# Intercept Form of the Equation of a Plane

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## Equation of Plane in Intercept Form

In order to find the equation of a plane in intercept form, it is important to get a grip about the concept of vectors. The idea of position vectors and the general equation of a plane are combined to understand the intercept form of a plane. A vector is a quantity having both magnitude and direction, existing in a three-dimensional space. In 3-D Geometry, position vectors are used to denote the position of a point in space. This acts as a reference to the point in question that has an origin in 3-D Geometry.

An infinite number of planes can be perpendicular to a vector. This means, there can be multiple planes passing through a single point. But taking into consideration a specific point only one specific plane exists which is perpendicular to the point, going through the given area. There is only one plane that passes through a given point and is perpendicular to a given vector. The vector equation of such a plane is,

($\vec{r}$ - $\vec{a}$) . $\vec{N}$ = 0

In the above equation, $\vec{r}$ and $\vec{a}$ are position vectors. $\vec{N}$ is a normal vector. Such a vector is perpendicular to the plane in the given question. In order to write the same equation in the Cartesian form, it is important to know the direction ratios of the given plane. The equation of a plane whose direction ratios are by A, B and C respectively can be represented as:

A (x – x1) + B (y – y1) + C (z – z1) = 0

### Equation of a Plane Based on Non-Collinearity

Now let us try to write the same equation of a plane that passes through three non-collinear points. Non-collinear points refer to the points that do not all lie on the same line. The three points can be denoted as (x1, y1), (x2, y2) and (x3, y3). The vector equation of a plane passing through the above three non-collinear points is:

($\vec{r}$ - $\vec{a}$) [($\vec{b}$ - $\vec{a}$) х ($\vec{c}$ - $\vec{a}$)] = 0

The three non-collinear points when referred from the origin, have position vectors of $\vec{a}$, $\vec{b}$ and $\vec{c}$ respectively. The Cartesian form of the above equation is represented as:

(x - x$_{1}$)(y - y$_{1}$)(z - z$_{1}$)(x$_{2}$ - x$_{1}$)(y$_{2}$ - y$_{1}$)(z$_{2}$ - z$_{1}$)(x$_{3}$ - x$_{2}$)(y$_{3}$ - y$_{2}$)(z$_{3}$ - z$_{2}$) = 0

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

### The Equation of a Plane in Intercept Form

According to the formula, the general equation of a plane is:

Ax + By + Cz + D = 0 , where D ≠ 0

The coordinates of the vector normal to the plane are represented by A, B, C. The plane passes through any point that has the coordinates (x, y, z) in a three-dimensional plane. The plane is considered to be having intercepts in three-dimensional space at points A, B, and C respectively on the x, y, and z-axes respectively. The coordinates of the points A, B and C on the x, y and z-axes respectively are (a, 0, 0), (0, b, 0) and (0, 0, c). On representing the above plane in three-dimensional space, we get the following projection:

### By: Dimensions

The points A (a, 0, 0), B (0, b, 0) and C (0, 0, c) are cut by the plane on the x-axis, y-axis and z-axis respectively. This shows that the plane also passes through each of these three points. Now substituting the points individually into the general equation of the plane and we have,

• Aa + D = 0

• Bb + D = 0

• Cc + D = 0

Thus, the vectors normal to the plane can be represented by the general equation as,

A = - $\frac{D}{a}$, C =  - $\frac{D}{c}$, B =  - $\frac{D}{b}$

On substituting these values of A, B, and C in the general equation of the plane, we shall get the equation of a plane in intercept form, which is,

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

The above equation is the requisite equation of the plane that forms intercepts on the three coordinate axes in the Cartesian system. Thus, it is easy to obtain the equation of a plane in its intercept form if the general equation of the same plane is known and the above concept is clear to you. Just making substitutions using the intercepts help in the derivation of the equation of a specific plane.

### Did You Know

A straight line equation also called a linear equation can be represented as y = mx + b. This is a universal formula where b is the y-intercept and m is the slope of the straight line. b is the value of the y-intercept at the point where the y axis crosses the x-axis.

1. A Plane Passes Through Three Non-collinear Points A(-1,1,4), B(2,4,3) and C(-3,4,1). Find the Equation of a Plane in Intercept Form.

Ans: The general equation of the plane is determined by the direction cosines i.e. components of the normal vector: N =Ai+Bj+Ck, and coordinates of any point through which the plane passes.

The normal vector is perpendicular to the plane determined by the given points.

The vector AB =  = (2i + 4j + 3k) – (-i + j + 4k) = 3i + 3j – k

The vector AC =   = (-3i + 4j + k) – (-i + j + 4k) =  -2i + 3j – 3k

Thus,

=  . [(3 × -3) – (3 × -1)] +  . [(3 × -3) – (-2 × -1)] +  . [(3 × 3) – (-2 × 3)]

= -6

The standard equation now stands at:

-6x + 11y + 15z +D =0

The point A (-1,1,4), when substituted in the above equation, we get:

-6 (-1) + 11 (1) + 15 (4) + D = 0

D = -77

Therefore, equation of the plane is

-6x + 11y + 15z -77 = 0

2. What is the Normal Form of the Equation of a Plane?

Ans:  A plane can be identified when the length and the direction of the perpendicular line from the origin to the plane are known. The normal form of the equation of a plane is represented as:

λx + μy + νz - p = 0

where λ, μ, and ν represent the position vectors of a normal to the plane heading away from the origin and p is the perpendicular distance from the origin to the plane.

According to the figure below, n is a unit vector rising from the origin and extending along line OQ. Let r is a position vector of any point (P) on the plane (c).

r • n = p