# Thread: HELP!? Vector Riddle, anybody up for solving this problem? I've given up.

1. ## HELP!? Vector Riddle, anybody up for solving this problem? I've given up.

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. Originally Posted by mfetch22
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 $g(x)$
What do you mean by "normal to f(x) pointing toward g(x)"? At any point on a graph, in two dimensions, there are two normals, point in opposite directions.

For $\displaystyle f(x)= x^2$, $\displaystyle f'= 2x$ so the tangent at point $\displaystyle (x_0, x_0^2)$ is of the form [tex]y= 2x_0(x- x_0)+ x_0^2[/itex]. That will cross the line y= 0 where $\displaystyle 2x_0(x- x_0)+ x_0^2= 0$ or $\displaystyle (x_0/2, 0)$. The vector you want is the one from $\displaystyle (x_0, x_0^2)$ to $\displaystyle (x_0/2, 0)$.

. 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

3. Okay I understand how to get this vector that ends on the x or y axis, I just need its respective componants to be zero, but what if the line is g(x) = (1/3)x and we want to find the general normal from f(x) [the normal that points towards g(x), not pointing the other way] at (x, f(x)) that terminates exactly on g(x) when added to P=<x, f(x)>? So a general vector at any point (x, f(x)) that terminates on the set of vectors of the form k<3, 1>.

4. Let M be the graph of your f(x,y) and P be the graph of g(x,y). Given any point p = (x,y,f(x,y)) in M, you want to calculate a vector Vn, such that Vn is normal to M and ends up on P.

Obviously the direction of Vn is V = (fx, fy, -1), where fx, fy are the partial derivatives.
Let Vn = kV and we need to calculate k.
Let pk be the end point of Vn, pk = p + Vn = (x,y, f(x,y)) + k*(fx, fy, -1) = (x+k*fx, y+k*fy, f(x,y)-k)
Since pk is on the graph of g, g(x+k*fx, y+k*fy)=f(x,y)-k
Solve this equation you'll get k.

5. Originally Posted by xxp9
Let M be the graph of your f(x,y) and P be the graph of g(x,y). Given any point p = (x,y,f(x,y)) in M, you want to calculate a vector Vn, such that Vn is normal to M and ends up on P.

Obviously the direction of Vn is V = (fx, fy, -1), where fx, fy are the partial derivatives.
Let Vn = kV and we need to calculate k.
Let pk be the end point of Vn, pk = p + Vn = (x,y, f(x,y)) + k*(fx, fy, -1) = (x+k*fx, y+k*fy, f(x,y)-k)
Since pk is on the graph of g, g(x+k*fx, y+k*fy)=f(x,y)-k
Solve this equation you'll get k.
Okay I think I get what your saying, so for a line with a slope of $\displaystyle \frac{1}{3}$, the general vector along this line (which I will call $\displaystyle G$) is: $\displaystyle a<3,\;1>$

And the general form of $\displaystyle P = <x,\;f(x)>$. And $\displaystyle V = <x,\;-1>$. I need $\displaystyle V_N$ to be in the same direction as $\displaystyle V$ but terminating on $\displaystyle g(x)$ when originating at $\displaystyle (x, f(x))$. In otherwords I need it such that:

$\displaystyle P+aV=bG$

Is there a simpler way then solving this equation for a to find out where the vector V intersects (i.e. terminates) exactly on G. More specifically, what the form of $\displaystyle V_N$ is?