Math Problem ...
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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 :)
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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
Oh spanner! I think you're right. P.S: Please don't tell my boss I got a question like this wrong! :-O
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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
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
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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.
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
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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
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But 20% = 0.2. I hope. :^) -- modified at 13:03 Tuesday 13th June, 2006 ;)
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
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gotcha now ... thanks ... Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
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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
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 --
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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
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
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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
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."
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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
just look at that happy happy math! whee! x2 cos(θ4) :) Cleek | Image Toolkits | Thumbnail maker
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But 20% = 0.2. I hope. :^) -- modified at 13:03 Tuesday 13th June, 2006 ;)
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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
You get a division by zero if you are dealing with vertical lines.
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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."
Though both are correct, I'd go with the simpler way since there is no worries in regards to division by zero. PJC
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You get a division by zero if you are dealing with vertical lines.
Paul Conrad wrote:
You get a division by zero if you are dealing with vertical lines.
You get division by zero if you deal with horizontal lines. Since tanθ = opposite / adjacent (i.e. vertical over horizontal). But it's only a couple of ifs to sort this out. To be honest I'm suprised that so many people seem to be accepting the incorrect versions here. All you need to know is Pythagoras (a²+o² = h²) to prove that the simpler formulae they've come up with are nonsense.
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 15:56:59 --
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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."
Jun Du wrote:
This is a perfect example that how people make things more complicated than they should be
Maybe it's more complicated but it works. Here's the proof using simple Pythagorean trig and his theorem.
Start with a 3,4,5 triangle with height 3 and width 4 (and hypotenuse 5).
If A is the bottom left-corner, located over an "origin" of (0,0) then it follows that
A = (0,0) (bottom-left corner)
B = (4,3) (top-right corner)
C = (4,0) (bottom-right corner)
θ = Angle BAC (i.e. bottom-left)
Also it can be shown (through Pythagorean Trigonometry) that:
sin(θ)=3/5
cos(θ)=4/5
tan(θ)=3/4So you require 20% of the line AB (in this case it is (0.2*5) = 1).
We can let the co-ordinates of the end point of this line (from A along AB for 20% of AB) be X=(x,y).
It then follows that since we have a hypotenuse equal to 1 and the angle has remained the same that:x = 1*cos(θ) = 4/5
y = 1*sin(θ) = 3/5So from A the new point X which is 20% of the line AB is A+X, i.e: X = (0 + 4/5, 0 + 3/5) = (4/5, 3/5).
If you take you're equations of:
x = a + 0.2 * (c - a)
y = b + 0.2 * (d - b)
You will get the following answers for this case:
x = 0 + 0.2 * (4 - 0) = 0.2 * 2 = 2/5
y = 0 + 0.2 * (3 - 0) = 0.2 * 3 = 9/10
Which is not the same as the ones above.The simplest proof that the previous equations are wrong can be shown quite simply by Pythagoras' Theorem.
If we assume that the triangle formed is of height and width 1 then, your equations show that the height and width of the "new" triangle which provides 20% of AB are both 0.2.
If you use Pythagoras on these you get a hypotenuse of √(0.2²+0.2²) which gives √0.08 which is ~0.2828, not 0.2 as you require.
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
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Paul Conrad wrote:
You get a division by zero if you are dealing with vertical lines.
You get division by zero if you deal with horizontal lines. Since tanθ = opposite / adjacent (i.e. vertical over horizontal). But it's only a couple of ifs to sort this out. To be honest I'm suprised that so many people seem to be accepting the incorrect versions here. All you need to know is Pythagoras (a²+o² = h²) to prove that the simpler formulae they've come up with are nonsense.
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 15:56:59 --
Ed.Poore wrote:
You get division by zero if you deal with horizontal lines. 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))
Check your one post because when c-a = 0, the line is vertical :) -- modified at 17:02 Tuesday 13th June, 2006
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Ed.Poore wrote:
You get division by zero if you deal with horizontal lines. 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))
Check your one post because when c-a = 0, the line is vertical :) -- modified at 17:02 Tuesday 13th June, 2006
Oops, maybe your right, did things in a bit of rush because I've got annoyed that people can't see that the "complex" formula is the correct one, you can't get any simpler. Anyway I'm not going to argue just yet, I'm off to have some supper, I'll come back later and see if people have finally accepted it.
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
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Though both are correct, I'd go with the simpler way since there is no worries in regards to division by zero. PJC
How can both be correct, I've just posted the proof that they aren't.
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
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Oops, maybe your right, did things in a bit of rush because I've got annoyed that people can't see that the "complex" formula is the correct one, you can't get any simpler. Anyway I'm not going to argue just yet, I'm off to have some supper, I'll come back later and see if people have finally accepted it.
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
Ed.Poore wrote:
the "complex" formula is the correct one
The solution by Jun Du is also correct. Tried both your solution and his for several different points and I get the same answers from both. Consider if this problem was being programmed on some weird machine where no square root functions or trig functions were available via hardware, then Jun Du's solution would be best.
