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Quick Math Question (Algebra/Geometry)

Ultra_Kool_Dude

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I'm having a mental lapse :? :

If I want to find the distance between two parallel lines, do I just take the absolute value of the cross product?
 
if you know the length of the lines, you can connect the edges, so you get a square triangle (with a 90? corner, don't know how it's called)

.... _______
....|\
A..|......\ C........... (<-- supposed to be a triangle :p)
....|_______\
......B

and than you can apply the rules from 'soscastoacao" (dutch). you do need a scientific calculator though

sinus = overlaying side / tiled side (not border at the 90? corner)
cosinus = bordering side / tilted side
tangens = border side / opposite side
cotangens = opposite side / bordering side

if you know the length of B, you have to use the tangens rule on [ab], and that'll give you the length of C
if you have this, you apply the cosinus rule on [ac], and that'll give you the length of A

problem solved :)
 
Thanks, bone. :thumbsup: I really should have been more specific, I don't know the lengths of the lines.

Here's what I have:
3x+y-2z=2
3x+y-2z=10
Find the distance between the two lines.

It's not urgent or anything, so no one sweat it. 8)
 
Do you have correct answer? If it's 8/sqrt(10) (~=2.53), then I'll write down a solution, 'couse it's quite long the way I did it :)

EDIT: I looked at this solution again, and it's obviosly wrong, so nevermind.
 
I don't have the right answer, but I think I figured it out.

I have a formula to find the distance between a point and a line, so I should be able to just sub in a point from one of the lines and use the formula.
 
From my Linear algebra book:
For u and v in Rn, the distance between u and v, written as dist(u,v), is the length of the vector u-v. That is,
dist(u,v)=||u-v||
 
that would be a little bit too easy, wouldn't it :lol:

i vaguely remember a formula from a few years back, but that was for only 2 coordinates

(y1 - y2) = ((y - y1)/(x - x1)) * (x1 - x2)
sth like that

i think you'll have to solve it with something similar for 3 coordinates given
 
For 2D distance is abs((x2-x1)*(y3-y1)-(y2-y1)*(x?3-x1)) / sqrt((x2-x1)^2+(y2-y1)^2)

x1,y1 and x2,y2 are on one line, and x3,y3 is on another. I guess you can scale this to 3-dimensions.
 
I believe you add them. Multiply the top or bottom by -1. Then you will be able to figure out Y.

Not really sure though.

It's funny how you don't remember what should be easier things even though your in Calculus.
 
Ultra_Kool_Dude said:
Thanks, bone. :thumbsup: I really should have been more specific, I don't know the lengths of the lines.

Here's what I have:
3x+y-2z=2
3x+y-2z=10
Find the distance between the two lines.

It's not urgent or anything, so no one sweat it. 8)


Too easy. I'll call 'r' the distance between the lines; x,y,z as the distances between the lines.

Here's what you do:

project the lines into the x,y plane, and find the distances (treating it as a 2D line, I take it you know how to do that, yes?). This should get you x and y. Then project it into the x,z or y,z plane (your preferacne) to get z.

r = SQRT (x^2 + y^2 + z^2)
 
Yep, that's the way to do it without any vectors and dot/cross products. My answer was indeed the distance between the lines of projection to x,y
 
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