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.

For understanding the equation of a plane in the intercept form, it is necessary to first familiarize ourselves with 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 a 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 vector 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,

\[A(x - x_{1}) + B(y - y_{1}) + C(z - z_{1})\] = 0

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:

\[(\overrightarrow{r} - \overrightarrow{a}).[(\overrightarrow{b} - \overrightarrow{a})\times (\overrightarrow{c} - \overrightarrow{a})]\] = 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.

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.

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

Solution 1) Considering the above equation and dividing it by 5, we will get:

\[(\frac{2}{5}) x + (\frac{y}{5}) - (\frac{z}{5})\] = 1 (1)

=> \[\frac{(x)}{(\frac{5}{2})} + (\frac{y}{5}) + (\frac{z}{-5})\] = 1 (2)

known that the equation of a plane in intercept form is, \[\frac{x}{x} + \frac{b}{y} + \frac{c}{z} = 1\], where a, b, c are the intercepts that are cut off by the plane at x, y, z axes respectively.

Therefore a = \[(\frac{5}{2})\], b = 5, and c = -5.

Thus the intercepts that are cut off by the plane are \[(\frac{5}{2})\], (5), and (-5).

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

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

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

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:

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

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

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

6x + 4y + 3z = 12.

FAQ (Frequently Asked Questions)

Question 1) What are the Different Types of Equations of a Line?

Answer 1) Different types of equation of a line are:

Equations of horizontal and vertical lines: Horizontal line equations are in the form of y = b, where b represents the y-intercept. Vertical line equations are in the form of x = a where a represents the shared x coordinate of all points.

Point slope form of equation of a line: Point slope form equations are represented in the form y-y₁=m(x-x₁).

Two-point form equation of a line: Two-point form equation can be represented in the form (x₁, y₁) and (x₂, y₂).

Slope intercept form equation of a line: Slope intercept form of an equation can be represented as y = mx + b

Intercept form: Intercept form of an equation of a line can be represented as xa + yb = 1

Normal form: Normal form of an equation can be represented as x cos α + y sin α = p