I with this question about vectors

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The discussion revolves around solving a vector problem involving two boats, P and Q, with specific initial positions and velocities. Participants emphasize the importance of using vector diagrams to clarify the problem, particularly in understanding the direction "northeast." The calculations for the positions of both boats over time are discussed, with a focus on determining when Q is northeast of P. There are multiple requests for clarification and corrections regarding arithmetic mistakes in the calculations, highlighting the complexity of the problem. Ultimately, the resolution hinges on correctly expressing the relative positions of the boats as functions of time to find the desired intersection point.
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Homework Statement


At noon two boats P and Q have a position vector (i+7j)km and (3i+8j)km respectively to the origin O.
i and j are unit vectors in direction to East and north respectively
P is moving south east at 5√2 km/h and Q is moving at a constant velocity of (6i+5j) km/h
At time t after noon the position vectors of P and Q with respect to O are p km and Q km
a) show that the velocity of P is ( 5i + 5j)
b) find the time to the nearest minute when Q is north east of P

Homework Equations


Phythogorus theorum a^2 = b^2 + c^2

The Attempt at a Solution


I have done part a)
Where if the velocity is (5i+5j), then if you use phythogorus theorum where √(5^2+5^2) I get 5√2

I don't know how to do part b)
 
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Have you drawn a diagram?
 
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lioric said:

Homework Statement


At noon two boats P and Q have a position vector (i+7j)km and (3i+8j)km respectively to the origin O.
i and j are unit vectors in direction to East and north respectively
P is moving south east at 5√2 km/h and Q is moving at a constant velocity of (6i+5j) km/h
At time t after noon the position vectors of P and Q with respect to O are p km and Q km
a) show that the velocity of P is ( 5i + 5j)
The red quotes seem inconsistent.
 
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PeroK said:
Have you drawn a diagram?
Yeah. What he said.

@lioric discussing this kind of problem without a vector diagram is a waste of time.
 
phinds said:
Yeah. What he said.

@lioric discussing this kind of problem without a vector diagram is a waste of time.
Yes I have drawn a diagram
 
lioric said:
Yes I have drawn a diagram
Uh ... and do you plan to share it with US? You ARE asking us to discuss it after all.
 
But my main problem is figuring out the wording of the question
Is the question asking me to find the time when Q is at a bearing of 45 degrees of P?
 
lioric said:
But my main problem is figuring out the wording of the question
Is the question asking me to find the time when Q is at a bearing of 45 degrees of P?
What does "North East" mean?
 
phinds said:
What does "North East" mean?
East is 90 degrees from North
North east is the bisection of that
It's the line equidistant from the East and the north
The angle is 45 degrees from north
 
  • #10
lioric said:
East is 90 degrees from North
North east is the bisection of that
It's the line equidistant from the East and the north
The angle is 45 degrees from north
There are two lines that are 45 degrees from North. Yes, I DO know what you mean but my point is that without a vector diagram, again, this discussion is useless.
 
  • #11
@lioric: Are you ever going to respond to post #3 with a correct statement of the problem?
 
  • #12
LCKurtz said:
@lioric: Are you ever going to respond to post #3 with a correct statement of the problem?
AND a vector diagram !
 
  • #13
LCKurtz said:
@lioric: Are you ever going to respond to post #3 with a correct statement of the problem?
What can I say about what the red quotes say? It's what the question said. Word to word
 
  • #14
phinds said:
AND a vector diagram !
I have drawn the diagram
And I can solve the problem when I'm sure about the north east part
I just need to know the exact meaning of that
Just clearing my doubts
 
  • #15
lioric said:
What can I say about what the red quotes say? It's what the question said. Word to word
Do you understand that the direction and vector I quoted in red do not agree?
 
  • #16
LCKurtz said:
Do you understand that the direction and vector I quoted in red do not agree?
He might have figured that out if he had done a correct vector diagram, which is one reason I keep telling him to do one (AND show it to us so we can help)
 
  • #17
Sorry that's a typo
It's 5i-5j
Sorry
 
  • #18
At noon two boats P and Q have a position vector (i+7j)km and (3i-8j)km respectively to the origin O.
i and j are unit vectors in direction to East and north respectively
P is moving south east at 5√2 km/h and Q is moving at a constant velocity of (6i+5j) km/h
At time t after noon the position vectors of P and Q with respect to O are p km and Q km
a) show that the velocity of P is ( 5i - 5j)
b) find the time to the nearest minute when Q is north east of P

This is the correct one
 
  • #19
IMG_20171225_171312.jpg
 

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  • #20
lioric said:
Sorry that's a typo
It's 5i-5j
Sorry
That and the now changed original position of Q. So now you can show us how you calculate Q and P's positions at time t. Then think about what must be true about Q-P for Q to be northeast of P.
 
  • #21
LCKurtz said:
That and the now changed original position of Q. So now you can show us how you calculate Q and P's positions at time t. Then think about what must be true about Q-P for Q to be northeast of P.
Q must be at a 45 degrees bearing to P
 
  • #22
I posted: "So now you can show us how you calculate Q and P's positions at time t". Which you ignored.
 
  • #23
LCKurtz said:
I posted: "So now you can show us how you calculate Q and P's positions at time t". Which you ignored.
We use r=r0 + vt
 
  • #24
lioric said:
We use r=r0 + vt
LCKurtz's question was more specific than this. In other words, what were your calculations for Q and P at an arbitrary time t?
 
  • #25
Mark44 said:
LCKurtz's question was more specific than this. In other words, what were your calculations for Q and P at an arbitrary time t?
Let's say at time t the position of P is p and the position of Q is q so in terms of p and q and t I can give the new position of P as
r=r0+vt
= (i+7j) + (5i-5j)t
=i+7j + 5it-5jt
=(1+5t)i + (7+5t)j

And new position for Q as
r=r0+vt
=(3i-8j) + (6i+5j)t
=3i-8j + 6it+5jt
=(3+6t)i + (5t-8)j
 
  • #26
Is there some reason you don't simplify those? Anyway, now you have them, what next? There was more in post #20. Tell us what you are thinking or why you are stuck.
 
  • #27
LCKurtz said:
Is there some reason you don't simplify those? Anyway, now you have them, what next? There was more in post #20. Tell us what you are thinking or why you are stuck.
Well you can see how I have drawn the vector diagram I drew it true to the i j to x y vector coordinates. I just don't know how to go about the question. Some hints would be helpful
 
  • #28
You have had some hints. In post #20 I asked you about Q - P. Have you even calculated that yet? Are you going to show us what you get? Have you thought about whether it points Northeast? Or how you would tell? Neither I nor anyone else on this board is going to work it for you.
 
  • #29
You don't need any diagrams. The key is to subtract the vector velocity and position of P from Q. This makes P the new origin and gives you a new location and new vector velocity for Q. You can now use those velocity components to figure out when the two position components will be equal which is when the position of Q is north-east of P.
 
  • #30
jimkris69 said:
You don't need any diagrams. The key is to subtract the vector velocity and position of P from Q. This makes P the new origin and gives you a new location and new vector velocity for Q. You can now use those
velocity components to figure out when the two position components will be equal which is when the position of Q is north-east of P.
Thank you all for your help
So the position of
P = (i+7j) with v= (5i-5j)
Q=(3i-8j) with v=(6i+5j)
Already Q is on the east side of P and the rate at with Q moves in x-axis is faster than P
So therefore Q will always be on the east side of P
When we look at the y-axis it seems that P is moving down at the same rate as Q is moving up
So the moment these two meet in y plan would be the time that Q is exactly on the east of P
So using speed = distance /time
Distance= 7-(-8)= 15
To find mid position 15/2 =7.5
Speed = 5
Time=distance/speed
=7.5/5=1.5hrs
At this time the two ships will be is line in the
With a gap in the x axis
Anything more than 1.5 hours and Q will be in the NE section of P

How's this?
Ok?
 
  • #31
How's that? Not good. You haven't even used ##t##.

If you had actually read and responded to my posts you would have had this problem solved a week ago. You haven't simplified your answer in post #25 or responded to post #28. Until you do you will hear nothing more from me.
 
  • #32
LCKurtz said:
How's that? Not good. You haven't even used ##t##.

If you had actually read and responded to my posts you would have had this problem solved a week ago. You haven't simplified your answer in post #25 or responded to post #28. Until you do you will hear nothing more from me.
I'm sorry about that
Was away for some time and no internet there
But I'll start working on this now.
Ok so simplifying the new position of P
(1+5t)i + (7+5t)j
P= i+5ti + 7j+5tj
New position of Q
(3+6t)i + (5t-8)j
Q= 3i+6ti + 5tj-8j

Q-P
(3i-8j)-(i+7j)
=(2i-15j)
 
  • #33
Make sure you are using the correct numbers as given in post #18.
When you "simplify" a vector it should look like (...)i + (...)j, one i term and one j term.
Your final answer for Q-P at time t must have a "t" in it.
Then we can talk about whether it points NE.
 
  • #34
Post 18

At noon two boats P and Q have a position vector (i+7j)km and (3i-8j)km respectively to the origin O. i and j are unit vectors in direction to East and north respectively P is moving south east at 5√2 km/h and Q is moving at a constant velocity of (6i+5j) km/h
 
  • #35
LCKurtz said:
Make sure you are using the correct numbers as given in post #18.
When you "simplify" a vector it should look like (...)i + (...)j, one i term and one j term.
Your final answer for Q-P at time t must have a "t" in it.
Then we can talk about whether it points NE.
I m very confused
In post #25 it seems that the vectors are in the format that you asked (...)i+(...)j and you asked me to simply them. So I expanded them and now your asking me to revert them back again.
 
  • #36
Are you asking me to subtract the new coordinates of P and Q or the initial coordinates given in the question?
 
  • #37
lioric said:
I m very confused
In post #25 it seems that the vectors are in the format that you asked (...)i+(...)j and you asked me to simply them. So I expanded them and now your asking me to revert them back again.
OK, never mind about the i and j comments.
This thread is getting very jumbled. There are multiple arithmetic errors and/or typos. In post #25 you have an arithmetic mistake in the new position of P. Then in post #32 you carried the same mistake through. Also in post #32 you dropped the ##t## in the calculation of Q-P, as I told you in post #33. Fix those. Once you finally get Q-P correct, with a ##t## in it, it is a simple one more step to finish your problem.
 
  • #38
LCKurtz said:
OK, never mind about the i and j comments.
This thread is getting very jumbled. There are multiple arithmetic errors and/or typos. In post #25 you have an arithmetic mistake in the new position of P. Then in post #32 you carried the same mistake through. Also in post #32 you dropped the ##t## in the calculation of Q-P, as I told you in post #33. Fix those. Once you finally get Q-P correct, with a ##t## in it, it is a simple one more step to finish your problem.
That's what I was clarifying in post 36
The initial P and Q position vectors given in the question has no t in it
So which P and Q are you referring to?
Is it the answers on post 25?
 
  • #39
lioric said:
That's what I was clarifying in post 36
The initial P and Q position vectors given in the question has no t in it
So which P and Q are you referring to?
Is it the answers on post 25?
Think about the problem you are trying to solve. You are given original positions P and Q and their velocities. They are moving. Then you are asked at what time ##t## is Q northeast of P. So you need their positions at time ##t## don't you? So you need to know for what ##t## does Q-P point northeast. Fix your arithmetic and show Q-P as a function of ##t## and you are almost done.
 
  • #40
LCKurtz said:
Think about the problem you are trying to solve. You are given original positions P and Q and their velocities. They are moving. Then you are asked at what time ##t## is Q northeast of P. So you need their positions at time ##t## don't you? So you need to know for what ##t## does Q-P point northeast. Fix your arithmetic and show Q-P as a function of ##t## and you are almost done.

Ok
Let's think here
I'm just trying to clear my confusion please do tell if I'm thinking right

Let's keep the vector situation and go to a simpler linear one
Let's say P is on (1,1) on a grid
And Q is on ( 5,-5) on the same grid
P is moving on positive x-axis (to the right) at let's say 1 unit/s
And Q is moving in the positive y-axis (upward) 1 unit/s
Let's say the question is asking to find the time that P moves east of Q

Is this the same or similar sort of thing?
 
  • #41
lioric said:
Ok
Let's think here
I'm just trying to clear my confusion please do tell if I'm thinking right

Let's keep the vector situation and go to a simpler linear one
Let's say P is on (1,1) on a grid
And Q is on ( 5,-5) on the same grid
P is moving on positive x-axis (to the right) at let's say 1 unit/s
And Q is moving in the positive y-axis (upward) 1 unit/s
Let's say the question is asking to find the time that P moves east of Q

Is this the same or similar sort of thing?
I am not going to go off on another tangent with you. You have understood the problem well enough to draw a picture long ago in post #19. Just finish the problem as I have suggested to you.
 
  • #42
LCKurtz said:
I am not going to go off on another tangent with you. You have understood the problem well enough to draw a picture long ago in post #19. Just finish the problem as I have suggested to you.

If I understood I wouldn't ask

But from what I have gathered and from the things you have asked me to do this is what I came up with
P initial position (i +7j) velocity (5i-5j)
Q initial position (3i-8j) velocity (6i+5j)

P new position
(i +7j)+(5i-5j)t.
(1+5t)i + (7-5t)j

Q new position
(3i-8j)+(6i+5j)t
(3+6t)i +(5t-8)j

Now Q-P

(3+6t)i + (5t-8)j - [ (1+5t)i + (7-5t)j ]
(3+6t)i + (5t-8)j - (1+5t)i - (7-5t)j
(3+6t)i + (5t-8)j +(-1-5t)i +(-7+5t)j
(3-1+6t-5t)i + (5t+5t-8-7)j
(2+t)i + (10t-15)j

Is this what you asked?
 
  • #43
Yes. Finally! Now, for what ##t## does that point northeast? What do you require about the components for a vector to point northeast?
 
  • #44
LCKurtz said:
Yes. Finally! Now, for what ##t## does that point northeast? What do you require about the components for a vector to point northeast?
It should point to somewhere in the first quadrant meaning it should not have a bearing exceeding 90 degrees
 
  • #45
I would assume in this problem that the meaning of northeast means exactly northeast, that is midway between north and east.
 
  • #46
LCKurtz said:
I would assume in this problem that the meaning of northeast means exactly northeast, that is midway between north and east.
So that means as I said before bearing of 45 degrees
 
  • #47
lioric said:
So that means as I said before bearing of 45 degrees
Yes. And the question asks at what time does that happen? So...
 
  • #48
LCKurtz said:
Yes. And the question asks at what time does that happen? So...

We have to find the time taken to make 45 degrees between P and Q
But I don't know how
 
  • #49
Do you know how to tell if any vector points NE? Which, if any, of the following vectors point NE?
3i - 5j
2i +3j
-i - 2j
3i + 3j
i - j
-2i + 2j
Explain how you can tell.
 
  • #50
LCKurtz said:
Do you know how to tell if any vector points NE? Which, if any, of the following vectors point NE?
3i - 5j
2i +3j
-i - 2j
3i + 3j
i - j
-2i + 2j
Explain how you can tell.
Ya the 3i+3j
It means the position is 3,3 on an x,y axis
The gradient of that would be 1 and that gives a 45 degrees
 

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