## ganeshie8 one year ago 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 $$ax+by=C$$

1. Owlcoffee

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: $r_1 + k(r_2)=0$

2. Empty

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.

3. ikram002p

give her a square and divide it to slices with same ax+by=c equation, so she would figure out its not unique representation.

4. ikram002p

if you got what i mean |dw:1438078875765:dw|

5. ikram002p

u can also use the book itself and papers as slices

6. anonymous

the equation $$ax+by=C$$ suggests that we're looking for points $$(x,y)$$ that have the same inner product with respect to $$(a,b)$$ -- $$(x,y)\cdot(a,b)=C$$

7. anonymous

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 $$(x-x_0,y-y_0)\cdot(a,b)=0$$

8. anonymous

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*

9. anonymous

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) $$y-y_0=a(x-x_0)\\z-z_0=b(x-x_0)$$

10. anonymous

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 $$x(t)=x_0+tu$$

11. anonymous

@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

12. anonymous

in $$\mathbb{R}^2$$ these two agree because a hyperplane is just a standard line : $$(x-x_0,y-y_0)\cdot (a,b)=0\\a(x-x_0)+b(y-y_0)=0\\y-y_0=-\frac{a}b(x-x_0)$$

13. anonymous

$$\frac{y-y_0}a=\frac{x-x_0}{-b}$$

14. Jaynator495

Explosive Diarrhea ? I dunno im in a weird mood today xD

15. Empty

$$\color{blue}{\text{Originally Posted by}}$$ @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.