 # Equation of a Plane

A plane is any flat and two-dimensional surface that can extend infinitely in terms of distance. It can also be considered as a two-dimensional analogue of a point that has zero dimensions, a line that has one dimension, and a space that has 3 dimensions.

This plane with a three-dimensional space has the following equation

ax + by + cz + d=0, ax+by+cz+d=0,

In this plane, there should be at the minimum one of the numbers a, b,a,b, and cc must have a value which is non-zero. A plane in this coordinate space can be determined by the use of a point along with a vector that is at a 90-degree angle or perpendicular to the plane.

### Equation of Plane in 3d

A plane in 3D coordinate space is established by a point and a vector that is at the angle of 90 degrees to the plane. Let P0​=(x0​,y0​,z0​) be the point given, and n as the orthogonal vector. Also, consider P=(x,y,z) as any point in the plane, and r and r0 be the position vectors of points P and P0​, respectively. Now if we let, n=(a,b,c), then, since we already know that  P0​P​ is at 90 degrees to the n in position, we get

P0​P​⋅n​=(r−r0​​)⋅n

=(x−x0​,y−y0​,z−z0​)⋅(a,b,c)

= a(x-x0) + b(y-y0) + c(z-z0

=0

The above equation of the plane can also be written as

ax+by+cz+d = 0,

where d= -(ax0} + by0 + cz0.d=−(ax0​+by0​+cz0​).

This equation, however, does not hold true if one of a, b, c is zero. Wherein, the vector is parallel to either one of the points of coordinate planes. Let’s Say c = 0, in this case, the vector is parallel to the XY plane and the equation of this plane will be as a(x-x0) + b(y-y0) = 0 which is in a straight line in the plane of XY and z is clear. Similar opinions affect if two of a, b, c is zero.

An Alternative way to contemplate the equation of the plane is as a flattened parallelepiped. A flattened parallelepiped is constructing by the use of three vectors a=⟨x1​,y1​,z1​⟩,b=⟨x2​,y2​,z2​⟩,c=⟨x3​,y3​,z3​⟩, has a definite volume of 0.  This volume can also be calculated using the scalar triple product.

0=a⋅(b×c), which gives the vector located as normal to the plane.

Let's now assume that the endpoints of (b×c) are ( x, y, z )(x,y,z) and (x_0, y_0, z_0 )(x0​,y0​,z0​) and the constituents of a= (a,b,c) Then by considering the dot product, the equation we get is,

0=a(x−x0​)+b(y−y0​)+c(z−z0​).

### Parallel to the Coordinate Planes

The equation of planes which are parallel to each of the xy-, yz-, and xz-planes and passing through a point A=(a,b,c)A=(a,b,c) is considered as follows:

1) The equivalence of the plane which is parallel to the xy plane is z=c.
2) The equivalence of the plane which is parallel to the yz plane is x=a
3) The equivalence of the plane which is parallel to the xz plane is y=b.

Let’s try a problem for better understanding:

What is the equation of the plane which passes through the point B=(4,1,0)B=(4,1,0) and is parallel to the yz-plane?

Since the xx-coordinate of B is 4, the equation of the plane passing through B parallel to the yz-plane is

Let us determine the equation of plane that will pass through given points

(-1,0,1) parallel to the xz plane.

### Normal Vector and a Point

The equation of a plane is easily established if the normal vector of a plane and any one point passing through the plane is given.

Thus, the equation of a plane through a point A=(x_{1}, y_{1}, z_{1} )A=(x1​,y1​,z1​) whose normal vector is N = (a,b,c) is

a(x-x_{1}) + b(y- y_{1} )+ c(z-z_{1}) = 0 .a(x−x1​)+b(y−y1​)+c(z−z1​)=0.

Check out the Following Examples:

If a plane is passing through the point A=(1,3,2) and has a normal vector N = (3,2,5), then what is the equation of the plane?

The equation of the plane which passes through A=(1,3,2) and has normal vector N = (3,2,5) is

3(x-1) + 2(y-3) + 5(z-2) =0

3x - 3 + 2y - 6 + 5z - 10 = 0

3x + 2y + 5z - 19 =0.

Q. If a plane is passing through the point A=(5,6,2) and has a normal vector n = (-1,3,-7),  then what is the equation of the plane?

The equation of the plane which passes through the point A=(5,6,2)A=(5,6,2) and has normal vector n= (-1,3,-7) is

-1(x-5) + 3(y-6) -7(z-2) = 0

-x+5+3y-18-7z+14= 0

-x+3y-7z+1 =0

How to find the equation of a plane in 3d when three points of the plane are given?

When we know three points on a plane, we can find the equation of the plane by solving simultaneous equations.

Let ax+by+cz+d=0 be the equation of a plane on which there are the following three points: A=(1,0,2), B=(2,1,1), and C=(-1,2,1). Then the equation of the plane is established as follows:

We already have the equation of the plane with 4 unknown constants:

ax + by + cz +d = 0………… (1)

We also get the following 3 equations by substituting the coordinates of A, B, and C into (1):

a⋅1+b⋅0+c⋅2+d=0

a⋅2+b⋅1+c⋅1+d=0

a⋅(−1)+b⋅2+c⋅1+d​=0​

Substituting (2)into (1) we have

ax + 3ay + 4az -9a= 0

x + 3y + 4z - 9 =0.

Hence, the equation of the plane passing through the three points

A=(1,0,2), B=(2,1,1), and C=(-1,2,1)  is

x + 3y + 4z - 9 =0 .x+3y+4z−9=0.

Let’s try on some problem questions to understand the concept better.

A plane is passing through, say, 3 points. The three points are as follows:

points A= (0,0,2),

point B= (1,0,1), and

Point C=(3,1,1),

Keeping these points in the equation what equation of plane can be formed?

Consider an equation of plane as ax+by+cz+d=0……..Take the following equation as 1

Now, since the plane is dealing with 3 points, The equation can further be progressed as

a⋅0+b⋅0+c⋅2+d=0

a⋅1+b⋅0+c⋅1+d=0

a⋅3+b⋅1+c⋅1+d​=0

This leaves us with =-2a, c=a, d=-2a. ......Take the following equation as 2

The next step is to substitute equation (2) into equation(1), which give us,

ax + -2ay + az -2a = 0

x -2y + z - 2 =0. ​

Hence, the equation of the plane passing through the three points A= (0,0,2), B=(1,0,1) and C=(3,1,1) is

x -2y + z - 2 =0. ​

1. What is the Equation of a Line?

Definition: To define a straight line on the coordinate plane, the most convenient way is to denote it with the equation y = mx+b. In this equation m is considered to be the slope of the line and b as the intercept. The equation of a line  can most conveniently be written as y=mx+b where m represents the slope and b is taken as the y-intercept.

Equation from 2 points using Slope-Intercept Form

Calculate the slope from 2 points.

Substitute either point into the equation.

When we Substitute b, -1, into the equation from step 2.

2. What are the Examples of Planes in Geometry?

In the subject of geometry, any flat surface ranging for forever in just two dimensions is known to be a plane. This plane has length but lacks any thickness.

A few examples could be the shape of  triangles, squares, rectangles, lines, circles, points, pentagons, the road stop signs which are generally in the shape of octagon, boxes which are prism shaped. A few more examples to add on would be would be a computer, the black board, any piece of paper, a TV set, house windows, the four walls of a chamber, or a door.