Plane:

Definition:  

Consider the locus of point P(x, y, z). If x, y, z are allowed to vary without any restriction for their different combinations, we have a set of points such as P. The surface on which these points lie, is called as locus of P. It may be a plane or any curved surface. If Q be any other point on it is locus and all the points of straight line PQ lie on it, it is a plane. Or we can say that if the straight line PQ, however the small and in any direction it may be, lies completely on locus, it is a plane, else any curved surface.

Equation of Plane in Different Forms:

• General equation of the plane is ax + by + cz + d = 0
• Equation of plane in Normal form is lx + my + nz = p where p is length of the normal from the origin to the plane and (l, m, n) be the direction cosines of normal.
• The equation to plane passing through P(x₁, y₁, z₁) and with direction ratios
  (a, b, c) for its normal is  a(x - x₁) + b(y - y₁) + c (z - z₁) = 0
• The equation of plane passing through 3 non collinear points (x₁, y₁, z₁),
  (x₂, y₂, z₂)  and (x₃, y₃ , z₃) is    = 0
• The equation of plane whose intercepts are a, b, c on the x, y, z axes respectively is 
• The equation of YZ plane is x = 0,           equation of plane parallel to YZ plane is x = d.
• The equation of ZX plane is y = 0,           equation of plane parallel to ZX plane is y = d.
• The equation of XY plane is z = 0,           equation of plane parallel to XY plane is z = d.
• Four points A (x₁, y₁, z₁), B (x₂, y₂, z₂), C (x₃, y₃, z₃) and D (x₄, y₄, z₄) will be coplanar if one point lies on plane passing through other 3 points.

Example:  Find out the equation to the plane passing through the point (2, -1, 3) which is foot of perpendicular drawn from origin to the plane.           

Solution :     The direction ratios of normal to the plane are 2, -1, 3.

                        The equation of the required plane is 2(x -2) -1 (y + 1) + 3 (z -3) = 0

                        => 2x - y + 3z -14 = 0

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