plane equation from 2 points

[markraz]

Hi, If I have a 2 points: P (x,y,z) and O <0,0,0>(Origin) as shown above, is it possible to find a plane equation from this limited info?
Can O<0,0,0> and P<x,y,z> represent a normal to the plane? I need to fond the plane perpendicular to this vector shown

any help would be appreciated



thanks

 

[Ishuda]

View attachment 5235
Hi, If I have a 2 points: P (x,y,z) and O <0,0,0>(Origin) as shown above, is it possible to find a plane equation from this limited info?
Can O<0,0,0> and P<x,y,z> represent a normal to the plane? I need to fond the plane perpendicular to this vector shown

any help would be appreciated



thanks

The equation for a plane can be written as
a (x-x₀) + b (y-y₀) + c (z-z₀) = 0
where (x, y, z) and (x₀, y₀, z₀) are points on the plane. The vector (a, b, c) is just a vector normal to the plane. Notice that a, b, and, c are not unique but if we normalize the vector [that is divide a, b, and c by \(\displaystyle \sqrt{a^2\, +\, b^2\, +\, c^2}\) so that its norm is 1], any normal vector can be written as \(\displaystyle \alpha\) times the normal vector. You might want to look at
http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx

For your particular problem, if what you have is two points on the plane then there is an infinite number of planes which would satisfy this [think of a line in a plane and rotating the plane around this line]. If you have a point and a normal vector, then you can form a unique plane.

 

[markraz]

thanks... I checked out that site pretty cool thanks.. He says you can use a point and a normal vector to find the plane

so in my graphic, can I use point P (x,y,z) and point O <0,0,0> to generate a the normal vector?? and then also use P as my single point on the plane? or do I need a different point because P would then be redundant and not unique?


thanks for your help

 

[Ishuda]

thanks... I checked out that site pretty cool thanks.. He says you can use a point and a normal vector to find the plane

so in my graphic, can I use point P (x,y,z) and point O <0,0,0> to generate a the normal vector?? and then also use P as my single point on the plane? or do I need a different point because P would then be redundant and not unique?


thanks for your help

First, let's get some notation straight. When you are saying a point P (x,y,z) do you mean a specific point like x is equal to a particular fixed value, y is equal to a particular fixed value and z is equal to a particular fixed value or do you mean a general point in three space with co-ordinates x, y, and z? If you do mean that it is a particular fixed point, what are you calling the the general three space vector?

 

[markraz]

First, let's get some notation straight. When you are saying a point P (x,y,z) do you mean a specific point like x is equal to a particular fixed value, y is equal to a particular fixed value and z is equal to a particular fixed value or do you mean a general point in three space with co-ordinates x, y, and z? If you do mean that it is a particular fixed point, what are you calling the the general three space vector?

hi

I guess what I'm trying to do is, find a plane perpendicular to the line in my graphic. Can I make a vector from the 2 points I have? The startpoint O <0, 0, 0> and end point P is just some xyz point.
So for example let's just say point P is 4,3,2. So to create a new vector could I use the formula (X2-X1) + (Y2-Y1) + (Z2-Z1) ?? so (4 - 0) + (3 - 0) + (2 - 0 ) ... so would my vector be
OP = 4i +3j+2k ??...... Can this be considered a normal vector? And then can I use this to find a plane equation? Would my plane just be perpendicular to this vector?

thanks

 

[Ishuda]

hi

I guess what I'm trying to do is, find a plane perpendicular to the line in my graphic. Can I make a vector from the 2 points I have? The startpoint O <0, 0, 0> and end point P is just some xyz point.
So for example let's just say point P is 4,3,2. So to create a new vector could I use the formula (X2-X1) + (Y2-Y1) + (Z2-Z1) ?? so (4 - 0) + (3 - 0) + (2 - 0 ) ... so would my vector be
OP = 4i +3j+2k ??...... Can this be considered a normal vector? And then can I use this to find a plane equation? Would my plane just be perpendicular to this vector?

thanks

O.K., let's use (x₀, y₀, z₀) for that second point P in the plane and (x,y,z) as the general point in \(\displaystyle \mathbb{R}^3\). That point and the origin [point (0,0,0)] form the line
L(t) = t * (x₀, y₀, z₀)
in \(\displaystyle \mathbb{R}^3\). You could have many different planes containing that line. You could arbitrarily choose any other point Q equal to (x₁, y₁, z₁) not on L to define your plane [see the link I gave you before]

If you would like to use (x₀, y₀, z₀) as the normal vector to define a plane containing the point (0,0,0), you could do that as pointed out in that same link. Just use (x₀, y₀, z₀) as the normal vector \(\displaystyle \overrightarrow{n}\)

 

[Steven G]

hi

I guess what I'm trying to do is, find a plane perpendicular to the line in my graphic. Can I make a vector from the 2 points I have? The startpoint O <0, 0, 0> and end point P is just some xyz point.
So for example let's just say point P is 4,3,2. So to create a new vector could I use the formula (X2-X1) + (Y2-Y1) + (Z2-Z1) ?? so (4 - 0) + (3 - 0) + (2 - 0 ) ... so would my vector be
OP = 4i +3j+2k ??...... Can this be considered a normal vector? And then can I use this to find a plane equation? Would my plane just be perpendicular to this vector?

thanks

You have been told that two points are not enough for a plane. IF you could do something to those two points and then eventually get a plane then you would have been told that two points make a plane.
Consider the following (very similar to what has already been written): You have a sheet of paper attached to a pencil. Think of the paper as a plane. Now given two points in space place the pencil so that it passes through those two points. Since you can rotate the pencil in such a way that the pencil still passes through the two points you will have a new plane. In fact, you have an infinite number of planes if you continue to rotate the pencil. If you pick one more point then you will have a unique plane. You need to picture that this will happen, MUST.