Vector problems

[Patrick 22]

Hi all, I need some help on some vector questions.

does anyone have any idea about how to go about answering these questions. I’m totally stuck

Many thanks

 

[Deleted member 4993]

Hi all, I need some help on some vector questions.

does anyone have any idea about how to go about answering these questions. I’m totally stuck

Many thanks

Hint:

Assuming the grid is made of parallelograms:

2p = OA + AB = 2 * OA

4q = 4 * OF

continue....

 

[Steven G]

Hint:

Assuming the grid is made of parallelograms:

2p = OA + OB = 2 * OA

4q = 4 * OF

continue....

I am sorry but I do not agree with your hint--corner time!

I suspect that you meant that 2p = OA + AB = 2*OA. However I do not agree with that either--more corner time.

Looking at the diagram it looks to me that p = .5*OA or 2p = OA.

 

[Otis]

… I suspect that you meant that 2p = OA + AB = 2*OA. However I do not agree with that either … [on] the diagram it looks to me that p = .5*OA …

Hi Jomo. I agree that Subhotosh meant to type OA+AB, but I'm going with Subhotosh's interpretation that OA is the horizontal unit-vector and OF is the diagonal unit-vector. Otherwise, obtaining those vectors' magnitudes from the diagram would be subjective. For example, I see p=0.60·OA and q=0.75·OF.

I'm puzzled about the apostrophe, where they wrote the vector HC'.

PS: I'm not sure, but the author may have used a different notation (where the arrowhead is centered on the vector) because it appears that the entire grid line OA is bold. Same for OF. If that's not a different notation, then the diagram is just sloppy.

?

 

[Otis]

… I need some help on some vector questions …

Hi Patrick. Your answer for 6(a) needs fixing. You wrote that vector -p is the same as HF, and that vector p is the same as OC. Those are not correct. You need to add multiples of vectors p and q.

?

 

[pka]

Hi all, I need some help on some vector questions.

does anyone have any idea about how to go about answering these questions. I’m totally stuck

Many thanks

View attachment 17614

8a) \(\vec{HC}=\vec{p}-\vec{q}\)

8b) \(2\vec{p}-4\vec{q}=\vec{SB}\)

I am too lazy to do 8c).

 

[Otis]

8b) … = SB … I am too lazy to do 8c).

That's 6(b) and 6(c), and there is no SB on the diagram. ? Try the zoom control.

As an alternative to completing the assignment, you could have explained something about the answers you did post (if you're not too lazy for that). Patrick said he's "totally stuck".

\(\;\)

 

[Steven G]

Hi Jomo. I agree that Subhotosh meant to type OA+AB, but I'm going with Subhotosh's interpretation that OA is the horizontal unit-vector and OF is the diagonal unit-vector. Otherwise, obtaining those vectors' magnitudes from the diagram would be subjective. For example, I see p=0.60·OA and q=0.75·OF.

I'm puzzled about the apostrophe, where they wrote the vector HC'.

PS: I'm not sure, but the author may have used a different notation (where the arrowhead is centered on the vector) because it appears that the entire grid line OA is bold. Same for OF. If that's not a different notation, then the diagram is just sloppy.

?

I too thought that p was more than .5*OA but certainly not OA. After enlarging the image I know see that OA is in bold and yes p = OA

 

[pka]

That's 6(b) and 6(c), and there is no SB on the diagram. ? Try the zoom control.

Otis are you using a normal size screen, more then a cell phone or small tablet?
There are clearly both S & B in the grid. Moving from S to B is two to the right, \(2\vec{p}\), and four down, \(-4\vec{q}\), going from S to B in the grid.
Of course you may not be accustomed to using vectors as equivalence classes.

 

[Otis]

… There are clearly both S & B in the grid …

I stand corrected, thank you.

… using vectors as equivalence classes.

I don't know what that means, so I don't know whether I'm accustomed to it.

?

 

[pka]

I don't know what that means, so I don't know whether I'm accustomed to it.

In the grid to move from \(O\) to \(W\) we must move four to the right and four up.
So \(4\vec{p}+4\vec{q}=\vec{OW}\). But if asked, moving only right or up how many different paths are the from \(O\) to \(B\) then the answer is \(\dfrac{8!}{(4!)^2}\)