Kobe Dixon

2023-03-16

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

Kieran Orozco

Beginner2023-03-17Added 9 answers

If we have a plane $ax+by+cz=d$, the vector normal to the plane is given by $(a,b,c)$

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

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

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

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

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

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

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

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

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

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

Chestonky1a

Beginner2023-03-18Added 5 answers

Using the outcome, we can get a perpendicular vector.

that $\varphi (x,y,z)=0\Rightarrow \nabla \varphi (x,y,z)$ is a vector perpendicular to the plane

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

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

keeping in mind that we treat the other variables as constants when we partially discriminate with respect to a variable

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

$\therefore \nabla \varphi (x,y,z)=$

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

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

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

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

call this $\overrightarrow{n}$

a unit vector is then

$\hat{\overrightarrow{n}}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}$

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

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

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

that $\varphi (x,y,z)=0\Rightarrow \nabla \varphi (x,y,z)$ is a vector perpendicular to the plane

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

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

keeping in mind that we treat the other variables as constants when we partially discriminate with respect to a variable

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

$\therefore \nabla \varphi (x,y,z)=$

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

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

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

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

call this $\overrightarrow{n}$

a unit vector is then

$\hat{\overrightarrow{n}}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}$

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

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

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

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