| FragKrag United States. August 13 2008 12:46. Posts 11068 | Profile Blog # | |
| | *TL CJ Entusman #40* "like scissors does anything to paper except MAKE IT MORE NUMEROUS" -paper | |
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| Pellucidity Netherlands. August 13 2008 12:50. Posts 373 | Profile Blog # |
| The answer is love. Love is the answer to everything. |
| | "NO MUCH. WHY ARE YOUR SCARABS SO STUPID" - Tasteless |
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| BottleAbuser Korea (South). August 13 2008 12:50. Posts 1873 | Profile Blog # |
For 1, decompose the vectors into horizontal and vertical components. You can do this by multiplying their lengths by the sine(angle) and cos(angle). Then it's simple addition and subtraction.
For example, the red vector is composed of
12 * sin(37 degrees)i + 12 * cos(37 degrees)j
=7.22178028i + 9.58362612j
where i and j are unit vectors in X and Y axes, respectively.Last edit: 2008-08-13 12:59:55 |
| | Compilers are like boyfriends, you miss a period and they go crazy on you. |
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| FragKrag United States. August 13 2008 12:56. Posts 11068 | Profile Blog # |
I decomposed and got:
Ax = 9.58m
Ay = 7.22m
Bx = 11.49m
By = 9.6m
Cx = 3.0m
Cy = 5.1m
I think Cy, By, and Cx should be negative. Am I right?
Last edit: 2008-08-13 12:56:49 |
| | *TL CJ Entusman #40* "like scissors does anything to paper except MAKE IT MORE NUMEROUS" -paper | |
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| Kwidowmaker Canada. August 13 2008 12:56. Posts 976 | Profile Blog # |
Basically just pick a positive direction. Up (in the y) and right (in the x) is easiest, but different problems can make other combinations easier. The important thing is that you stay consistent throughout the question. Then get the components of the vectors (A would be 12cos37 for y and 12sin37 x). Then add the components (if you go with up and right as positive, then left and down would be negative). The root of the sum of the net x component squared and the y component squared is the magnitude of R. To find the direction you create a triangle out of the net components and use trigonometry to find the angle from the horizontal or vertical, whichever you or your teacher prefers.
Second question is basically the same thing.
-edit- Beaten by two other people >_>
-edit2- yes you're right, they should be negative.Last edit: 2008-08-13 12:59:38 |
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| BottleAbuser Korea (South). August 13 2008 12:59. Posts 1873 | Profile Blog # |
Oh, be careful with the sign.
To generalize, you should figure out the angle from the origin (positive X-axis is 0 degrees, and increment counter-clockwise). And then you can use cos(theta)i + sin(theta)j for every angle. But they're not expressed that way in the diagram, so you'd have to convert it first. So the blue vector would be at 320 degrees, or -40 degrees, not 40 degrees, and the red one would be 240 degrees.Last edit: 2008-08-13 13:00:22 |
| | Compilers are like boyfriends, you miss a period and they go crazy on you. |
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| Saracen United States. August 13 2008 13:00. Posts 5127 | Profile Blog # |
On August 13 2008 12:56 FragKrag wrote:
I decomposed and got:
Ax = 9.58m
Ay = 7.22m
Bx = 11.49m
By = 9.6m
Cx = 3.0m
Cy = 5.1m
I think Cy, By, and Cx should be negative. Am I right?
yes you're right :D
then manipulate the X and Y component separately and pythag back to get the answer
O_O |
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| BottleAbuser Korea (South). August 13 2008 13:03. Posts 1873 | Profile Blog # |
For the first part of problem 2:
If |C|^2 = |A|^2 + |B|^2, then that means the lengths are a pythagorean triplet (pretty much the definition of a pythagorean triplet). Which means they can be used to construct a right triangle. Trivial from there.
Dunno exactly how I'd approach the second and third part - I suck at proofs too. |
| | Compilers are like boyfriends, you miss a period and they go crazy on you. |
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| Saracen United States. August 13 2008 13:12. Posts 5127 | Profile Blog # |
i don't understand problem 2 (i think you mistyped something as well... "If C^2 = A^2 + B^2 show that angle is greater than 90.")
C=A+B
C^2 = A^2+2AB+B^2
thus
C^2 > A^2+B^2
as long as both vectors A and B exist
and i don't understand where angles come about in this question... you just draw 2 vectors arbitrarily... then compare it with some arbitrary variable C? if C is supposed to be a vector that connects A and B, then the angle between A and B would have to be 180
EDIT: I'm a dumbass, you can ignore like all of that...Last edit: 2008-08-13 13:15:49 |
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| Grobyc Canada. August 13 2008 13:14. Posts 16976 | Profile Blog # |
| damn, i didnt know TL helps with homework lmao |
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| FragKrag United States. August 13 2008 13:15. Posts 11068 | Profile Blog # |
ha, my bad. It was supposed to be C^2 > A^2 + B^2. I fixed the mistake. It's mainly about the Pythagorean theorem (I think). I just really suck at proofs.
Also: Thanks for all the help!! <3 for allLast edit: 2008-08-13 13:15:43 |
| | *TL CJ Entusman #40* "like scissors does anything to paper except MAKE IT MORE NUMEROUS" -paper | |
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| riotjune United States. August 13 2008 13:19. Posts 546 | Profile # |
| Component vectors are your friend. |
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| BottleAbuser Korea (South). August 13 2008 13:31. Posts 1873 | Profile Blog # |
Right. Express A and B in terms of component vectors. Then you have C (and |C|) in terms of A, B, and theta.
If theta is 90 degrees, then solve it out and you'll be able to show that |C|^2 exactly equals |A|^2 and |B|^2. |
| | Compilers are like boyfriends, you miss a period and they go crazy on you. |
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| FragKrag United States. August 13 2008 13:35. Posts 11068 | Profile Blog # |
On August 13 2008 13:31 BottleAbuser wrote:
Right. Express A and B in terms of component vectors. Then you have C (and |C|) in terms of A, B, and theta.
If theta is 90 degrees, then solve it out and you'll be able to show that |C|^2 exactly equals |A|^2 and |B|^2.
It gives me C^2 = A^2 + B^2, and it wants me to prove that the angle difference is 90. :/
I think the C = A + B part just means it's a triangle, whereas the A, B, and C in the next part are magnitudes.Last edit: 2008-08-13 13:38:44 |
| | *TL CJ Entusman #40* "like scissors does anything to paper except MAKE IT MORE NUMEROUS" -paper | |
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| BottleAbuser Korea (South). August 13 2008 13:49. Posts 1873 | Profile Blog # |
C = A + B indicates that you can construct C by adding the vectors A and B (a la problem 1). Which means that if you turned it by 90 degrees, yes, you could construct a triangle with those vectors.
Remember, you don't know the angle theta, but if you plug in 90 degrees (for the angle offset between vectors A and B) then you find that A^2+B^2=C^2 and you're done.
(phew, should remember to use proper notation..)Last edit: 2008-08-13 13:50:18 |
| | Compilers are like boyfriends, you miss a period and they go crazy on you. |
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| FragKrag United States. August 13 2008 13:57. Posts 11068 | Profile Blog # |
I can't simply assume 90 degrees can I? It gave me A^2+B^2=C^2, not 90 degrees ;_;
I don't even understand anymore :/ |
| | *TL CJ Entusman #40* "like scissors does anything to paper except MAKE IT MORE NUMEROUS" -paper | |
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| Zortch Canada. August 13 2008 14:06. Posts 635 | Profile Blog # |
Heres the proof:
I hope you know the dot product, which I will denote with *.
% is the angle between a and b.
Consider c^2=a^2+b^2, c=1+b:
then
c^2=(a+b)^2=a^2+b^2
a^2+2(a*b)+b^2=a^2+b^2
2(a*b)=0
a*b=0 So, a and b are orthogonal.
If c^2>a^b+b^2 then similarly we get:
a*b>0
|a||b|cos(%)>0
|a| and |b| are greater than 0 so:
cos(%)>0
This means that % is in the interval (90,180).
If c^2<a^b+b^2 similarly:
a*b<0
|a||b|cos(%)<0
|a| and |b| are greater than 0 so:
cos(%)<0
This means that % is in the interval (0,90).
Hope that helps ^^
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| Saracen United States. August 13 2008 14:09. Posts 5127 | Profile Blog # |
okay so:
C=A+B
and
Ccos(thetac) = Acos(thetaa) + Bcos(thetab)
Csin(thetac) = Asin(thetaa) + Bsin(thetab)
if mAB = 90 then thetab = thetaa+90
so
Ccos(thetac) = Acos(thetaa) + Bcos(thetaa+90)
Csin(thetac) = Asin(thetaa) + Bsin(thetaa+90)
square both equations and add them together
C^2(cos^2(thetac)+sin^2(thetac)) = A^2cos^2(thetaa) + B^2cos^2(thetaa+90) + 2Acos(thetaa)Bcos(thetaa+90) + A^2sin(thetaa) + B^2sin(thetaa+90) + 2Asin(thetaa)Bsin(thetaa+90)
getting rid of cos^2+sin^2=1:
C^2 = A^2 + B^2 + 2Acos(thetaa)Bcos(thetaa+90) + 2Asin(thetaa)Bsin(thetaa+90)
and 2Acos(thetaa)Bcos(thetaa+90) + 2Asin(thetaa)Bsin(thetaa+90) = 0 because for any theta,
cos(theta)cos(theta+90) = sin(theta)sin(theta+90)
therefore, C^2=A^2+B^2 if mAB = 90
um i'll let you do the rest
this just reminded me about what horrible things summer does to your brain
god i feel so stupid now
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| BottleAbuser Korea (South). August 13 2008 14:12. Posts 1873 | Profile Blog # |
| You could arbitrarily decide that A's direction is in line with your X axis and just use a single theta o.O but yeah that works. |
| | Compilers are like boyfriends, you miss a period and they go crazy on you. |
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| Saracen United States. August 13 2008 14:12. Posts 5127 | Profile Blog # |
On August 13 2008 14:06 Zortch wrote:Heres the proof: I hope you know the dot product, which I will denote with *. % is the angle between a and b. Consider c^2=a^2+b^2, c=1+b: then c^2=(a+b)^2=a^2+b^2 a^2+2(a*b)+b^2=a^2+b^2 2(a*b)=0 a*b=0 So, a and b are orthogonal. If c^2>a^b+b^2 then similarly we get: a*b>0 |a||b|cos(%)>0 |a| and |b| are greater than 0 so: cos(%)>0 This means that % is in the interval (90,180). If c^2<a^b+b^2 similarly: a*b<0 |a||b|cos(%)<0 |a| and |b| are greater than 0 so: cos(%)<0 This means that % is in the interval (0,90). Hope that helps ^^
yeah dot product helps >_<
god i forgot all of that ... dot product, cross product  i hate vectors
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