Find an equation for the set of all points that are equidistant from the planes 3x – 4y + 12z = 6 and 4x + 3z = 7.
Let the set of points be denoted by (x1,y1,z1)
Distance of (x1,y1,z1) from 3x – 4y + 12z = 6
We know, the distance of point (x1,y1,z1) from the plane
is given by:
⟹
⟹
Similarly,
Distance of (x1,y1,z1) from 4x + 3z = 7:
According to the question,
⟹
⟹
⟹ or
⟹ 37x1 + 20y1 – 21z1 –61 = 0 or 67x1 –20 y1 + 99z1 –121 = 0
∴ Equations of set of points equidistant from planes 3x – 4y + 12z = 6 and 4x + 3z = 7 is 37x1 + 20y1 – 21z1 –61 = 0 or 67x1 –20 y1 + 99z1 – 121 = 0