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Require Help.

by AraProdieur
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AraProdieur
#1
Aug28-07, 09:57 PM
P: 27
1. The problem statement, all variables and given/known data
Vector A has a magnitude of 11 and points in the positive x-direction. Vector B has a magnitude of 22 and makes an angle of 32 degrees with the positive x-axis.
What is the magnitude of Vector A minus Vector B?


2. Relevant equations



3. The attempt at a solution
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learningphysics
#2
Aug28-07, 10:00 PM
HW Helper
P: 4,124
Can you describe where you're getting stuck and what you tried?
AraProdieur
#3
Aug28-07, 10:01 PM
P: 27
I also have another question whereas it states: Use the method of components to find the magnitude and direction of the vector sum R1 where R1= A + B. The Vector A= 15.2 m at an angle alpha= 180 degrees from the positive horizontal axis, and Vector B = 17.2 m at an angle Beta= 41.3 degrees from the positive horizontal axis. Answer in meters. Answer in units of m.

What is the angle, theta 1, from the positive horizontal axis of the vector sum, R 1?

What is the magnitude of the vector difference R 2 where R 2= Vector A - Vector B?

What is the angle, theta 2, of the resulting vector?

What is the magnitude of the vector difference R 3 where R 3= B - A?

What is the angle, theta 3, of the resulting vector?

AraProdieur
#4
Aug28-07, 10:02 PM
P: 27
Require Help.

Quote Quote by learningphysics View Post
Can you describe where you're getting stuck and what you tried?
I've only sketched it to get a graphical representation of the vectors, but other than that I have only gotten the resultant magnitude of A + B, but don't know how to get the resultant magnitude for A - B.
rootX
#5
Aug28-07, 10:05 PM
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P: 1,294
draw a llgm with two vectors A and B,
you have found one diagonal, just find other one, and it would be A - B

or,
reverse the direction of vector B
learningphysics
#6
Aug28-07, 10:07 PM
HW Helper
P: 4,124
Quote Quote by AraProdieur View Post
I've only sketched it to get a graphical representation of the vectors, but other than that I have only gotten the resultant magnitude of A + B, but don't know how to get the resultant magnitude for A - B.
A - B is the resultant of A and (-B). Use the same approach as before except use -B instead of B...

another idea... draw both vectors originating from the same point... what is the vector joining the arrow ends of B and A?
AraProdieur
#7
Aug28-07, 10:08 PM
P: 27
Quote Quote by rootX View Post
draw a llgm with two vectors A and B,
you have found one diagonal, just find other one, and it would be A - B

or,
reverse the direction of vector B

Yes, I know what you are referring to about the -B pertaining to its opposite direction, but the problem is I only know how to find the resultant magnitude for A + B and don't know how to find it for A - B because it's different since it's only two vectors and not more.
AraProdieur
#8
Aug28-07, 10:10 PM
P: 27
Quote Quote by learningphysics View Post
A - B is the resultant of A and (-B). Use the same approach as before except use -B instead of B...

another idea... draw both vectors originating from the same point... what is the vector joining the arrow ends of B and A?
Ok, so but isn't finding the magnitude of the resultant vector using basically the pythagorean theorem? If I use -B it will come out to the same answer. The thing that I'm confused about is that when I draw it graphically the A + B resultant vector looks shorter than the A - B resultant vector. So the problem is that I don't know how to find that A - B resultant vector, I only know the A + B resultant vector.
learningphysics
#9
Aug28-07, 10:11 PM
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Quote Quote by AraProdieur View Post
Yes, I know what you are referring to about the -B pertaining to its opposite direction, but the problem is I only know how to find the resultant magnitude for A + B and don't know how to find it for A - B because it's different since it's only two vectors and not more.
So you have drawn A - B, but are just having trouble calculating the magnitude?
learningphysics
#10
Aug28-07, 10:13 PM
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P: 4,124
Quote Quote by AraProdieur View Post
Ok, so but isn't finding the magnitude of the resultant vector using basically the pythagorean theorem? If I use -B it will come out to the same answer. The thing that I'm confused about is that when I draw it graphically the A + B resultant vector looks shorter than the A - B resultant vector. So the problem is that I don't know how to find that A - B resultant vector, I only know the A + B resultant vector.
No, the pythagorean theorem is only for right triangles... this isn't a right triangle... pythagorean theorem won't work for A-B or A + B.
AraProdieur
#11
Aug28-07, 10:14 PM
P: 27
Quote Quote by learningphysics View Post
So you have drawn A - B, but are just having trouble calculating the magnitude?
Yes, the way I depicted it was with Vector A going towards the right with a magnitude of 11 along the x-axis, while the -B is at 212 degrees going in the southwest direction with a magnitude of 22.
AraProdieur
#12
Aug28-07, 10:15 PM
P: 27
Quote Quote by learningphysics View Post
No, the pythagorean theorem is only for right triangles... this isn't a right triangle... pythagorean theorem won't work for A-B or A + B.
? But I've used it before to determine the magnitude value of three vectors with their x and y coordinates provided.
learningphysics
#13
Aug28-07, 10:20 PM
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Quote Quote by AraProdieur View Post
? But I've used it before to determine the magnitude value of three vectors with their x and y coordinates provided.
Yes... from the x and y components, you can calculate the magnitude using the pythagorean theorem... That is because the x component is a vector in the x direction... the y component is a vector in the y direction... they are perpendicular and represent a right triangle...

if you calculate the x component (call it x) and y component (call it y) of A+B... then you can use the pythagorean theorem to calculate the magnitude... so maginitude = [tex]\sqrt{x^2 + y^2}[/tex] but this is not equal to [tex]\sqrt{A^2 + B^2}[/tex]

ie: A and B are not the x and y components of A+B...
AraProdieur
#14
Aug28-07, 10:22 PM
P: 27
Quote Quote by learningphysics View Post
Yes... from the x and y components, you can calculate the magnitude using the pythagorean theorem... That is because the x component is a vector in the x direction... the y component is a vector in the y direction... they are perpendicular and represent a right triangle...

if you calculate the x component (call it x) and y component (call it y) of A+B... then you can use the pythagorean theorem to calculate the magnitude... so maginitude = [tex]\sqrt{x^2 + y^2}[/tex] but this is not equal to [tex]\sqrt{A^2 + B^2}[/tex]
Oh, I see, so then how do I find the resultant magnitude of A - B?
learningphysics
#15
Aug28-07, 10:23 PM
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P: 4,124
Have you studied the law of cosines and the law of sines for triangles (non-right triangles)?

EDIT: You can also calculate the x-component of A-B, and the y-component of A-B... then use pythogrean theorem.

Both approaches are good... it's up to you...
AraProdieur
#16
Aug28-07, 10:28 PM
P: 27
Quote Quote by learningphysics View Post
Have you studied the law of cosines and the law of sines for triangles?
Yes, but I'm not very familiar with using it with the vectors.
AraProdieur
#17
Aug28-07, 10:32 PM
P: 27
Quote Quote by learningphysics View Post
Have you studied the law of cosines and the law of sines for triangles (non-right triangles)?

EDIT: You can also calculate the x-component of A-B, and the y-component of A-B... then use pythogrean theorem.

Both approaches are good... it's up to you...
So the x/y component of Vector A would be <11,0>? and the vector component to find -B would be cos(212)= x/22? By which I would then use tan to find the y-component value?
learningphysics
#18
Aug28-07, 10:34 PM
HW Helper
P: 4,124
Quote Quote by AraProdieur View Post
Yes, but I'm not very familiar with using it with the vectors.
Once you see the triangle you need... just approach it like triangles now... at this point don't think about them being vectors...

you've got the triangle made up of |A|, |-B|, and |A-B|... You already know |A|, |-B| and the angle between these two...

So you've got a triangle where you know two sides, and the angle between the two sides... you want the third side... do you see how to use the cosine rule here to calculate the third side?


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