Equation of plane and
its different forms: A plane is a
surface such that if any two points are taken on it, the line segment joining
them lies completely on the surface. In other words, every point on the line
segment joining any two points lies on the plane.
Every
first degree equation x, y and z represents a plane i.e., ax +by + cz + d = 0
is the general equation of a plane.
Equations of a Plane passing
through given Point: The general
equation of a plane passing through a point (x₁, y₁, z₁)
is
a
(x  x₁) + b (y  y₁) + c (z  z₁) =0.
Three Point Form of a
Plane: Let ax + by + cz + d = 0 … (i)
Be
the equation of a plane passing through three points P (x₁, y₁,
z₁), Q (x₂, y₂, z₂), R (x₃,
y₃, z₃).
If
four points P (x₁, y₁, z₁), Q (x₂,
y₂, z₂), R (x₃, y₃, z₃) and
S (x₄, y₄, z₄) are coplanar, then the plane
through three points also passes through fourth point.
Thus,
the condition of coplanarity of four points is
Intercept Form of a
Plane: The equation of a plane intercepting
lengths a, b and c with xaxis, yaxis and zaxis respectively is x/a +
y/b + z/c = 1.
To
determine the intercepts made by a plane with the coordinate axes we proceed as
follows:
For x  intercept: put
y = 0, z = 0 in the equation of the plane and obtain the value of x. The value
of x is the intercept on x  axis.
For y  intercept: put
x = 0, z = 0 in the equation of the plane and obtain the value of y. The value
of y is the intercept on y  axis.
For z  intercept: put
x = 0, y = 0 in the equations of the plane and obtain the value of z. The value
of z is the intercept on z  axis.
Normal Form of a Plane: The
equation of a plane at a distance p form the origin and having direction
cosines l, m, n of a normal to it is lx + my + nz = p.
Equation of a Parallel Plane: The
equation of a plane parallel to the plane ax + by + cz + k = 0.
Equation of Planes
Bisecting the Angle between Two Given Planes: The equation of the planes bisecting the
angles between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0.
Are
given by (a₁x + b₁y + c₁z + d₁)/√ (a₁² + b₁² + c₁²) = ± (a₂x + b₂y + c₂z)/ √ (a₁²
+ b₁² + c₁²)
Equation of a Plane
Passing through the Intersection of two Planes: The equation of a plane passing through the
intersection of a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is (a₁x +
b₁y + c₁z + d₁) + λ (a₂x + b₂y + c₂z
+ d₂) = 0, where λ is a constant.
Example: Find
the equation of the plane containing the line of intersection of the plane x + y
+ z  6 = 0 and 2x + 3y + 4y + 5 and passing through the point (1, 1, 1).
Solution: The equation of the plane through the line of
intersection of the given planes is
(x
+ y + z  6) + λ (2x + 3y + 4y + 5)
= 0 … (i)
If
(i) passes through (1, 1, 1), then

3 + 14 λ = 0 => λ = 3/14.
Putting
λ = 3/14 in (i), we obtain the equation o
the required plane as
(x
+ y + z  6) + λ = 3/14 (2x  3y + 4y + 5) = 0
20x
+ 23y + 26z  69 = 0.