Solve the following ??

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)

1 Answer
Oct 6, 2017

See below.

Explanation:

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