How do you find the vector #C# that is perpendicular to #A-> -3x+9y-z=0# and which vector #C# has a magnitude of #1#?

1 Answer
Aug 18, 2016

Answer:

#{{-3,9,-1}}/sqrt(91) #

Explanation:

One of the usual plane representations is

#Pi-> << p - p_0, vec n >> = 0#

where

#p = {x,y,z}# a generic point pertaining to #Pi#
#p_0 = {x_0,y_0,z_0}# a given point in #Pi#
#vec n = {a.b.c}# a normal vector to #Pi#

So

#Pi -> (x-x_0)a+(y-y_0)b+(z-z_0)c = 0#

or

#a x + b y + c z = a x_0+b y_0 + c z_0#

After that we can recognize

#vec n = {-3,9,-1}#
#p_0 = {0,0,0}#

Now, normalizing #vec n# we have

#hat n = (vec n)/norm vec n = {{-3,9,-1}}/sqrt(3^2+9^2+1) = {{-3,9,-1}}/sqrt(91) #