how do i explain this to a 10th grader who was just introduced to lines and planes explain why a line in 3-space cannot be represented by a scalar equation like \(ax+by=C\)

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how do i explain this to a 10th grader who was just introduced to lines and planes explain why a line in 3-space cannot be represented by a scalar equation like \(ax+by=C\)

Mathematics
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Because the "z" component is equal to zero: \[ax+by+(0)z=C\] This means that it has "z" coordinate of zero, meaning that it only defines a plane, and that plane will be created by the family of lines: \[r_1 + k(r_2)=0\]
In some sense you can, the problem is you have to use a system of linear equations to represent a line. So let's just say the z component depends on x and y, this is true for a line. \(z(x,y)=ax+by+c\) but the problem is this lets us pick any point in the xy plane to a height, which will give us an entire surface! We need to restrict what values we can pick for x and y, so we need to go further and define \(y(x)=mx+n\) which will be the projection of our actual line onto the xy plane, since we are going to map only points from a line to a line on the plane represented by \(z(x,y)\) to get our line.
give her a square and divide it to slices with same ax+by=c equation, so she would figure out its not unique representation.

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if you got what i mean |dw:1438078875765:dw|
u can also use the book itself and papers as slices
the equation \(ax+by=C\) suggests that we're looking for points \((x,y)\) that have the same inner product with respect to \((a,b)\) -- \((x,y)\cdot(a,b)=C\)
if we have some point \((x_0,y_0)\) such that \((x_0,y_0)\cdot(a,b)=C\) then it follows we want to find other points \((x,y)\) such that the vector from \((x_0,y_0)\) to \((x,y)\) are orthogonal to \((a,b)\), since \((x-x_0,y-y_0)\cdot(a,b)=0\)
in general, if we have \(x,x_0,n\in\mathbb{R}^n\) we have \((x-x_0)\cdot n=0\) singles out a subspace of one less dimension, a *hyperplane*
if you want to describe a line in n-dimensional euclidean space, we actually need to show that changes in our n dimensions are all proportional, giving a set of at least n-1 equations in our n variables of the form: $$t=x-x_0=\frac{y-y_0}a=\frac{z-z_0}b=\dots$$in the standard plane you commonly see this in the form \(x-x_0=\frac{y-y_0}b\Leftrightarrow y-y_0=a(x-x_0)\) in three dimensions we have $$x-x_0=\frac{y-y_0}a=\frac{z-z_0}b$$ which takes at least two equations (necessary since a line has one degree of freedom) $$y-y_0=a(x-x_0)\\z-z_0=b(x-x_0)$$
you can also introduce a parameter \(t\) (essentially giving us a chart of our line in space) like i wrote above to write \(n\) parametric equations in \(n+1\) variables: $$x-x_0=t\implies x=x_0+t\\\frac{y-y_0}a=t\implies y=y_0+at\\\frac{z-z_0}b=t\implies z=z_0+bt\\\dots$$ this can be written in more concise form by using vectors \(x,x_0,u\in\mathbb{R}^n\) so $$x(t)=x_0+tu$$
@Empty you were close, but you need to define \(y,z\) *both* in terms of \(x\) (or any of the three in terms of hte other two, really) to get the dimension right
in \(\mathbb{R}^2\) these two agree because a hyperplane is just a standard line : $$(x-x_0,y-y_0)\cdot (a,b)=0\\a(x-x_0)+b(y-y_0)=0\\y-y_0=-\frac{a}b(x-x_0)$$
$$\frac{y-y_0}a=\frac{x-x_0}{-b}$$
Explosive Diarrhea ? I dunno im in a weird mood today xD
\(\color{blue}{\text{Originally Posted by}}\) @oldrin.bataku @Empty you were close, but you need to define \(y,z\) *both* in terms of \(x\) (or any of the three in terms of hte other two, really) to get the dimension right \(\color{blue}{\text{End of Quote}}\) I did, just not in such a convoluted way as you.

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