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#3776 Re: Help Me ! » Parallel line equations. » 2005-12-04 02:52:46
Ah, so you did mean a slope of -3/2 after all.
"y= -1.5x - 6"
Try plugging in the point (2, -3) into this. That would be: -3 = -1.5 * 2 - 6, so -3 = -9. That is, of course, wrong. The line can't possibly pass through that point.
It's much easier way to do this problem is called point-slope form. That is, it's the equation that occurs naturally when you have a point and a slope, and you want to form a line (exactly what you have 
).
The form is: Slope: m, point: (x1, y1), equation: (y - y1) = m(x - x1). So inputing your values, m = -3/2, x1 = 2, and y1 = -3, you get:
(y - (-3) ) = -3/2 * (x - (2) ), which is y = -3/2x + 3 - 3, which is y = -3/2x. This can then become 2y = -3x, and so 3x + 2y = 0.
No visualization required.
#3777 Re: Help Me ! » calculate position c based on position a and b » 2005-12-04 02:06:08
You have a points A and B, (ax, ay) and (bx, by). The vector between these two points is <ax - bx, ay - by>, going from B to A. Now what you need is the unit vector. To get this, we divide that vector by it's length. So let L = √( (ax - bx)² + (ay - by)² ), which is the length. <(ax - bx) / L, (ay - by) / L> is the unit vector. Then multiply this by the distance you want the camera, D: <D*(ax - bx) / L, D*(ay - by) / L>.
So Cx = bx + D*(ax - bx) / L and Cy = by + D*(ay - by) / L. This is because that vector will be the distance you have to travel from b to get to c, so to find the position of c, you just add the vector to that of b.
For all three planes, all you do is add in a z value:
A: (ax, ay, az)
B: (bx, by, bz)
Vector: <ax - bx, ay - by, az - bz>
L: √( (ax - bx)² + (ay - by)² + (az - bz)² )
Unit vector: <(ax - bx) / L, (ay - by) / L>
So you get Cx = bx + D*(ax - bx) / L, Cy = by + D*(ay - by) / L, and Cz = bz + D*(az - bz) / L.
#3778 Re: Help Me ! » Parallel line equations. » 2005-12-04 01:49:51
The book is wrong.
3x + 2y = 0 becomes 2y = -3x, and so y = -(3/2)x, which of course has a slope of -3/2, not 3/2.
#3779 Re: Help Me ! » Integer system » 2005-12-04 01:39:40
a, b, x, and y = 0 is a solution, though uninteresting.
Other than that, I can only come up with the restrictions:
0 ≤ y² ≤ x² ≤ a²
0 ≤ b² ≤ x² ≤ a²
By using the fact that squares have to be positive, so x² and y² must be positive, and thus b² has to be less than or equal to x², since y² is at least 0, and at most, x².
#3780 Introductions » Hi » 2005-12-04 00:32:35
- Ricky
- Replies: 11
I'm a CS and math major at Virginia Tech, currently a sophomore. I am looking for a forum where I can ask some advanced questions. Right now I am taking differential equations, multi variable calc, and intro to proofs, and next semester will be taking advanced calc, modern algebra, and linear algebra (two of which sound deceptively easy). I tried looking through some posts on the forum, and it seems to be mostly lower level math, but there were a few higher calc questions.
For the most part, I am into theory based math, proofs. I have been thinking about making a "request-a-proof" site, or something of the like, but if it turns to be like anything else I've created, it would be an udder flop (mostly due to lack of motivation). So right now, I just do the ones that interest me. For example, I never understood why x^0 (where x <> 0) was 1 until I proved it. So if you have a proof you want to see, by all means, I'll give it my best shot.
I also started my programming career on a ti-83 doing basic, now becoming quite a master at it, so if anyone needs any help in making math programs, I could be of assistance.
Anyways, that's me in a nutshell.