Distance to a plane question?

Mathemagician

New member
Joined
Jun 3, 2014
Messages
1
Find all points with a distance of 5 units away from the plane defined by the equation 2x - 3y + 4z + 5 = 0.

My solution was to use: |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2) = 5. The result I got were:

|2x - 3y + 4z + 5| = 5sqrt(29).

Answer 1: 2x - 3y + 4z + 5 = 5sqrt(29).
Answer 2: -2x + 3y - 4z - 5 = 5sqrt(29).

Are these correct? If not, how would one go about solving this question?
 
you're correct. two planes parallel to the plane 2x-3y+4z+5=0.
 
Top