Equations of a Plane

1. The Normal Vector: The Essence of a Plane

The single most important concept for defining a plane is the normal vector. This is a vector that is perpendicular (at a 90° angle) to every line or vector lying within the plane.

Imagine a flagpole standing perfectly upright on a flat courtyard. The courtyard is the plane, and the flagpole is the normal vector.

This diagram illustrates a plane in a 3D coordinate system. The vector $\vec{n}$ is the normal vector, perpendicular to the plane. Any vector formed by connecting two points within the plane, like $\overrightarrow{PQ}$, will be orthogonal to $\vec{n}$.

This orthogonality is the key. Since the dot product of two perpendicular vectors is zero, if we take any two points P and Q on the plane, the vector $\overrightarrow{PQ}$ lies in the plane, and its dot product with the normal vector $\vec{n}$ must be zero.

This simple equation is the foundation for all forms of a plane's equation.

For a great intuitive explanation of the normal vector (which the instructor calls a "झंडा" or flag) and its central role, watch this short introductory clip.

2. Form 1: The Point-Normal Form

This is the most fundamental way to write the equation of a plane. We need two ingredients:

1. A specific point $A(x_1,y_1,z_1)$ that lies on the plane.
2. A normal vector $\vec{n}=A\hat{i}+B\hat{j}+C\hat{k}$ perpendicular to the plane.

Vector Equation

Let $\vec{a}$ be the position vector of the known point A, and $\vec{r}$ be the position vector of any general point P on the plane. The vector connecting A to P is $\overrightarrow{AP}=\vec{r}-\vec{a}$. Since this vector lies in the plane, it must be perpendicular to the normal vector $\vec{n}$.

Therefore, their dot product is zero:

This is the vector equation of the plane in point-normal form. It can also be written as $\vec{r}\cdot \vec{n}=\vec{a}\cdot \vec{n}$.

Cartesian Equation

We can convert the vector form to Cartesian coordinates.
Let $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$, $\vec{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$, and the normal vector's direction ratios be $(A,B,C)$, so $\vec{n}=A\hat{i}+B\hat{j}+C\hat{k}$.

Substituting into the vector equation:

This is the Cartesian equation in point-normal form. Notice that the coefficients of $x,y,z$ in the general equation $Ax+By+Cz+D=0$ are simply the direction ratios of the normal vector.

The following resource provides a clear walkthrough of these derivations.

3. Form 2: The Normal Form (and Intercept Form)

This form is special because it directly involves the perpendicular distance of the plane from the origin.

Normal Form

• Vector Equation: $\vec{r}\cdot \hat{n}=d$
  Here, $\hat{n}$ is the unit normal vector, and $d$ is the perpendicular distance from the origin to the plane.
• Cartesian Equation: $lx+my+nz=d$
  Here, $l,m,n$ are the direction cosines of the normal vector.

To convert a general equation $Ax+By+Cz+D=0$ to normal form, you divide the entire equation by $\sqrt{A^2+B^2+C^2}$.

Intercept Form

This is a simple and useful special case of the Cartesian equation. If a plane makes intercepts of length $a$, $b$, and $c$ on the x, y, and z axes respectively, its equation is:

This form is very handy for quickly sketching a plane or finding where it crosses the axes.

4. Form 3: The Three-Point Form

What if you are not given a normal vector, but instead three points that lie on the plane? As long as the points are non-collinear (i.e., they don't all lie on the same straight line), they uniquely define a plane.

Let the three points be $A(x_1,y_1,z_1)$, $B(x_2,y_2,z_2)$, and $C(x_3,y_3,z_3)$.

1. Find two vectors in the plane: We can create two vectors from these points, for example, $\overrightarrow{AB}$ and $\overrightarrow{AC}$.
   • $\overrightarrow{AB}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$
   • $\overrightarrow{AC}=(x_3-x_1)\hat{i}+(y_3-y_1)\hat{j}+(z_3-z_1)\hat{k}$
2. Find the normal vector: Your knowledge of the cross product is crucial here. The cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ will produce a vector perpendicular to both, which is exactly the normal vector $\vec{n}$ we need.
   • $\vec{n}=\overrightarrow{AB}\times \overrightarrow{AC}$
3. Use the Point-Normal Form: Now you have a normal vector $\vec{n}$ and a point on the plane (you can use A, B, or C). You can plug these into the point-normal form from Section 2.

Cartesian (Determinant) Form

This entire process can be condensed into a single formula using the scalar triple product, which is most efficiently calculated with a determinant. For any point $(x,y,z)$ on the plane, the vectors $\overrightarrow{AP}$, $\overrightarrow{AB}$, and $\overrightarrow{AC}$ must be coplanar. This means their scalar triple product is zero.

This determinant form is extremely useful and is a very common question type in exams like the MHT-CET.

This video demonstrates exactly how to apply this method, emphasizing its importance for your exam.

5. Summary of Plane Equations

The image below provides an excellent visual summary of the main forms of the equation of a plane.

*This image displays four key scenarios for finding the equation of a plane:

1. Normal Form: $\vec{r}\cdot \hat{n}=d$, defined by the unit normal and distance from origin.
2. Point-Normal Form: $(\vec{r}-\vec{a})\cdot \vec{N}=0$, defined by a point and a normal vector.
3. Three-Point Form: $(\vec{r}-\vec{a})\cdot [(\vec{b}-\vec{a})\times (\vec{c}-\vec{a})]=0$, using the scalar triple product.
4. Intersection of Two Planes: This is a topic we will cover later.*

Conclusion

Today we've moved from lines to planes, and you've learned how to describe these 2D surfaces in 3D space using both vector and Cartesian equations. The underlying logic is consistent and builds upon your prior knowledge of vector operations.

Here are the key takeaways:

• The normal vector $\vec{n}$ is the most crucial element in defining a plane's orientation.
• Point-Normal Form: The most fundamental form, requiring a point $\vec{a}$ and a normal $\vec{n}$, is $(\vec{r}-\vec{a})\cdot \vec{n}=0$.
• Three-Point Form: Given three points, you can find the normal vector using the cross product of two vectors formed by the points, then use the point-normal form. The determinant method is the most direct way to solve this.
• The coefficients $A,B,C$ in the Cartesian equation $Ax+By+Cz+D=0$ are the direction ratios of the normal vector.

In our next lesson, we will use these equations to solve more complex geometric problems. We will learn how to calculate the angle between two lines, two planes, or a line and a plane, which directly applies the concepts of direction and normal vectors we've mastered.