M1 vector question. Watch

Core
Badges: 6
Rep:
?
#1
Report Thread starter 7 years ago
#1
Hey this question has taken me awhile and i still haven't solved it, ive tried finding a connection between each point and still nothing, heres the question plus its diagram associated with it.

OABC is a square. M is the mid-point of OA, Q divides BC in the ratio 1:3. AP and MQ meet at P. If OA =a and OC= c, express OP in terms of a and c.
Attached files
0
reply
In One Ear
Badges: 17
Rep:
?
#2
Report 7 years ago
#2
Hey man, i don't study m1 so i have no idea of the techniques you may use. However i came to the conclusion OP(-->)= 2/5a + 3/5c.

How i did this?

Well i set up a pair of axis and created a theoretical square of sides 5 so that the vertices are (0,0), (0,5), (5,5) and (5,0). I then set up equations for the lines MQ and AC, being;

MQ=1/4x+2.5
AC=-x+5

Which allowed me to find the point of intersection (P) which is (2,3). That is to say, 2/5 of vector a, and 3/5 of vector c. Inspection of the sketch that you provided suggests that if this is not correct (which i believe it is) then it is at least very close to the truth.

Sorry i couldn't help more, im only studying c1-c2 and d1, s1, s2 atm. Goodluck with the question!
0
reply
Core
Badges: 6
Rep:
?
#3
Report Thread starter 7 years ago
#3
(Original post by In One Ear)
Hey man, i don't study m1 so i have no idea of the techniques you may use. However i came to the conclusion OP(-->)= 2/5a + 3/5c.

How i did this?

Well i set up a pair of axis and created a theoretical square of sides 5 so that the vertices are (0,0), (0,5), (5,5) and (5,0). I then set up equations for the lines MQ and AC, being;

MQ=1/4x+2.5
AC=-x+5

Which allowed me to find the point of intersection (P) which is (2,3). That is to say, 2/5 of vector a, and 3/5 of vector c. Inspection of the sketch that you provided suggests that if this is not correct (which i believe it is) then it is at least very close to the truth.

Sorry i couldn't help more, im only studying c1-c2 and d1, s1, s2 atm. Goodluck with the question!
Its cool how you were able to apply or knowledge of those subjects to an m1 question without any experience in that unit.

I am however unaware of how you come about your coordinates, and this prevents me from understanding your explanation beyond this point, please explain how you found these coordinates.

Also why did you choose 5?
0
reply
Core
Badges: 6
Rep:
?
#4
Report Thread starter 7 years ago
#4
I understand the coordinates now, Q is 3/4 of a, therefore it is 3/4 of 5, when x =0, y=2.5=1/2a=M
0
reply
In One Ear
Badges: 17
Rep:
?
#5
Report 7 years ago
#5
5? Well idk :P. Reminds me of equations of circles in c1 where 5 came in a lot of questions as it was a practical number. I could have chosen any value, though for the purposes of manipulation a smaller number is easier .

Consequently once ive chosen to create this imaginary square of sides 5, i want to place its lower right corner onto the origin of my axis (0,0). From this point its probably easy to see that the lower right corner must be (0,0+5)=(0,0), the top right corner (0+5,0+5)=(5,5) and so on.

I could set up my equations because i knew M was the midpoint of OA and therefore had coordinates of (0,2.5) and Q split BC 1:3, so was 3/4 of CB up from C so had the coordinates (5,3.75). I used simultaneous equations to find the point of intersection.

Hope this helps . Do you have the answer btw? And if you do, was i right?

I should say, rather than pick an arbitrary side length 5, i could have solved this algebraicly, calling the side length of the square z or something. However, when i was younger i always used theoretical numbers to help me visualise the solution when i hadn't a learnt technique to apply, i guess i havn't really gotten out of that habbit yet. Some numbers to reason with quantitively is still easier for me when deciding if my logic has been correct.
0
reply
Core
Badges: 6
Rep:
?
#6
Report Thread starter 7 years ago
#6
(Original post by In One Ear)
5? Well idk :P. Reminds me of equations of circles in c1 where 5 came in a lot of questions as it was a practical number. I could have chosen any value, though for the purposes of manipulation a smaller number is easier .

Consequently once ive chosen to create this imaginary square of sides 5, i want to place its lower right corner onto the origin of my axis (0,0). From this point its probably easy to see that the lower right corner must be (0,0+5)=(0,0), the top right corner (0+5,0+5)=(5,5) and so on.

I could set up my equations because i knew M was the midpoint of OA and therefore had coordinates of (0,2.5) and Q split BC 1:3, so was 3/4 of CB up from C so had the coordinates (5,3.75). I used simultaneous equations to find the point of intersection.

Hope this helps . Do you have the answer btw? And if you do, was i right?

I should say, rather than pick an arbitrary side length 5, i could have solved this algebraicly, calling the side length of the square z or something. However, when i was younger i always used theoretical numbers to help me visualise the solution when i hadn't a learnt technique to apply, i guess i havn't really gotten out of that habbit yet. Some numbers to reason with quantitively is still easier for me when deciding if my logic has been correct.

Your creative with your methods of solving problems, are very comfortable in the knowledge of your units? I was not not able to understand your second equation AC=-x+5
0
reply
In One Ear
Badges: 17
Rep:
?
#7
Report 7 years ago
#7
Well to me it was obvious the line was the same as x=y but reflected in the y axis and moved up 5.

To do it by the book; use grad of line= difference in ys/ difference in xs and use the the y and x intercepts as known points. This gives (0-5)-(5-0)=-1. You can see that the y intercept is 5 just by looking at the graph, but otherwise, when x=0, y=0+c, or 5=0+c, c=5 (form y=mx+c).

I don't need to be comfortable with units here . Usually my near absolute lack of revision leads me into a pickle with things like units in physics exams, but luckily many questions in physics exams can also be worked out with thinking styles similar to this. Over the past few days though i've decided to get my act together with regards to my ASs so i've been doing some maths revision . I come on here to see if i can help people or learn some neat tricks in observing others.

EDIT: Can i ask, if you have the answer, what is it?
0
reply
Core
Badges: 6
Rep:
?
#8
Report Thread starter 7 years ago
#8
 \frac{3}{5}a+\frac{2}{5}c
0
reply
In One Ear
Badges: 17
Rep:
?
#9
Report 7 years ago
#9
(Original post by Core)
 \frac{3}{5}a+\frac{2}{5}c
Huh?
0
reply
Core
Badges: 6
Rep:
?
#10
Report Thread starter 7 years ago
#10
(Original post by In One Ear)
Huh?
this is the answer in vector form, 3/5 vector a, 2/5 vector b.
0
reply
In One Ear
Badges: 17
Rep:
?
#11
Report 7 years ago
#11
Ah yes ok, sorry when i quoted saying 'huh' it was showing some weird half number :P.

And yes this is correct, and indeed what i meant to say! I accidently put the wrong coordinate to each vector. For some reason i thought a was the OC :P.

Can you see how i got the the answers now though?

Was this an exam question? And if so, worth how many marks? I have a feeling that there is probably a much simpler way of doing this...maybe someone else could step in and explain?
0
reply
Core
Badges: 6
Rep:
?
#12
Report Thread starter 7 years ago
#12
No I cant, i mean i can understand how what you did got you two those numbers, but how you were able to formulate it in relation to the question is my main issue.
0
reply
In One Ear
Badges: 17
Rep:
?
#13
Report 7 years ago
#13
Well this is a skill hard to teach unfortunately. In terms of passing an exam, a good way to learn this skill for this style of question would be just to do loads of them, but then you are pretty much in quick sand as soon as the style changes.

Maybe we can continue this in personal message? If you have another similar problem, you can try solving it as i have solved this one, and if you can't then send me as far as you have gotten, and i can try and show you how to set it up for the rest. Alternatively i could create a similar problem for you to this one, and see how far you can get solving it.

I'm going to bed now because its getting really late, so cya tomorrow maybe .
0
reply
Core
Badges: 6
Rep:
?
#14
Report Thread starter 7 years ago
#14
(Original post by In One Ear)
Well this is a skill hard to teach unfortunately. In terms of passing an exam, a good way to learn this skill for this style of question would be just to do loads of them, but then you are pretty much in quick sand as soon as the style changes.

Maybe we can continue this in personal message? If you have another similar problem, you can try solving it as i have solved this one, and if you can't then send me as far as you have gotten, and i can try and show you how to set it up for the rest. Alternatively i could create a similar problem for you to this one, and see how far you can get solving it.

I'm going to bed now because its getting really late, so cya tomorrow maybe .
Ok thank you for your help.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Cranfield University
    Cranfield Forensic MSc Programme Open Day Postgraduate
    Thu, 25 Apr '19
  • University of the Arts London
    Open day: MA Footwear and MA Fashion Artefact Postgraduate
    Thu, 25 Apr '19
  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (427)
37.85%
No - but I will (87)
7.71%
No - I don't want to (76)
6.74%
No - I can't vote (<18, not in UK, etc) (538)
47.7%

Watched Threads

View All