Equation of a Plane in Normal Form Made Easy
The three-dimensional geometry is a nightmare for some students who are appearing for their board exams. Nothing can be scarier for a student than to know a question, which is a sure shot will come in an exam, and they don’t know how to solve it. What makes geometry so tricky is the concept of planes, which is an integral part of 3D geometry. Today we are going to slay this behemoth named equation of the normal form of a plane and make it easy for students to understand it. Please make no mistake even we have to churn the gears of our mind to understand it fully, so we know how difficult this concept is for students to learn.
There are two ways to find out the general equation of a plane, the first one is by using the standard form, and the other way of doing it is by using the Cartesian form. The first method is quite essential, and once you get to know how to solve the normal vector of a plane with it, you can derive the Cartesian form on your own. So in this article, we are covering the first method only.
Theorem For Equation Of Plane In Normal Form
First, you need to think about a plane that is perpendicular to the origin, and its distance is D. In addition to this, D is not equal to 0.
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Here D ≠ 0
\[\overrightarrow{ON}\] ---> Normal from the origin
\[\widehat{n}\] ---> is the unit vector, which is normal along \[\overrightarrow{ON}\]
Now we have \[\overrightarrow{ON}\] = d\[\widehat{n}\]
Step 2:- Now here, you need to take any point (P) on the plane. Thus making vector \[\overrightarrow{NP}\] perpendicular to \[\overrightarrow{ON}\]
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As a result, the dot product of these two vectors will come out to be zero, meaning \[\overrightarrow{NP}\] . \[\overrightarrow{ON}\] = 0
Step 3:- Here we will take a line from the origin (0) and connects it with point (P)
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This will be our position vector \[\overrightarrow{r}\] of point (P).
Step 4:- From here, you take out the triangle ∆OPN, and we have the given equation.
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\[\overrightarrow{NP}\] + \[\overrightarrow{ON}\] = \[\overrightarrow{OP}\]
As a result, we can say \[\overrightarrow{NP}\] = \[\overrightarrow{OP}\] - \[\overrightarrow{ON}\]
We know \[\overrightarrow{OP}\] = \[\overrightarrow{r}\] and\[\overrightarrow{ON}\] = d\[\widehat{n}\] .
Now we will put these values in our given equation i.e \[\overrightarrow{NP}\] = \[\overrightarrow{OP}\] - \[\overrightarrow{ON}\]
\[\overrightarrow{NP}\] = \[\overrightarrow{OP}\] - \[\overrightarrow{ON}\]
\[\overrightarrow{NP}\] = \[\overrightarrow{r}\] - d\[\widehat{n}\]
Now here comes a twist, you need to remember the very first equation that we showed you, that product of two products will be zero.
i.e \[\overrightarrow{NP}\] . \[\overrightarrow{ON}\] = 0
nowhere, in this equation, we will put the value of \[\overrightarrow{NP}\] = \[\overrightarrow{r}\] - d\[\widehat{n}\]
now it looks like this,
( \[\overrightarrow{r}\] - D\[\widehat{n}\] ) . d\[\widehat{n}\] = 0
Step 5:- In this step, we are just going to simplify the equation to reach the final answer.
\[\overrightarrow{r}\] - d\[\widehat{n}\] - \[d^{2}\] ( \[\widehat{n}\] . \[\widehat{n}\] ) = 0
d [ \[\overrightarrow{r}\] . \[\widehat{n}\] - d ] = 0
[ \[\overrightarrow{r}\] . \[\widehat{n}\] - d ] = 0/d
\[\overrightarrow{r}\] . \[\widehat{n}\] - d = 0
\[\overrightarrow{r}\] . \[\widehat{n}\] = dThis is your equation of Plane in normal form.
Now for solving problems, you need to know about the Cartesian form, which is Ax + By + Cz = d. Where (A, B, C) are direction cosines of n and (x,y,z) is the distance of point P from the origin.
It might look difficult at the start, but once you start solving it, the answer will come on its own.
Solved Example
Now we know how to get to the equation; let’s try to apply it by solving some problems so you can better understand its usage.
Question. The distance of a given plane from the origin O is \[\frac{10}{\sqrt{36}}\], the normal vector given to us is 4\[\widehat{i}\] + 3\[\widehat{j}\] - 2\[\widehat{k}\] You have to find out the vector equation for the plane?
Answer. First, you need to find out the unit vector of a normal vector of a plane.
\[\widehat{n}\] = \[\overrightarrow{n}\] / | \[\overrightarrow{n}\] |
Thus, putting in the values we have
\[\widehat{n}\] = \[\frac{4\widehat{i} + 3\widehat{j} - 2\widehat{k}}{\sqrt{16+9+4}}\]
= \[\frac{4\widehat{i} + 3\widehat{j} - 2\widehat{k}}{\sqrt{29}}\]
Now, you need to substitute the vector equation in order to find out the required equation of the plane.
\[\overrightarrow{r}\] . \[\widehat{n}\] = d
\[\overrightarrow{r}\] . \[(\frac{5}{\sqrt{29}}\widehat{i}\; + \;\frac{3}{\sqrt{29}}\widehat{j}\; + \;\frac{-2}{\sqrt{29}}\widehat{k}\;)\] = \[\frac{10}{\sqrt{36}}\] your final answer
There you have it, the equation of a plane in normal form with its solved theorem and example. Now, you might have gone through this article and think you have learned it, well, that’s where you are wrong. Mathematics is not something that you can learn without solving. Hence, you must practice and try to solve some of these problems on your own, and if you have any issues, we are here to help you out.
FAQs on Equation of a Plane in Normal Form Made Easy
1. What is the equation of a plane in normal form, and what does it represent?
The equation of a plane in normal form provides a unique description of a plane's position and orientation in 3D space. It is defined by its perpendicular distance from the origin and the direction of the normal vector. It has two main representations:
- Vector Form: The equation is given by r ⋅ n̂ = d, where r is the position vector of any point on the plane, n̂ is the unit vector normal (perpendicular) to the plane from the origin, and 'd' is the perpendicular distance of the plane from the origin.
- Cartesian Form: The equation is lx + my + nz = d, where (l, m, n) are the direction cosines of the normal vector to the plane, and 'd' is the perpendicular distance from the origin.
2. What is the significance of the components 'd', 'l', 'm', and 'n' in the normal form equation lx + my + nz = d?
Each component in the Cartesian normal form equation has a precise geometric meaning:
- d: This represents the perpendicular distance of the plane from the origin. It must always be a non-negative value.
- l, m, n: These are the direction cosines of the normal vector to the plane. They define the orientation of the plane in space because they specify the direction of the line perpendicular to it. The condition l² + m² + n² = 1 is always satisfied.
3. Why is this specific equation called the 'normal' form?
It is called the normal form because it is defined using the vector that is normal (perpendicular) to the plane. Unlike other forms that can use any point on the plane or any vector parallel to it, the normal form relies on two unique properties tied to the origin: the direction of the perpendicular line from the origin to the plane (the normal) and the length of that perpendicular line (the distance 'd'). This makes it a fundamental way to define a plane's position in space.
4. How can you convert the general equation of a plane, Ax + By + Cz + D = 0, into the normal form?
To convert the general equation Ax + By + Cz + D = 0 to the normal form (lx + my + nz = d), you need to find the direction cosines of the normal and the perpendicular distance. The steps are:
- First, move the constant term to the other side: Ax + By + Cz = -D.
- Calculate the magnitude of the normal vector's direction ratios: √(A² + B² + C²).
- Divide the entire equation by this magnitude. This gives: (A/√(A² + B² + C²))x + (B/√(A² + B² + C²))y + (C/√(A² + B² + C²))z = -D/√(A² + B² + C²).
- This is now in the form lx + my + nz = d, where l, m, n are the direction cosines and d is the distance. Important: The distance 'd' must be positive. If -D is negative, you must multiply the entire equation by -1.
5. How does the normal form equation differ from the equation of a plane passing through a given point with a given normal vector?
While both forms use a normal vector, they differ in their specifics and what they directly reveal:
- Normal Form (r ⋅ n̂ = d): This form is very specific. It requires a unit normal vector (n̂) and directly gives the perpendicular distance (d) from the origin. It's a positional definition.
- Point-Normal Form ((r - a) ⋅ n = 0): This is a more general form. It uses any normal vector (n), not necessarily a unit vector, and a position vector (a) of any point on the plane, not necessarily the one closest to the origin. It doesn't directly tell you the perpendicular distance from the origin.
Essentially, the normal form is a standardized version of the point-normal form.
6. What is the equation of a plane in normal form if it passes through the origin?
If a plane passes through the origin, its perpendicular distance from the origin is zero. Therefore, the value of 'd' is 0. The normal form equation simplifies to:
- Vector Form: r ⋅ n̂ = 0
- Cartesian Form: lx + my + nz = 0
This shows that for any point (x, y, z) on such a plane, its position vector is always perpendicular to the plane's normal vector.
7. Can you determine the normal form equation of a plane using three non-collinear points?
Yes, you can. Although it's an indirect process, you would follow these steps:
- Let the three points be A, B, and C. Form two vectors lying in the plane, such as vector AB and vector AC.
- Find a normal vector n to the plane by taking the cross product of these two vectors (n = AB × AC).
- Use this normal vector n and one of the points (say, A) to write the general vector equation of the plane: (r - a) ⋅ n = 0.
- Finally, convert this general equation into the normal form by finding the unit normal vector n̂ and calculating the perpendicular distance 'd', as explained in the conversion process.
