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A bit of summer work that I can't figure out.
#1: a) Find the magnitude and direction of Vector R that is sum of the three vectors A, B, and C.
b) Find the magnitude and direction of the vector S=C-A-B.
Diagram:
Code:
A = Red
B = Blue
C = Green
2. Vectors A and B are drawn from a single point and C = A + B. If C^2 = A^2 + B^2 show that the angle between A and B is 90 degrees. If C^2 > A^2 + B^2 show that angle is greater than 90. If C^2 < A^2 + B^2 show that angle is smaller than 90.
The second one is just because I just suck at proofs ;; I tend to skip steps, and confuse myself.
 
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The answer is love. Love is the answer to everything.
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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.
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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?
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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.
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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.
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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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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.
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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...
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damn, i didnt know TL helps with homework lmao
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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 all
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Component vectors are your friend.
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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.
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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.
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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..)
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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 :/
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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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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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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.
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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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Dot product is god 
Its so awesome in linear algebra ^^
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On August 13 2008 14:13 Zortch wrote:Dot product is god Its so awesome in linear algebra ^^
its the basis of fourier transforms which is the basis of all technology
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On August 13 2008 14:13 Zortch wrote:Dot product is god Its so awesome in linear algebra ^^
you sure he doesn't need the proof for why if A(dot)B = 0, A and B are othogonal, though? ^^
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No. It's not. My computer does the ffts, which makes computers the basis of all technology.
Don't argue; I'm not open to reason right now.
Nuh! Uh! Ah! Quiet!
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On August 13 2008 14:20 Saracen wrote:Show nested quote +On August 13 2008 14:13 Zortch wrote:Dot product is god Its so awesome in linear algebra ^^ you sure he doesn't need the proof for why if A(dot)B = 0, A and B are othogonal, though? ^^
Mmm, its not hard to not assume that though.
If we have a*b=0 then |a||b|cos(%)=0 and since a and b are non-zero (by assumption) cos(%)=0, which implies that %=0 or 90.
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United States24768 Posts
Hm I find it interesting people are referring to 'decomposing' the vectors as opposed to 'resolving' them... new way of talking about it for me iirc.
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sry I think I used that term first here, didn't know the correct terminology.
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Russian Federation4235 Posts
What the heck has this to do with physics?
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United States24768 Posts
On August 13 2008 18:30 BluzMan wrote:
What the heck has this to do with physics?
Most likely, he was assigned this work in preparation for a physics class.
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On August 13 2008 18:30 BluzMan wrote:
What the heck has this to do with physics?
Just because it's easy compared to say calculating the cross section of a radiative association reaction doesn't mean it's not physics :/
Just as if someone is asking for help with the equation 5*x=3*y, 10*x-5*y=30. He is still asking for help with maths just as much as if he was asking for help with a forier transform.
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On August 13 2008 19:09 KlaCkoN wrote:Show nested quote +On August 13 2008 18:30 BluzMan wrote:
What the heck has this to do with physics? Just because it's easy compared to say calculating the cross section of a radiative association reaction doesn't mean it's not physics :/ Just as if someone is asking for help with the equation 5*x=3*y, 10*x-5*y=30. He is still asking for help with maths just as much as if he was asking for help with a forier transform.
What he meant is that it´s not physics but mathematics 
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On August 13 2008 18:30 BluzMan wrote:
What the heck has this to do with physics?
This stuff is pretty standard when learning about and understanding how a force pulls on ropes etc, how vectors work etc. I had this stuff too in the lower high school classes.
Anyway just use sin/cos/tan to split the arrows into horizontal and vertical vectors and then it's just a case of substracting and drawing the remaining vector.
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Russian Federation4235 Posts
Fourier transform is not physics as well. This thread is trivial school math.
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