Hi, Can some body tell me how to find out the plane equation. I have a slide which says that: I got this slide (#12) from the following link: http://www.mamcet.com/it/e-learning/7sem/gm/unit%20II/3D%20object%20representation.pdf Can somebody plz guide me with this? Zulfi.
So what is your understanding of the equation of a plane in 3 dimensions? In this particular case it is the infinite set of points that satisfy the condition that x = 1, and y and z are members of belong to R (the set of real numbers). In general the equation of a plane is ax + by + cz = d In this particular case b = c = 0, a = 1, and d = 1 You could have Googled "Equation of a Plane" you know.
Hi, I know this eq of plane but I cant understand why x=1? (for the shaded portion given in the figure). In the figure, there is no a, b, c, or d. This is the confusion. According to the eq your answer is correct but according to figure this cant be deduced because figure does not have these symbols. Kindly guide me. Zulfi.
In words of as few syllables as possible x - 1 = 0 is exactly the same statement as 1*x + 0*y + 0*z = 1 which is the general form of the equation of a plane. Using basic algebra if you add +1 to both sides of the original equation x = 1 is what you get. What's the problem?
Hi, I got some understanding after looking at the picture. The shaded portion is touching the x-axis (is it correct?). What would be the eq if the bottom portion of the cube is also shaded?? Kindly guide me. Zulfi.
Hi, Thanks for your response. If we shade the bottom portion, then it would be touching, x, y and z but it would be touching y at origin. So values would be x=z=1 & y=0. Kindly reply me about this answer. Zulfi.
I note this is slide 9 in your link. Note that y=0 and z=0 whatever the value of x. So taking papbravo's answer Edit : Where is x=0 for all y and z? Answer the yz plane. So the plane x-1=0 = the plane x=1 is a plane parallel to the yz plane. This is clearly shown in your slide 12, where all points in the yellow shaded side are exactly 1 unit along the x axis from the yz plane. If you extend this side in all directions it will become the plane x=1. There are an infinite number of such planes all parallel to the yz plane and all of the form x= a constant. Does this help.
First check my edit I was too hasty, sorry about that. The bottom portion (face) is sitting on the xz plane. So whatever the value of x and z and y=0, not 1 Note that the numbers x=1, y=0 z=1 refer to a point not a plane. The values of x and z vary from 0 to 1 over this face and of course the value of y is 0 whatever the values of x and z
Hi, Thanks for your response and answer to my Question. This means my answer was right. Considering the shaded portion again as shown in the figure. Its x=1 plane as you said. Its a vertical plane as visible from figure. It should have y-value. Why y is zero?? Lets consider the line x=1, y is not zero on this line. Sorry for keeping you busy. Zulfi.
I really am not having a good morning I have had to edit my first post again. Here is the correct version A small point The shaded portion is only part of the x=1 plane. Hopefully this makes proper sense now.
Hi, Thanks for your re-writings. Still the concept of plane is not clear to me specially for x=1 (shaded portion of )plane. This is because its a rectangular area. Why we dont have any y-component for this?? I am able to understand the additional information i.e.: Kindly guide me if you have time. Zulfi.
The specifications x = 1, y = 0, z = 1 defines the coordinates of a point, not a plane. What does not change on the bottom face? It is y = 0. That defines the entire bottom plane, not just the rectangular area. The rectangular bottom face has boundaries, along with y = 0, 0 <= x <= 1 0 <= z <= 1
Hi, Thanks. According to this logic, the eq of top face of cube would be: y=1?? (Yes or No) & eq of front surface would be: z=1 ??(Yes or No) & eq of back surface would be: z=0?? (Yes or No) Now kindly tell me why we dont have a y-component for the x=1 plane even though its rectangular in shape?? The standard plane equation has got all the three components: x, y, & z. Also tell me about the normal vector on the same slide? There is one another slide which i got from web which says: Kindly guide me about this normal vector N also. Plz provide answers to my (Yes or No) Questions so that i can correct my concepts. Zulfi.
x = 1 is the equation of the plane. The plane extends to infinity and has no shape. In order to define the rectangular shape you have to set the boundary conditions for y and z. y = 0 is the equation for the bottom plane, not the rectangular face. The intersection of two planes x = 1 and y = 0 will give you a line. This line is infinitely long, parallel to the z-axis.
Hi, Thanks for your help. I think we must not call it a point. Consistent with the slide we must call it a Normal vector N. We have to put the values of Normal vector into standard eq: x+z + 1= 0 according to post # 2. Kindly guide me if its correct or not. Zulfi.
Hi, I have found this at some other site: at: http://thejuniverse.org/PUBLIC/LinearAlgebra/LOLA/planes/std.html This means the normal vector k has a value: (0, 0, 1) & D is also 0. So Mr. Chips was right when he said that y=0 was the eq of xz plane. I was only confused how we can use the standard eq for finding this eq as shown in post#2. Zulfi.
You should distinguish carefully between a point and a vector. A point is just that, a point. A vector is a line. Both are specified by three numbers, and indeed you can regard the three numbers that specifiy a point as the vector or line from the origin to that point. So the triplet T = (a,b,c) specifies the point x=a, y=b, z=c. It also specifies the line or vector joing the point (0,0,0) to T. A normal vector is at right angles to something (usually a plane or area). There is no single equation for a line in three dimensions. A line needs three equations to specify it. The equation specified ax+by+cz=d specifies a plane. The general plane has a general normal vector at right angles to it. When the planes are parallel to the base planes of the coordinate system, the normal vector has a particularly easy fomat.
Hi, Thanks all of for providing me feed back related to important concepts of plane. I want to summarize that in order to find the eq of plane we need a normal and a point through which the plane passes. For instance in case of xz plane, the plane passes through origin (0,0,0) and has normal vector (0,1,0), so the eq according post# 2 is: 0(x-0)+1(y-0)+0(z-0)=0 so y=0. Note that D is not included in this eq. as from a slide: Now for the shaded portion: Normal vector is (1,0,0) and the point is (1,0,0) So the eq is: 1(x-1) + 0(y-0) + 0(z-0)=0 So x -1=0 Zulfi.