 \vec{i} - 2 \vec{j} + 3 \vec{k}}{\sqrt{{6}^{2} + {2}^{2} + {3}^{2}}}$

$= \frac{6 \vec{i} - 2 \vec{j} + 3 \vec{k}}{7}$

$= \frac{6}{7} \vec{i} - \frac{2}{7} \vec{j} + \frac{3}{7} \vec{k}$