A plane is at a distance 3 units from the origin. The direction ratios of normal to the plane are 1,-2,1. Find the vector and cartesian equation of the plane.

If the cartesian equation of the plane 2x-y+2z=3 find the vector equation of the plane in standard form.

Given the plane 5x-2y+4z-9=0 i.Find the coordinates of foot of the perpendicular from the origin to the plane. ii. Find the vector equation and the cartesian equation of this perpendicular .

The foot of the perpendicular drawn from origin to a plane is (4,-2,5). i.How far is the plane from the origin ? ii.Find a unit vector perpendicular to that plane. iii. Obtain the equation of the plane is general form.

If the Cartesian equation of a plane is 3x - 4y + 32 -8, find the vector equation of the plane in the standard form.

Find the unit normal vectors to the plane 2x+y-2z=5 .

Find the vector equation of the plane which is at a distance of 6/sqrt29 from the origin and its normal vector from the origin is 2hati-3hatj+4hatk . Also find its cartesian form.

Find the vector equation of a plane with is at a distance of 7 units from the origin and normal to the vector 3hati+5hatj-6hatj .

The foot of the perpendicular drawn from the origin in a plane is (8, -4, -3) . Find the equation of the plane.

The equation of the plane at a distance P from the origin and perpendicular to the unit normal vector hat(d) is