Q8 of 230 Page 29

Find the distance of the point (3, 3, 3) from the plane

Given:

* Points: A(3, 3, 3)

* Plane: , which in cartesian form is:

5x + 2y – 7z + 9 = 0

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

is given by:

Putting the values

⟹

⟹ units

∴ the distance of the point (3, 3, 3) from the plane

is units.

Find the equations of the planes parallel to the plane x – 2y + 2z – 3 = 0 and which are at a unit distance from the point (1, 1, 1).

Find the distance of the point (2, 3, 5) from the xy–plane.

If the product of distances of the point (1, 1, 1) from the origin and the plane x – y + z + λ = 0 be 5, find the value of λ.

Find an equation for the set of all points that are equidistant from the planes 3x – 4y + 12z = 6 and 4x + 3z = 7.