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

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Answer 1 · 2016-10-08T08:00:42

Unit vector perpendicular to plane #6x-2y+3z+8=0# is #(6/7,-2/7,3/7)#

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

Hence 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)=sqrt49=7#

Hence, unit vector perpendicular to plane #6x-2y+3z+8=0# is #(6/7,-2/7,3/7)#

Answer 2 · 2017-10-14T20:31:31

#6/7veci-2/7vecj+3/7veck#

we can find a perpendicular vector by using the result

that #phi(x,y,z)=0=>gradphi(x,y,z)#is a vector perpendicular to the plane

where #grad# is the del operator and is defined as:

#grad-=vecidel/(dx)+vecjdel/(dely)+veckdel/(delz)#

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

#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#