How to Derive and Apply the Intercept Form in Coordinate Geometry
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:
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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.
FAQs on Intercept Form of the Equation of a Plane
1. What is the intercept form of the equation of a plane as per the CBSE Class 12 syllabus?
The intercept form of the equation of a plane is a specific format that represents the plane in terms of the intercepts it makes on the coordinate axes. The standard equation is x/a + y/b + z/c = 1. Here, 'a' is the x-intercept, 'b' is the y-intercept, and 'c' is the z-intercept. This form is particularly useful for quickly sketching the plane in a 3D coordinate system.
2. How can you find the intercepts of a plane from its general equation, for example, Ax + By + Cz + D = 0?
To find the intercepts of a plane from its general equation, you can follow these steps:
- To find the x-intercept (a): Set y = 0 and z = 0 in the equation. You get Ax + D = 0, so x = -D/A. Thus, the x-intercept is 'a' = -D/A.
- To find the y-intercept (b): Set x = 0 and z = 0. You get By + D = 0, so y = -D/B. Thus, the y-intercept is 'b' = -D/B.
- To find the z-intercept (c): Set x = 0 and y = 0. You get Cz + D = 0, so z = -D/C. Thus, the z-intercept is 'c' = -D/C.
3. How do you convert the general equation of a plane to the intercept form?
To convert the general equation of a plane, Ax + By + Cz + D = 0, into the intercept form, follow this procedure:
1. Move the constant term 'D' to the right-hand side of the equation: Ax + By + Cz = -D.
2. Divide the entire equation by -D (assuming D is not zero) to make the right-hand side equal to 1: (Ax/-D) + (By/-D) + (Cz/-D) = 1.
3. Rearrange the terms to match the standard intercept form: x/(-D/A) + y/(-D/B) + z/(-D/C) = 1. This gives you the intercept form where a = -D/A, b = -D/B, and c = -D/C.
4. Why is the intercept form useful for visualising a plane in 3D geometry?
The intercept form is highly useful for visualisation because it directly gives three specific points on the plane: (a, 0, 0), (0, b, 0), and (0, 0, c). By plotting these three points on the x, y, and z axes respectively, you can easily sketch the triangular section of the plane in the first octant. This provides a clear and immediate mental picture of the plane's position and orientation in 3D space, which is often more intuitive than interpreting the normal vector and a point from other forms.
5. What does it mean if a plane's equation cannot be written in the standard intercept form?
A plane cannot be written in the standard intercept form x/a + y/b + z/c = 1 under two primary conditions:
- If the plane passes through the origin: In this case, the intercepts a, b, and c are all zero. Division by zero is undefined, so the intercept form is not applicable. The general equation for such a plane will have D=0 (i.e., Ax + By + Cz = 0).
- If the plane is parallel to any coordinate axis: For instance, if a plane is parallel to the x-axis, it will never intersect it, making the x-intercept infinite. An infinite value cannot be used in the denominator of the standard intercept form.
6. What is the key difference in the information provided by the intercept form versus the normal form of a plane?
The key difference lies in the geometric information they directly provide.
- The Intercept Form (x/a + y/b + z/c = 1) tells you exactly where the plane cuts the coordinate axes. Its parameters (a, b, c) are points of intersection.
- The Normal Form (lx + my + nz = p) tells you the plane's orientation and distance from the origin. Its parameters (l, m, n) are the direction cosines of the normal vector perpendicular to the plane, and 'p' is the perpendicular distance from the origin to the plane.
7. What would be the equation of a plane that makes intercepts of 2, -5, and 3 on the x, y, and z-axes respectively?
To find the equation, you directly substitute the given intercept values into the standard intercept form equation, x/a + y/b + z/c = 1.
Given:
- x-intercept (a) = 2
- y-intercept (b) = -5
- z-intercept (c) = 3
Substituting these values, we get the equation: x/2 + y/(-5) + z/3 = 1, which can also be written as x/2 - y/5 + z/3 = 1.
