Thread: Determining a codirectional vector that terminates on a line and for a plane

1. Determining a codirectional vector that terminates on a line and for a plane

This question involves 3 dimensions, but I will start in 2 dimension to get a better fit explaination. Its been bothering me for a couple days and I've been unsuccesful as getting an simple answer (I am able to derive an answer, but it seems to complicated to be reasonable). So, 2D first:

Suppose I have a line $g(x) = 0$ and a parabola $f(x) = \frac{1}{2}x^2$. I construct a vector$P$ from the origin to the point $(x, f(x))$ on $f(x)$. I want to determine the vector normal to $f(x)$ at $(x, f(x))$ pointing towards $g(x)$. I know from the derivitive of $f(x)$ that this vector, in general, is $V = $. Now, I need to determine a vector $V_N$ in the same direction as $V$ that when originating at the point $(x, f(x))$ will terminate exactly on the line $g(x)$ (i.e. the x-axis). For this I understand I simply need to find a general value $k$ such that:

$kV = V_N$

Lets say $D = P + V_N$. And remember that $P = $. I know that, since I need the vector $V_N$ (originating at point $(x, f(x))$ I remind you) to terminate on the x-axis, $D$ must be of the form $$ where $c$ is some general expression in terms of x So I need to find the value of $V$ such that:

$P + kV = $

Looking at P and V, I hope you can see that the value of $k$ would be $f(x)$, or I've done a poor job of explaining. This is because then we have:

$P +kV = + f(x)$

$= +$$$

$= $

This gives me the form of $V_N$ that assures it terminates on $g(x)$. But how would I determine a vector $V_N$ that terminates on$g(x)$ when $g(x)$ is not one of the axes? Say $g(x) = \frac{1}{3}x$, how do I determine $V_N$, originating at point $(x, f(x))$ such that it terminates on a line of this slope? Or a line of any slope?

And this question is the underlying question to the main one:

How would I determine this whole process for a parabaloid $f(x,y)$ and a plane $g(x, y)$? I know how to find the vector for $z=0$ but other wise I am lost, short of exiting vectors and determining line intersections, which will just defeat the purpose of my project. Anybody who can help, please do. I've been stumped on this one for some time. Thank you in advance

2. This explain is more complex than original problem. Is it derivative of nonlinear reflection ?

3. This is the manner in which I derived the reflection of the plane z=0 over the parabola f(x, y)=x^2+y^2+3. I simply dont know how to determine a vector that behaves in the way I need to in 3 dimensions. I am using vectors to define the reflection instead of complicated and bulky linear and explicit equations. I know matricies make things simpler, but I'm simply exploring.

4. 3-D computation is simple by modifying equation.
Demo is in attachment.

5. Originally Posted by math2009
3-D computation is simple by modifying equation.
Demo is in attachment.
Damn, I need to get some math software and take a crack at the 3D reflections. Just out of curiousity, although I know you are using matricies (atleast that what I though you were using) to generate these reflections, I was wondering if the underlying definition of reflection was still the same as the one I originally proposed? One that takes the lines normal to the "mirror", extends them out to the graph to be reflected and then takes all points where the normal intersects the graph of g(x) and reflects them over the corresponding tangent lines? I'm getting a trial of Mathematica today, maybe I'll be able to generate some reflections. Can you explain the matrix process generally? Or is it to complicated? Also, I've been trying to apply this vector method to 3-dimensions. Its much simpler then the bulky equations, but like I said I'm sure matricies are even simpler.

6. Tangent lines form plane in 3-D, plane is "mirror".
More explain will be programming issues. I'm using MATLAB. But Mathematica is also good. Syntax are different.
MATLAB is more close to math language of matrix.

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codirectional vector formula

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