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

In normal words if we say then this equation of plane is used to represent the shape on the Rheem dimensional area. There can be four different methods by which a general equation of plane can be derived.  If we talk about the equation of a plane, it can be expressed either in Cartesian form or vector form.

When you are available with different inputs then only you can do the work of computing the equation through the plane. The equation in a plane can be expressed in different ways but the different expression for expressing it are as follows

• Plane’s Equation at the perpendicular distance from the origin

• Plane’s Equation perpendicular to the given vector says it to be A vector.

• Plane’s Equation passing through three non-collinear points. 

• Plane’s Equation  passing through the intersection of two planes.

In this article you are going to learn about the plane, the general equation of the plane and in thud the details you are going to learn us about what is the plane, concepts of plane in the 3 dimensional Geometry, non Collinearity and the Equation of plane equation of plane on intercept form, then you are provided with the  you are provided with done solved Examples and frequently asked questions  these solved Examples will help you to get a good knowledge about the topic and about the applications of this topic while the frequently Asked Questions will keep you up to date and will clear your queries . Vedanth has spe illustrations designed this article as a package for the students where they will elaborate concepts knowledge along with it doubts will be cleared and they will have knowledge of how to solve these problems that use their application and finally the most  queries are also previously answered.

What is a Plane?

In mathematics, we describe a plane as a two-dimensional flat surface that extends infinitely far and has length and width but no thickness. It is a two-dimensional analog of a point or a line. 

An example of a plane would be a coordinate plane. An intercept is one of the several ways of writing an equation of a line. This form of the line equation is named as “slope-intercept form”.

The equation of a plane in the intercept form can be made simple by using the concepts of position vectors and the general equation of a plane. 

Concepts of a Plane in 3-Dimensional Geometry

For understanding the equation of a plane in the intercept form, it is necessary to first familiarise ourselves with a few important terms, which will help us to get a good grasp of this topic.

We already know what a vector is but let us quickly pull up our socks and rewind our memory a little to have a clear path. A vector is a quantity in the three-dimensional space that has both magnitude and direction. In 3-D Geometry, we usually use the position vectors to represent the position of a point in space which serves us as a reference to the point in any question. It refers to a point in the question with respect to the origin in 3-D Geometry.

We already know that there can be multiple planes passing through a single point but there can be only one plane that can pass through a given point which will be perpendicular to a given vector. The equation of such a plane will be,

[(overrightarrow{r} – overrightarrow{a}) . overrightarrow{N} = 0 ]

One thing that we must note here is that r and a are the position vectors but N is a normal vector i.e. the vector perpendicular to the plane in the question. This same equation can also be written in a Cartesian form as well. The equation of such a plane whose direction ratios are provided by A, B, and C respectively would be,

A(x − x1) + B(y − y1) + C(z − z1) = 0

Non – Collinearity and the Equation of a Plane

What if we were to write an equation of a plane that passes through three non-collinear points? The name already suggests that non-collinear points are the points that do not lie on the same line. The Vector equation of a plane passing through three non-collinear points can be written as:

[(overline{r} – overline{a}) . begin{bmatrix} (overline{b} – overline{a} times (overline{c} – overline{a} end{bmatrix} = 0]

Here, a, b and c are the position vectors of the three non-collinear points, when referred to from the origin. In Cartesian form, we can write it as:  

|[begin{vmatrix} (x − x_1) & (y − y_1) & (z − z_1) \ (x_2 − x_1) & (y_2 − y_1 ) & (z_2 − z_1) \  (x_3 − x_1) & (y_3 − y_1) & (z_3 − z_1)  end{vmatrix}] = 0|

The above equation is in the Cartesian form of an equation of a plane that passes through three non-collinear points in the three-dimensional space.

The Equation of a Plane in Intercept Form

It is not unknown that the general equation of a plane can be written as

 Ax + By + Cz = 0, where D ≠ 0

In the equation above A, B, C are the coordinates of a normal vector to the plane, whereas (x, y, z) are the coordinates of any point through which the plane can pass. Now, let us consider that this plane makes intercepts in the three-dimensional space at points A, B, and C on the x, y, and z-axis respectively. Therefore, the coordinates of these points can respectively be (a, 0, 0), (0, b, 0) and (0, 0, c). Thus we have:

A (a, 0, 0), B (0, b, 0), and C (0, 0, c) as the points which can be cut by the plane on the x-axis, y-axis, and z-axis respectively.

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Since the plane too passes through each of the three points, we can substitute them into the general equation of the plane and we will have:

• Aa + D = 0

• Bb + D = 0

• Cc + D = 0

In other words we can write them as: A = – D/a , B = -D/b , C = -D/c. Substituting these values of A, B, and C in the general equation of the plane, we can get the equation of a plane in intercept form, which can be written as:

[frac{x}{a} + frac{y}{b} + frac{z}{c} = 1]

The equation above is the required equation of the plane that cuts intercepts on three coordinate axes in the Cartesian system. Therefore, this is how we can simply obtain the intercept form of the equation of a plane that is if we are provided with the general equation of a plane. By making such simple substitutions with the help of the intercepts, the derivation of the equation of the plane can be achieved.

Solved Examples

1. Find the intercepts that are cut off by the plane 2x + y – z = 5.

Ans: 1) Considering the above equation and dividing it by 5, we will get: [ frac{2}{5} x +  frac{1}{5} y − frac{1}{5} z  = 1] ….. (1)

[Rightarrow frac{x}{frac{5}{2}} +  frac{y}{frac{5}{1}} − frac{z}{frac{5}{1}}  = 1 ] ….. (2)

known that the equation of a plane in intercept form is, [frac{a}{x} + frac{b}{y} + frac{c}{z} = 1]

where a, b, c are the intercepts that are cut off by the plane at x, y, z axes respectively.

Therefore  a = (5/2), b = 5, and c = -5.

Thus the intercepts that are cut off by the plane are (5/2), (5), and (-5).

2. Find the equation of a plane with intercepts 2, 3, and 4 on the x, y, and z respectively.

Ans: The equation of a plane with intercepts 2,3, and 4 on x, y, and z can be written as:

[frac{a}{x} + frac{b}{y} + frac{c}{z} = 1]

Given, 

Intercept on x – axis = 2 Therefore, a = 2, 

Intercept on y – axis = 3 Therefore, b = 3, 

Intercept on z-axis = 4, Therefore, c = 4

The equation of a plane would be:

[frac{x}{2} + frac{y}{3} + frac{z}{4} = 1]

[frac{6x}{12} + frac{4y}{12} + frac{3z}{12} = 1]

6x + 4y + 3z = 12