URGENT: jane1234's Specialist Questions...

[jane1234]

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. 

[brightsky]

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.

[jane1234]

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...

[brightsky]

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.

[jane1234]

Okay, I think I get it now Thanks heaps!

[jane1234]

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.

[xZero]

hmm you sure everything else is right? your solution to part c) and the answer to part d) doesnt match up

[jane1234]

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?

[xZero]

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

[xZero]

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

[jane1234]

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??

[xZero]

it should be gradient vector aka directional vector, my bad.

[jane1234]

it should be gradient vector aka directional vector, my bad.

Ah that makes sense... thanks heaps for all your help

[jane1234]

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

[b^3]

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).