Q3 of 230 Page 29

Find the distance of the point (2, 3, –5) from the plane x + 2y – 2z – 9 = 0.

Given:

* Point : A(2, 3, –5)

* Plane : π = x + 2y – 2z – 9 = 0

We know, the distance of point (x₁,y₁,z₁) from the plane

is given by:

Putting the necessary values

⟹ Distance of the plane from A

= 3 units

∴ the distance of the point (2, 3, –5) from the plane

x + 2y – 2z – 9 = 0 is 3 units

Show that the points and are equidistant from the plane

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

Show that the points (1, 1, 1) and (–3, 0, 1) are equidistant from the plane 3x + 4y – 12z + 13 = 0.