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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.
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