Is (y + z)x1 = (y1 + z1)x the equation of a straight line in 3D space?

[sachin]

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.

 

[vela]

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.

 

[sachin]

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

 

[vela]

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?

 

[sachin]

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 ?

 

[FactChecker]

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?)

 

[sachin]

2D line is a two dimensional line what is plotted in xy plane, we have y = mx + c and all as equations.

 

[sachin]

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,

 

[FactChecker]

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.

 

[Mark44]

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.

 

[sachin]

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 ?

 

[FactChecker]

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 .

 

[Mark44]

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.

 

[sachin]

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 ?

 

[FactChecker]

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.

 

[Mark44]

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.

 

[WWGD]

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.

 

[FactChecker]

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.