Math Problem ...

Ray Kinsella

Hey Guys, I have a math issue, I know how you like brain teasers so I thought I would post it here, say I have two points ab & cd, how do I find the point 20% of distance between these two points on the same slope ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch

HalfWayMan

Make a vector between the points [c-a,d-b], normalise the vector (divide it by it's magnitude) and then multiply that by 0.2. Then add this onto [a,b] and you should have that. Now watch me get that all wrong and look a fool :)

Red Stateler

You should change your name to OneFifthMan.

HalfWayMan

Urm... I don't get it :((

Red Stateler

0.5 = 1/2. 0.2 = 1/5. Get it???? :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :^)

HalfWayMan

But 20% = 0.2. I hope. :^) -- modified at 13:03 Tuesday 13th June, 2006 ;)

Eric Dahlvang

I'm no math whiz, but can't you just: .2(a+c)=e .2(b+d)=f ef=20% between ab & cd That was stupid. I thought about it, and I think this would work: a+(.2(c-a))=e b+(.2(d-b))=f ef=20% between ab & cd If that isn't right, then I'm too dumn to figure it out without more coffee. ---------- Some problems are so complex that you have to be highly intelligent and well informed just to be undecided about them. - Laurence J. Peters -- modified at 13:11 Tuesday 13th June, 2006

Rohde

1. Translate the point to the origin (remember the translation) 2. Get the position vector of the second point 3. Calculate the length of the vector 3. Normalize the vector 4. Multiply each component with 0.2 x the length from #3 5. Use the inverse translation -- modified at 13:06 Tuesday 13th June, 2006 Oh, I was too late

Jun Du

Ray Kinsella wrote:

the point 20% of distance between these two points on the same slope

If you mean 20% of the distance from the point (a, b), then (e, f)=(a+0.2*(c-a), b+0.2*(d-b)) Otherwise (e, f)=(a+0.8*(c-a), b+0.8*(d-b)) - It's easier to make than to correct a mistake.

dighn

If I'm reading the question right (ie 20% of the way from AB to CD or vice versa) I don't think the vector should be normalized, so it's just (CD-AB)*0.2+AB

HalfWayMan

Oh spanner! I think you're right. P.S: Please don't tell my boss I got a question like this wrong! :-O

Ray Kinsella

don't get it let a,b=2,3 c,d=6,6 2. vector = 4,3 3. magnitude |v| = 5 4. = <4, 3> huh ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch

Ray Kinsella

yep ... figured that out a minute ago .... brain hurts ... too long since school ... thanks to all. Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch

Rohde

Magnitude = 5 20% of 5 = 1. So in this case add the normalized vector to get (2.8, 3.6).

Ed Poore

HalfWayMan wrote:

20% = 0.2

20% = 0.2 = 2/10 = 1/5 ;P

Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9

Ray Kinsella

gotcha now ... thanks ... Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch

Ed Poore

That won't work since you're actually getting the distance that's Sqrt(2(0.2*0.2)) along the line. Think of it as a right-angled triangle and you're using getting 0.2 along the x and 0.2 along the y so you're actually getting closer to 0.3 along the line. Give me 5 mintues and I'll write out the solution (I hope I'm right :rolleyes:)

Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9

Last modified: 13 June 2006 13:25:47 --

Ed Poore

Ok, a reference image is provided @ http://ed.poore.googlepages.com/question.bmp[^] Basically: Let A=(a,b) Let B=(c,d) Let P=(x,y) (the point where you want to find) |AB| = √((c - a)²+(d - b)²)

θ = arctan((d - b) / (c - a))

x = a + (0.2*|AB|*cos(θ)) y = b + (0.2*|AB|*sin(θ))

Some basic trig and pythagoras, you can do this by vectors, but it essentially boils down to the same thing if I remember correctly (better remember correctly since I might need it for my exam on Friday).

Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9

Jun Du

This is a perfect example that how people make things more complicated than they should be:omg: This is how it works:

x = a + (0.2*|AB|*cos(θ))) = a + 0.2*(c-a);
y = b + (0.2*|AB|*sin(θ))) = b + 0.2*(d-b);

BTW, I like your "θ". How come It shows "θ" in my editing panel, but not in the post? ...maybe because of my Linux (I'm using RedHat 8 at work). Signature for today: "Don't discuss math problems with a methematician."

Chris Losinger

just look at that happy happy math! whee! x2 cos(θ4) :) Cleek | Image Toolkits | Thumbnail maker