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: $\frac{|Ax + By + Cz + D|}{\sqrt{A^2 + B^2 + C^2}} = 5$. The result I got were:
$|2x - 3y + 4z + 5| = 5\sqrt{29}$.
Answer 1: $2x - 3y + 4z + 5 = 5\sqrt{29}$.
Answer 2: $-2x + 3y - 4z - 5 = 5\sqrt{29}$.
Are these correct? If not, how would one go about solving this question?
Draw a perpendicular line (in both directions) of length 5 starting at your original plane. This describes the set of points at distance $5$ from your plane. Can you see it?