ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: jane1234 on July 08, 2011, 01:17:33 pm
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Question: Use a vector method to prove that the diagonals of a rhombus intersect at right angles.
Could someone please explain how you do this?
Thanks guys. :)
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let one side of the rhombus be a and an adjacent side be b such that the first diagonal is a+b.
we know from rhombus properties that the diagonals bisect each other, so the second diagonal can be given by 1/2(a+b) - a = 1/2b - 1/2a = 1/2(b - a).
(a+b).1/2(b-a) = 1/2(b.b -a.a) = 1/2(|b|^2 - |a|^2). since |b| = |a|, then this equals 0, completing the proof.
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Hmm ... get most of it but could you say that the second diagonal is simply a - b (because the vector opposite the a will be equivalent to a)? Or is that completely wrong?
I don't get how you found the second diagonal... ???
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oh woops...haha that was stupid of me. yes that's much easier, i was remembering the proof i did for a kite, which doesn't have parallel sides. but this one is just (a+b).(a-b) =0.
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Okay, I think I get it now :D Thanks heaps!
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Hey guys... another vector question. :( But it's not a proof (I think I finally get them haha)!
A radar station tracks a jet fighter flying with constant speed. If the radar station is considered to be at the origin, the fighter's starting position is 2i + 8j + k and 1 minute later it is at 8i - 4j + 13k. The units are in kms.
a) State the vector which indicates the path of the fighter. Got this, it's 6i - 12j + 12k.
b) State a unit vector in the direction of this path. Got this, it's 1/3(i - 2j + 2k).
c) Find a vector, in terms of m, which represents the position of the fighter at any time along the path. Got this, it's m/3(i - 2j + 2k).
d) Find the point along the path where the fighter is closest to the station (the origin).
Answer is 1/3(10i + 16j +11k).
No idea how to do part d) !! Any help appreciated. :)
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hmm you sure everything else is right? your solution to part c) and the answer to part d) doesnt match up
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Yes those are the answers from the back of the book as well. No idea how they got d) ... :-\
If you didn't know the answer to d) how would you work it out?
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Its basically the same as 'find the closest point on a graph to the origin' questions you see in methods, draw a line from the graph/path to origin, find the magnitude and differentiate it to find the closest point. I'll give the question a go later, still eating lunch :P
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okay so you know the direction of the path, it starts at (2,8,1) and the gradient would be m/3(1,-2,2). This gives us the actual equation of the path (2,8,1)+m/3(1,-2,2) or (2+m/3)i + (8-2m/3)j + (1+2m/3)k. Find the magnitude, diff it and find m.
If you ceebs working it out click this link http://www.wolframalpha.com/input/?i=sqrt%28+%282%2Bm%2F3%29^2+%2B+%288-2m%2F3%29^2+%2B+%281%2B2m%2F3%29^2+%29, it shows the global min value for m
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Why is the gradient considered m/3(1,-2,2)?? For a 2D vector xi + yj the gradient is just y/x... how does it work for 3D??
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it should be gradient vector aka directional vector, my bad.
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it should be gradient vector aka directional vector, my bad.
Ah that makes sense... thanks heaps for all your help :D
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Heffernan 2008 Exam 1:
Question 2:
A box of mass 5kg rests on a rough horizontal floor. Coefficient of friction is 0.1. A boy applies a horizontal dragging force D newtons to the box in an attempt to move it.
a) Find the values of D if the box is not at the point of moving across the floor.
I got D E [0,4.9)
Answers just say D < 4.9
Just wondering who was right...? Can you "apply" a force of a negative value?
Thanks in advance :)
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Heffernan 2008 Exam 1:
Question 2:
A box of mass 5kg rests on a rough horizontal floor. Coefficient of friction is 0.1. A boy applies a horizontal dragging force D newtons to the box in an attempt to move it.
a) Find the values of D if the box is not at the point of moving across the floor.
I got D E [0,4.9)
Answers just say D < 4.9
Just wondering who was right...? Can you "apply" a force of a negative value?
Thanks in advance :)
Logically you can't since it's negative, technically you can since it is a vector, it would just be in the opposite direction. But then they say it is a 'dragging' force, so that would have to be in the one direction. So I'd think you'd be correct, D E [0,4.9).
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Heffernan 2008 Exam 1:
Question 2:
A box of mass 5kg rests on a rough horizontal floor. Coefficient of friction is 0.1. A boy applies a horizontal dragging force D newtons to the box in an attempt to move it.
a) Find the values of D if the box is not at the point of moving across the floor.
I got D E [0,4.9)
Answers just say D < 4.9
Just wondering who was right...? Can you "apply" a force of a negative value?
Thanks in advance :)
Logically you can't since it's negative, technically you can since it is a vector, it would just be in the opposite direction. But then they say it is a 'dragging' force, so that would have to be in the one direction. So I'd think you'd be correct, D E [0,4.9).
Actually, I disagree with both the answers. However, the question is slightly ambiguous in my opinion. The word 'dragging' can mean two totally different things, so I just ignored it. Either way, Heffernan's answers seem wrong to me.
To my knowledge, the answer should be 'D is an element of (-4.9,4.9)\{0}. You will have to first take a specific direction as positive. The magnitude of the friction force has to be larger than the magnitude of the applied force. i.e. the applied force could be in either the positive or the negative direction. Hence, the resistance force (i.e. friction) will oppose this.
e.g. If D = -2. Then friction = 2 and hence, it is not at the point of moving.
If D = +2. Then friction = -2. Thus, it is not at the point of moving.
In order to show that Heffernan is wrong. If D = -5. Friction = 4.9. Hence, the box will move.
Also, this raises another question. If a question states a force is applied, can the value be 0 inclusive? I would think the terms imply that the force is larger than 0. However, I need clarification.
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I would say that the magnitude of force D will have to be between 0 and 4.9, so essentially what Luffy is saying i.e. (-4.9,4.9)
But consdering the question doesn't use vector notation in representing the letter "D" i'm assuming that the question implies that it is the magnitude, which means [0,4.9), but the restrictions that magnitudes have to be >= 0 applies, so technically you can say that D<4.9
Luffy, if it says a force is "applied" it logically can't be 0, because if it was 0, it contradicts the question because there is no force applied, but I personally would put 0 as inclusive to show that it is also not on the point of moving when no force is applied
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Heffernan 2008 Exam 1:
Question 2:
A box of mass 5kg rests on a rough horizontal floor. Coefficient of friction is 0.1. A boy applies a horizontal dragging force D newtons to the box in an attempt to move it.
a) Find the values of D if the box is not at the point of moving across the floor.
I got D E [0,4.9)
Answers just say D < 4.9
Just wondering who was right...? Can you "apply" a force of a negative value?
Thanks in advance :)
Logically you can't since it's negative, technically you can since it is a vector, it would just be in the opposite direction. But then they say it is a 'dragging' force, so that would have to be in the one direction. So I'd think you'd be correct, D E [0,4.9).
Actually, I disagree with both the answers. However, the question is slightly ambiguous in my opinion. The word 'dragging' can mean two totally different things, so I just ignored it. Either way, Heffernan's answers seem wrong to me.
To my knowledge, the answer should be 'D is an element of (-4.9,4.9)\{0}. You will have to first take a specific direction as positive. The magnitude of the friction force has to be larger than the magnitude of the applied force. i.e. the applied force could be in either the positive or the negative direction. Hence, the resistance force (i.e. friction) will oppose this.
e.g. If D = -2. Then friction = 2 and hence, it is not at the point of moving.
If D = +2. Then friction = -2. Thus, it is not at the point of moving.
In order to show that Heffernan is wrong. If D = -5. Friction = 4.9. Hence, the box will move.
Also, this raises another question. If a question states a force is applied, can the value be 0 inclusive? I would think the terms imply that the force is larger than 0. However, I need clarification.
I don't think that would be the answer. If you took the friction to be the positive end, then sure D would be negative, but it wouldn't be both. Let's say the force is applied in the positive direction, then it would be 4.9 before it begins to move. However, stating that D goes in the negative direction then you are still saying that you are applying a negative force. If you were going to be REALLY picky, then you'd probably have to say D E (-4.9,0) OR D E (0,4.9) for your answer to make sense.
But then again, I think the question implies that D is in the "positive" direction, as friction on a stationary surface with one mass is always against the direction of movement, and it is illogical to think that the friction force would ever be greater than the applied force... thus, the force will move in the D direction or not at all, making it silly to have D as the "negative" direction...
^ That probably doesn't make much sense haha but it just seems contradictory to say that D is in both the positive and negative direction at the same time, which is what you are saying with D E (-4.9,4.9)\{0}.
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Jane, it doesn't matter because the question would be talking about magnitudes anyway, not vectors
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@ Jane. Sorry, perhaps I didn't quite get what you meant. But, I'll just try to explain the question from my perspective. Though, I think the answers and the question are implying what Paul just said (i.e. D>=0 is implied and hence heffernan's solutions apply).
Nonetheless, in my opinion, the question is ambiguous and can be taken in many ways.
If you took the friction to be the positive end, then sure D would be negative, but it wouldn't be both. Let's say the force is applied in the positive direction, then it would be 4.9 before it begins to move. However, stating that D goes in the negative direction then you are still saying that you are applying a negative force. If you were going to be REALLY picky, then you'd probably have to say D E (-4.9,0) OR D E (0,4.9) for your answer to make sense.
Think about it this way. The question states that a force is applied, but it doesn't state a direction. Now, if we look at that object. Given that one particular direction is taken as positive, can we state a specific direction in which this force has been applied? Of course we can't. Hence, the force can take both positive and negative values. Yes, you can technically apply a negative force. All forces are technically "applied." It will just be in the direction opposite to which is considered positive.
However, I think Paul's statement about magnitudes is probably what the question was asking. Luckily, VCAA doesn't make such vague questions/answers.
If you were going to be REALLY picky, then you'd probably have to say D E (-4.9,0) OR D E (0,4.9) for your answer to make sense.
Haha. If we stated D E (-4.9,0) OR D E (0,4.9). Remember that the term "or" implies the union symbol. Hence, D E (-4.9,0) OR D E (0,4.9) is the same as D E (-4.9,0) U (0,4.9) which is again, the same as D E (-4.9,4.9)\{0}. So, you essentially just said the same thing as me :P
But then again, I think the question implies that D is in the "positive" direction, as friction on a stationary surface with one mass is always against the direction of movement, and it is illogical to think that the friction force would ever be greater than the applied force... thus, the force will move in the D direction or not at all, making it silly to have D as the "negative" direction...
I know what you mean here and it would be illogical to even set the D direction as negative. But, mathematically speaking, it is possible. Thus, negative values of D should be considered. But, as you said, the question is implying magnitude (i.e. D is considered positive).
^ That probably doesn't make much sense haha but it just seems contradictory to say that D is in both the positive and negative direction at the same time, which is what you are saying with D E (-4.9,4.9)\{0}.
I don't quite get what you meant here. If we have the equation, Area = x^2. That x can only take a single value at any one time. The same applies to D in this situation. The D value would either be positive or negative. Sorry if I confused you earlier, but that was never the point I was making.
Hope I clarified anything. Then again, I could be wrong and apologies, in advance, if I am.
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^ Haha yeah I got where you were coming from it was just that the solution D E (-4.9,0)U(0,4.9) implies that you are assigning D to both the positive and negative directions in the same situation, which can't be true.
I think paul was right in that it was talking about magnitude, and not specifically a vector quantity.
So if D>=0 is "implied", then am I wrong to say D E [0,4.9)?
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^ Haha yeah I got where you were coming from it was just that the solution D E (-4.9,0)U(0,4.9) implies that you are assigning D to both the positive and negative directions in the same situation, which can't be true.
I think paul was right in that it was talking about magnitude, and not specifically a vector quantity.
So if D>=0 is "implied", then am I wrong to say D E [0,4.9)?
Nah, you would be correct. But, I don't think D could equal 0. Hence, it would be D E (0,4.9). Could someone please clarify?
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Well technically people can apply a force of 0 magnitude, they just can't apply a force of negative magnitude... but I don't really know. I think because a magnitude CAN be 0 (whereas a magnitude can't be negative) it's included. But I might be wrong, I don't know. We'll have to wait for someone else to clarify...
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You can't apply a force of negative magnitude you can apply a force in the opposite direction, but it will still be of positive magnitude, that's why I'm trying to say, like I picked the right answer when I first did this question without even thinking about all this, hahah (A) damnn! :P
But I think a force = 0, it just means no force is applied, like I don't think that 0 should be excluded :S
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URGENT! SAC tomorrow!
Hi, I just wanted a bit of clarification on something:
You know the generic incline mechanics questions (ie body of mass x is on a plane of incline y, coefficient mu etc etc etc)?
I was just wondering, when it says "calculate the acceleration of the body", do you write
a = ( xi + 0j )m/s^2
~ ~ ~
OR
a = x m/s^2
OR
a = x m/s^2
~
I've seen it written all three ways...
This is assuming the question does not specify it wants the magnitude of the acceleration. Which option is safest?
Thanks! :D
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I know at times they draw the i vs j graph in the corner to show where they are aimed, but even so I've only ever seen the second one, a being a scalar/magnitude...
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I know at times they draw the i vs j graph in the corner to show where they are aimed, but even so I've only ever seen the second one, a being a scalar/magnitude...
But technically that's wrong - as acceleration is a vector quantity...?
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URGENT! SAC tomorrow!
Hi, I just wanted a bit of clarification on something:
You know the generic incline mechanics questions (ie body of mass x is on a plane of incline y, coefficient mu etc etc etc)?
I was just wondering, when it says "calculate the acceleration of the body", do you write
a = ( xi + 0j )m/s^2
~ ~ ~
OR
a = x m/s^2
OR
a = x m/s^2
~
I've seen it written all three ways...
This is assuming the question does not specify it wants the magnitude of the acceleration. Which option is safest?
Thanks! :D
I always try to write my answer in the first form. The third one is technically incorrect because the tilde indicates a vector, but you are merely giving a magnitude without direction.
I have seen the second one before, but I always like to assume that acceleration is a vector and hence, I put in the i and j components respectively. Though, I think both will be marked correct by VCAA.
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I always try to write my answer in the first form. The third one is technically incorrect because the tilde indicates a vector, but you are merely giving a magnitude without direction.
I have seen the second one before, but I always like to assume that acceleration is a vector and hence, I put in the i and j components respectively. Though, I think both will be marked correct by VCAA.
I would actually use the second one, because, as Luffy would know, I refrain from using vector notation unless absolutely necessary :)