Solve the following ??

Question

Find equation of the planes parallel to the plane #x + 2y - 2z + 8 = 0# which are at the distance of #2 units# from the point #(1, 1, 2)#

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Answer 1 · 2017-10-06T09:42:06

See below.

The plane

#Pi->x+2y-2z+8=0# can be equivalently represented as

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

where

#p = (x,y,z)#
#p_0 = (8,0,0)#
#vec n = (1,2,-2)#

The two parallel planes #Pi_1, Pi_2# are

#Pi_1-> << p - p_1, vec n >>#
#Pi_2-> << p - p_2, vec n >>#

such that given #q = (1,1,2)#

#<< q-p_1, vec n >> = d#
#<< q-p_2, vec n >> = -d#

or

#(1-x_1)1+(1-y_1)2+(2-z_1)(-2)=d = 2#
#(1-x_2)1+(1-y_2)2+(2-z_2)(-2)=-d=-2#

and thus

#p_1 = (-1,1,2)# and

#p_2=(3,1,2)#

or

#Pi_1->x+2y-2z+3=0#
#Pi_2->x+2y-2z-1=0#