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

Question

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

2 Answers

Impact of this question

23305 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-08T08:00:42

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

Explanation:

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

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

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

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

$\sqrt{6^{2} + {(- 2)}^{2} + 3^{2}} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$ $sqrt(6^2+(-2)^2+3^2)=sqrt(36+4+9)=sqrt49=7$

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

Answer 2 · 2017-10-14T20:31:31

$\frac{6}{7} \to i - \frac{2}{7} \to j + \frac{3}{7} \to k$ $6/7veci-2/7vecj+3/7veck$

Explanation:

we can find a perpendicular vector by using the result

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

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

$\nabla \equiv \to i \frac{\partial}{d x} + \to j \frac{\partial}{\partial y} + \to k \frac{\partial}{\partial z}$ $grad-=vecidel/(dx)+vecjdel/(dely)+veckdel/(delz)$

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

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

$:.gradphi(x,y,z)=$

$vecidel/(delx)(6x-2y+3z+8)$

$+vecjdel/(dely)(6x-2y+3z+8)$

$+veckdel/(delz)(6x-2y+3z+8)$

$=6veci-2vecj+3veck$

call this $vecn$

then a unit vector is

$hatvecn=vecn/|vecn|$

$:.hatvecn=(6veci-2vecj+3veck)/(sqrt(6^2+2^2+3^2))$

$=(6veci-2vecj+3veck)/7$

$=6/7veci-2/7vecj+3/7veck$

Science

Math

Humanities

... and beyond

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

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question