If your given the plane #-x+6y+5z=21#. How do you find the normal vector for that plane and a point on that plane?

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Answer 1 · 2016-08-31T16:44:01

#vec n = (-1,6,5)#
#p_0 =(0,0,21/5)#

A plane can be represented as

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

where

#p = (x,y,z) in Pi# is a generic point
#p_0 = (x_0,y_0,z_0) in Pi# is a given point
and #vec n = (n_x,n_y,n_z)# is a vector normal to #Pi#

Developping

#n_x(x-x_0)+n_y(y-y_0)+n_z(z-z_0) =#
#x n_x + y n_y + z n_z =-(x_0n_x+y_0n_y+z_0n_z)#

by comparisson we have

#n_x = -1#
#n_y = 6#
#n_z =5#

and

#-(-x_0+6y_0+5z_0) = 21#

choosing #x_0=y_0=0# we obtain

#z_0 = 21/5#

Finally

#vec n = (-1,6,5)#
#p_0 =(0,0,21/5)#