3,094 questions
The equation representing the set of points which are equidistant from the points (1, 2 , 3) and (3 , 2 , -1) is
Find the points on z-axis which are at a distance from the point (1, 2, 3).
The Cartesian form of the equation of the plane is:
If is the normal from the origin to the plane, and is the unit vector along . P(x, y, z) be any point on the plane and is perpendicular to . Then
The equation of the plane, which is at a distance of 5 unit from the origin and has 3i - 2j - 6k as a normal vector, is:
If l, m, n are the direction cosines of the normal to the plane and p be the perpendicular distance of the plane from the origin, then the equation of the plane is:
The distance between the planes 3x – 2y + 6z + 21 = 0 and – 6x + 4y – 12z + 35 = 0 is: