Kobe Dixon

2023-03-16

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

Kieran Orozco

If we have a plane $ax+by+cz=d$, the vector normal to the plane is given by $\left(a,b,c\right)$
Thus as given plane is $6x-2y+3z+8=0$ or $6x-2y+3z=-8$
and vector normal to this plane is $\left(6,-2,3\right)$
Now the magnitude of vector $\left(6,-2,3\right)$ is
$\sqrt{{6}^{2}+{\left(-2\right)}^{2}+{3}^{2}}=\sqrt{36+4+9}=\sqrt{49}=7$
Therefore, unit vector perpendicular to plane $6x-2y+3z+8=0$ is $\left(\frac{6}{7},-\frac{2}{7},\frac{3}{7}\right)$

Chestonky1a

Using the outcome, we can get a perpendicular vector.
that $\varphi \left(x,y,z\right)=0⇒\nabla \varphi \left(x,y,z\right)$ is a vector perpendicular to the plane
where $\nabla$ is the del operator and is defined as:
$\nabla \equiv \stackrel{\to }{i}\frac{\partial }{dx}+\stackrel{\to }{j}\frac{\partial }{\partial y}+\stackrel{\to }{k}\frac{\partial }{\partial z}$
keeping in mind that we treat the other variables as constants when we partially discriminate with respect to a variable
$\varphi \left(x,y,z\right)=6x-2y+3z+8=0$
$\therefore \nabla \varphi \left(x,y,z\right)=$
$\stackrel{\to }{i}\frac{\partial }{\partial x}\left(6x-2y+3z+8\right)$
$+\stackrel{\to }{j}\frac{\partial }{\partial y}\left(6x-2y+3z+8\right)$
$+\stackrel{\to }{k}\frac{\partial }{\partial z}\left(6x-2y+3z+8\right)$
$=6\stackrel{\to }{i}-2\stackrel{\to }{j}+3\stackrel{\to }{k}$
call this $\stackrel{\to }{n}$
a unit vector is then
$\stackrel{^}{\stackrel{\to }{n}}=\frac{\stackrel{\to }{n}}{|\stackrel{\to }{n}|}$
$\therefore \stackrel{^}{\stackrel{\to }{n}}=\frac{6\stackrel{\to }{i}-2\stackrel{\to }{j}+3\stackrel{\to }{k}}{\sqrt{{6}^{2}+{2}^{2}+{3}^{2}}}$
$=\frac{6\stackrel{\to }{i}-2\stackrel{\to }{j}+3\stackrel{\to }{k}}{7}$
$=\frac{6}{7}\stackrel{\to }{i}-\frac{2}{7}\stackrel{\to }{j}+\frac{3}{7}\stackrel{\to }{k}$

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