[Maths Class Notes] on Equation of a Plane Pdf for Exam

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  

(Image to be added soon)

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:

( be added soon)

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

​( be added soon)

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. ​