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# Intercept Form of the Equation of a Plane

What is the equation of a plane if it makes intercepts (a, 0, 0), (0, b, 0) and (0, 0, c) with the coordinate axes? The equation of a plane in intercept form is simple to understand using the concepts of position vectors and the general equation of a plane. A problem on how to calculate intercepts when the equation of the plane is at the end of the lesson.

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## Concepts of a Plane in 3-D Geometry

To be able to understand the equation of a plane in intercept form, it is important to familiarize ourselves with certain terms first, which shall help us learn this topic better.

A vector, like we know it, is a quantity in the three-dimensional space that has not only magnitude but also direction. In 3-D Geometry, we use position vectors to denote the position of a point in space which serves as a reference to the point in question. It refers to the point in question with respect to the origin in 3-D Geometry.

As discussed even in earlier lessons, there can be multiple planes passing through a single point but there is one and only one plane that passes through a given point and is perpendicular to a given vector. The equation of such a plane is,

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

You must note that here

$$\vec{r} \; and \; \vec{a} \; are \; position \; vectors. \; \vec{N} \; is \; the \; normal \; vector$$

i.e. the vector that is perpendicular to the plane in question. The same equation can be written in Cartesian form as well. The equation of such a plane whose direction ratios are given by A, B and C respectively would be,

$$A (x – x_1) + B (y – y_1) + C (z – z_1) = 0$$

## Non-collinearity and the Equation of a Plane

This case was simple. But what if we were to write the equation of a plane that passes through three non-collinear points? As the name suggests, non-collinear points are those points that do not all lay on the same line. The Vector equation of a plane passing through three non-collinear points is:

$$( \vec{r} – \vec{a} ) . [ ( \vec{b} – \vec{a} ) \times ( \vec{c} – \vec{a} ) ] = 0$$

You must note that $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ are the position vectors of the three non-collinear points, when referred to from the origin. To write this in Cartesian form would give us:

$$\begin{vmatrix}(x – x_1) & (y – y_1) & (z – z_1) \\ (x_2 – x_1) & (y_2 – y_1) & (z_2 – z_1) \\ (x_3 – x_1) & (y_3 – y_1) & (z_3 – z_1) \end{vmatrix} = 0$$

The above equation is 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

We know that the general equation of a plane is:

Ax + By + Cz + D = 0 , where D $$\neq$$ 0

Here, A, B, C are the coordinates of a normal vector to the plane, while (x, y, z) are the coordinates of any point through which the plane passes. Now let us suppose this plane is making intercepts in the three-dimensional space at points A, B, and C on the x, y, and z-axes respectively. Let the coordinates of these points 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 are cut by the plane on the x-axis, y-axis and z-axis respectively.

(Source: Wolfram MathWorld)

Since the plane also passes through each of these three points, we can substitute them into the general equation of the plane and we have,

• Aa + D = 0
• Bb + D = 0
• Cc + D = 0

In other words, A = – D/a , B = -D/b , C = -D/c. 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 required equation of the plane that cuts intercepts on the three coordinate axes in the Cartesian system. Thus you can see how we can simply obtain the intercept form of the equation of a plane if the general equation of a plane is known. By making simple substitutions using the intercepts, the derivation of the equation of the plane can be arrived at.

## Solved Example for You

Question 1: What intercepts are made by the plane and the three axes of the coordinate system when the equation of the plane is given by 2x + y- z = 5?

Answer: This question already gives us the equation of the plane and we are to find out the intercepts. The equation of the said plane is –

2x + y- z = 5

Dividing the equation by 5 on both sides shall give us,

$$\frac{x}{5/2} + \frac{y}{5/1} + \frac{z}{5/(-1)} = 1$$

Now, if we go by the general form of the equation of a plane that cuts intercepts on the coordinate axes, we have –

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

Comparing this with the equation we had obtained gives us,

a = 5/2 , b = 5 and c = -5

Thus the coordinates of the required intercepts will be:

(5/2, 0 , 0); (0, 5, 0) and (0, 0 , -5)

Question 2: Explain what is intercept in maths?

Answer: The intercept of a line refers to the y-value of the point where its crossing takes place of the y-axis.

Question 3: Explain the formula for Y intercept?

Answer: Linear equation refers to the equation of any straight line whose writing can take place as y = mx + b. Here, m is the representative of slope of the line while b represents the y-intercept. The y-intercept of this line happens to be the value of y at the point. This particular point is where the line crosses the y axis.

Question 4: Explain what is meant by slope intercept form?

Answer: The graph of the equation y = mx + b is as line with slope, m. Furthermore, the y-intercept is b. Moreover, m and b represent real numbers. This form of the equation of a line is referred to as the slope-intercept form. The slope of a line, m, happens to be a measure of its steepness.

Question 5: Explain how one can solve slope intercept?

Answer: One can solve the slope intercept with the following two steps:

• Identification of the slope, m.
• Finding the y-intercept.

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