# General Equation of Plane 3D

Here you will learn what is a plane and what is the general equation of a plane.

Let’s begin –

## What is Plane ?

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 on a plane lies on the plane.

## General Equation of the Plane

The general equation of a plane is ax + by + cz + d = 0.

Theorem : Prove that every first degree equation in x, y and z represents a plane.

Proof : Let ax + by + cz + d = 0 be a first degree in x, y and z.

In order to prove that the equation ax + by + cz + d = 0 represents a plane. it is sufficient to show that every point on the line segment joining any two points on the surface represented by it lies on it.

Let P($$x_1, y_1, z_1$$) and Q($$x_2, y_2, z_2$$) be two points on the surface represented by the equation ax + by + cz = 0. Then,

$$ax_1 + by_1 + cz_1 + d$$ = 0                           ………….(i)

$$ax_2 + by_2 + cz_2 + d$$ = 0                           ……………(ii)

Let R be an arbitrary point on the line segment joinining P and Q. Suppose R divides PQ in the ratio $$\lambda$$ : 1. Then, the coordinates of R are

($$x_1 + \lambda x_2\over \lambda + 1$$, $$y_1 + \lambda y_2\over \lambda + 1$$ $$z_1 + \lambda z_2\over \lambda + 1$$), where 0 $$\le$$ $$\lambda$$ $$\le$$ 1.

We have to prove that R lies on the surface represented by the equation ax + by + cz + d = 0 for all values of $$\lambda$$ satisfying 0 $$\le$$ $$\lambda$$ $$\le$$ 1. For this it is sufficient to show that the coordinates of R satisfy this equation.

Putting  x = $$x_1 + \lambda x_2\over \lambda + 1$$, y = $$y_1 + \lambda y_2\over \lambda + 1$$ and z = $$z_1 + \lambda z_2\over \lambda + 1$$ in ax + by  + cz + d = 0, we get

a$$x_1 + \lambda x_2\over \lambda + 1$$ + b$$y_1 + \lambda y_2\over \lambda + 1$$ + c$$z_1 + \lambda z_2\over \lambda + 1$$ + d

= $$1\over \lambda + 1$$ { ($$ax_1 + by_1 + cz_1 + d$$) + $$\lambda$$ ($$ax_2 + by_2 + cz_2 + d$$) }

= $$1\over \lambda + 1$$ ( 0  + $$\lambda$$ 0) = 0

Thus, the point R ($$x_1 + \lambda x_2\over \lambda + 1$$, $$y_1 + \lambda y_2\over \lambda + 1$$ $$z_1 + \lambda z_2\over \lambda + 1$$) lies on the surface represented by ax + by + cz + d = 0. Since R is arbitrary point on the line segment joining P and Q. Therefore, every point on PQ lies on the surface represented by equation ax + by + cz + d = 0.

Hence, the equation ax + by + cz + d = 0 represents a plane.

Remark : To determine a plane satisfying some given condition we will have to find the values of constants a, b, c and d. It seems that are four unknowns viz a, b, c and d in the equation ax + by + cz + d = 0. But, there are only three unkowns, because the equation ax + by + cz + d = 0 can be written as

($$a\over d$$ x + ($$b\over d$$) y + ($$c\over d$$) z + 1 = 0 or,

Ax + By + Cz + 1 = 0

Thus, to find a plane we must have three conditions to find the values of A, B and C.