Exploring Equation of a Straight Line in 3D Space

In summary: No, your assumption is not correct. Cross multiplying the terms in the equation of a line does not result in an equation of a plane. The equation (x-x1)/a = (y-y1)/b = (z-z1)/c is still an equation of a line, but in a different form. It is not an equation of a plane. To get an equation of a plane, you need three linear equations in three variables, not just two equations as in the case of a line. So your approach is not correct.
  • #1
sachin
60
7
Homework Statement
Can (y+z)x1=(y1+z1)x be an equation of a straight line in 3 dimensional space?
Relevant Equations
(x-x1)/a = (y-y1)/b = (z-z1)/c
Can we say,

(y + z ) x1 = (y1 + z1) x is also an equation of a straight line in 3 dimensional space,
where (x1,y1,z1) and (x,y,z) are the coordinates of a given point and a variable point respectively on a 3D line that passes through the origin,
have seen equation of a straight line in 3 dimensional space as,

(x-x1)/a = (y-y1)/b = (z-z1)/c

where (x1,y1,z1) is a known point in the line and a,b,c are the direction ratios (drs) of the line,

I am trying to modify the equation into another form which one will look like the equation of a line in a 2D case like ax + by + c = 0,

My approach,

Let (x1,y1,z1) and (x,y,z) be the coordinates of any given and a variable point respectively on a 3D line that passes through the origin, as direction ratios are just the length of the bases of the triangles formed in the space,

a = x1, b = y1 , c = z1,

now, instead of working on direction ratios measured by the cosine functions, I am taking the tan functions and using tan θ = opposite side/ base, along the y axis,

tan β = y1/x1 = y/x,

along the z axis,

tan γ = z1/x1 = z/x,

where β and γ are the angles made by the line with y and z axes respectively, so,we are getting,

y = (y1/x1) x,
z = (z1/x1) x,
y + z = (y1/x1) x + (z1/x1) x,

y + z = (y1+ z1) x/x1,
(y + z ) x1 = (y1 + z1 ) x,

which resembles the equation of a 2D line passing through the origin ax + by = 0,
my approach is to simply the equation of a 3D line into simple linear equation of three variables,
also in books it is given ax + by + cz + d = 0 is not the equation of a straight line but is the equation of a plane what I am finding is not the case as there a,b,c are not related to the plane but on the properties of a line what is perpendicular to the plane,
so I can conclude that ax + by + cz + d = 0 will represent a straight line in space but with the above modification what I got as (y + z ) x1 = (y1 + z1 ) x.
Just wanted to know if my approach is correct ?
thanks.
 
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  • #2
sachin said:
my approach is to simply the equation of a 3D line into simple linear equation of three variables,

also in books it is given ax + by + cz + d = 0 is not the equation of a straight line but is the equation of a plane what I am finding is not the case as there a,b,c are not related to the plane but on the properties of a line what is perpendicular to the plane,

so I can conclude that ax + by + cz + d = 0 will represent a straight line in space but with the above modification what I got as (y + z ) x1 = (y1 + z1 ) x.

Just wanted to know if my approach is correct ?
No. A single linear equation of three variables defines a plane.

A line has one degree of freedom. That means you can choose the value of one variable, but once you do that, the other two variables are determined.

A plane has two degrees of freedom. You can choose the value of two variables, but the third is then determined.

In your case, say you set ##x=0##. Assuming ##x_1 \ne 0## for simplicity, then as long as ##y=-z##, the equation is satisfied, so choosing ##x=0## did not determine unique values of ##y## and ##z##. So you don't have a line.
 
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  • #3
vela said:
No. A single linear equation of three variables defines a plane.

A line has one degree of freedom. That means you can choose the value of one variable, but once you do that, the other two variables are determined.

A plane has two degrees of freedom. You can choose the value of two variables, but the third is then determined.

In your case, say you set ##x=0##. Assuming ##x_1 \ne 0## for simplicity, then as long as ##y=-z##, the equation is satisfied, so choosing ##x=0## did not determine unique values of ##y## and ##z##. So you don't have a line.
i have just cross multiplied the terms in the equation of a line in (x-x1)/a = (y-y1)/b = (z-z1)/c , how can i get a plane from there ? as that will mean (x-x1)/a = (y-y1)/b = (z-z1)/c was the equation of a plane, is my assumption correct
 
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  • #4
i have just cross multiplied the terms in the equation of a line in (x-x1)/a = (y-y1)/b = (z-z1)/c , how can i get a plane from there ? as that will mean (x-x1)/a = (y-y1)/b = (z-z1)/c was the equation of a plane, is my assumption correct
No. In the first sentence, you said you had a line, and in the second, you said the same equations represented a plane. It's one or the other. It can't be both.

When you write
$$\frac{x-x_1}{a} = \frac{y-y_1}b = \frac{z-z_1}c,$$ it's shorthand for two equations, e.g.,
\begin{align}
\frac{x-x_1}{a} &= \frac{y-y_1}b \\
\frac{x-x_1}{a} &= \frac{z-z_1}c
\end{align} You can solve the first equation for ##y## in terms of ##x## and the second for ##z## in terms of ##x##. Once you choose a value for ##x##, you have no freedom to choose ##y## or ##z##. Their values are determined by the equations. You have one degree of freedom, so you have a line.

I've answered your original question, but it seems you have a different one in mind. Could you please state the problem you're trying to solve?
 
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  • #5
vela said:
No. In the first sentence, you said you had a line, and in the second, you said the same equations represented a plane. It's one or the other. It can't be both.

When you write
$$\frac{x-x_1}{a} = \frac{y-y_1}b = \frac{z-z_1}c,$$ it's shorthand for two equations, e.g.,
\begin{align}
\frac{x-x_1}{a} &= \frac{y-y_1}b \\
\frac{x-x_1}{a} &= \frac{z-z_1}c
\end{align} You can solve the first equation for ##y## in terms of ##x## and the second for ##z## in terms of ##x##. Once you choose a value for ##x##, you have no freedom to choose ##y## or ##z##. Their values are determined by the equations. You have one degree of freedom, so you have a line.

I've answered your original question, but it seems you have a different one in mind. Could you please state the problem you're trying to solve?
my concern is to give equation of a line a simpler form than bothering the ratios, what i found out from u is a single equation is always a plane, so y = x is not a 2d line but a plane ?
 
  • #6
sachin said:
my concern is to give equation of a line a simpler form than bothering the ratios, what i found out from u is a single equation is always a plane, so y = x is not a 2d line but a plane ?
More specifically, a single linear equation is a plane in 3D. Yes, y=x is a plane in 3D. (what is a 2D line?)
 
  • #7
2D line is a two dimensional line what is plotted in xy plane, we have y = mx + c and all as equations.
 
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  • #8
sachin said:
2D line is a two dimensional line what is plotted in xy plane, we have y = mx + c and all as equations.
Where did i go wrong in my assumption, is the mistake started when i added the equations or before that also in the post

y = (y1/x1) x,
z = (z1/x1) x,
y + z = (y1/x1) x + (z1/x1) x,
 
  • #9
sachin said:
Where did i go wrong in my assumption, is the mistake started when i added the equations or before that also in the post

y = (y1/x1) x,
z = (z1/x1) x,
y + z = (y1/x1) x + (z1/x1) x,
If you added equations, that was probably wrong. You can not turn one linear equation into three independent linear equations.
 
  • #10
sachin said:
2D line is a two dimensional line what is plotted in xy plane, we have y = mx + c and all as equations.
There's no such thing as a "2D line." A line is one dimensional object. The equation y = mx + c represents a line (one dimensional) in two dimensional space.
 
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  • #11
Mark44 said:
There's no such thing as a "2D line." A line is one dimensional object. The equation y = mx + c represents a line (one dimensional) in two dimensional space.
did not understand fully, is it like y = x fulfills the equation of a plane in 3d space and a line in 2d space ?
 
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  • #12
sachin said:
did not understand fully, is it like y = x fulfills the equation of a plane in 3d space and a line in 2d space ?
Yes. In general, one linear equation in N-dimensional space defines an N-1 dimensional subspace subset.
And n simultaneous linearly independent equations in N-dimensional space define an N-n dimensional subspace subset.
CORRECTION: These will not be subspaces unless they include the zero point. Thanks @Mark44 .
 
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  • #13
FactChecker said:
In general, one linear equation in N-dimensional space defines an N-1 dimensional subspace.
Technically, that would be an N-1 dimensional subset of the space it's embedded in, unless that subset happens to contain the origin.
 
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  • #14
Mark44 said:
Technically, that would be an N-1 dimensional subset of the space it's embedded in, unless that subset happens to contain the origin.
so if a standard 12 child asks if y = x is a line or a plane, what should be the answer ?
 
  • #15
Mark44 said:
Technically, that would be an N-1 dimensional subset of the space it's embedded in, unless that subset happens to contain the origin.
Thanks. I stand corrected.
 
  • #16
sachin said:
so if a standard 12 child asks if y = x is a line or a plane, what should be the answer ?
It depends. In ##\mathbb R^2## the equation defines a line through the origin. In ##\mathbb R^3## the equation defines a plane through the origin. In the first case, the equation represents a subspace of dimension 1; in the second case, it defines a subspace of dimension 2.
 
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  • #17
sachin said:
2D line is a two dimensional line what is plotted in xy plane, we have y = mx + c and all as equations.
not sure of your use of terms, but lines are intrinsically 1-dimensional objects, i.e., they are fully determined by 1 parameter.
 
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  • #18
sachin said:
so if a standard 12 child asks if y = x is a line or a plane, what should be the answer ?
Suppose we assume that "a standard 12 [year old] child" is only familiar with 2D graphs that were taught in school. Then the answer would be a line.
Suppose we assume that "a standard 12 [year old] child" is trying to develop/modify/learn code for a video game in 3D. Then the answer would be a plane.
 

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