The Student Room Group

Vectors

In Figure 5, vector OA = a and vector OB = b
The point C divides OB in the ratio 1: 3
The point D is the midpoint of AC
(a) Find, as a simplified expression in terms of a and b
The point E is such that OE = x 0A
Given that E, D and B are collinear
(b) find the value of x
Given that (Area triangle OAC)/(Area triangle OEB)= u
(c) find the value of u

I don’t know how to insert an image but I really hope someone can help with (c)
(edited 1 year ago)
Original post by shannonrhamilton
In Figure 5, vector OA = a and vector OB = b
The point C divides OB in the ratio 1: 3
The point D is the midpoint of AC
(a) Find, as a simplified expression in terms of a and b
The point E is such that OE = x 0A
Given that E, D and B are collinear
(b) find the value of x
Given that (Area triangle OAC)/(Area triangle OEB)= u
(c) find the value of u

I don’t know how to insert an image but I really hope someone can help with (c)


Can you put an image of the question / your working on another site and link it?
Alternatively reply below and click on the Camera icon (upload image) on the toolbar.
It keeps saying error uploading, try again later :frown: I’ve tried restarting etc but I don’t know how to fix it
Original post by shannonrhamilton
It keeps saying error uploading, try again later :frown: I’ve tried restarting etc but I don’t know how to fix it


Can you upload to a different site and link or is the question available on line or ...?
Original post by shannonrhamilton
I’ve found a link that has an image of the question, it’s just being approved by the moderation team now.
Thank you


Assuming thats the question, can you give a brief summary of what you did for a) and b)? If the upload has the working as well, dont worry.
Original post by mqb2766
Assuming thats the question, can you give a brief summary of what you did for a) and b)? If the upload has the working as well, dont worry.

For (a) I got:
AC = 1/4b +a
OD = 1/2a + 1/8b
BD= 1/2a - 7/8b

For (b)
I said BE = x(1/2a - 7/8b) because it’s an extension of BD
I also said BE = -b + ha
(h being mew)
This would make -7/8x =-1 because they’re both coefficients of b. Hence x= 8/7
Doing the same for h, h=1/2x, and we know x so h=4/7

I just don’t know how to do c
Original post by shannonrhamilton
For (a) I got:
AC = 1/4b +a
OD = 1/2a + 1/8b
BD= 1/2a - 7/8b

For (b)
I said BE = x(1/2a - 7/8b) because it’s an extension of BD
I also said BE = -b + ha
(h being mew)
This would make -7/8x =-1 because they’re both coefficients of b. Hence x= 8/7
Doing the same for h, h=1/2x, and we know x so h=4/7

I just don’t know how to do c


Guessing the picture ...

Pretty much agree with a and b). There is a sign typo on line 1 in a, but corrected on the next line. For b) youre asked to find x, but you redefine it has h. Agree that the value of h is the required value of x, but it may be confusing to do this?

For c) you know the ratio of two of the sides of the triangles and the angle is common so you can get the area ratio fairly easily just using the usual absin(C)/2 formula (where the a and b are related to the sides)
(edited 1 year ago)
Original post by mqb2766
Guessing the picture ...

Pretty much agree with a and b). There is a sign typo on line 1 in a, but corrected on the next line. For b) youre asked to find x, but you redefine it has h. Agree that the value of h is the required value of x, but it may be confusing to do this?

For c) you know the ratio of two of the sides of the triangles and the angle is common so you can get the area ratio fairly easily just using the usual absin(C)/2 formula (where the a and b are related to the sides)


Hi, im really sorry but do you have the working out for part C and how you got to an answer, i know i am 4 months late but i just got this question for hw and i am so confused on part c, if you could help i'd be rlly grateful. Thanks!
Original post by evsc1234
Hi, im really sorry but do you have the working out for part C and how you got to an answer, i know i am 4 months late but i just got this question for hw and i am so confused on part c, if you could help i'd be rlly grateful. Thanks!


As in the last post on the thread, use the area formula
1/2 ab sin(C)
where C is the common angle and a and b are the corresponding sides for each of the two triangles.
Reply 10
Original post by flo23323
Hi, im really sorry but do you have the working out for part C and how you got to an answer, i know i am 4 months late but i just got this question for hw and i am so confused on part c, if you could help i'd be rlly grateful. Thanks!

You posted this 4 months ago in #9 and I replied in #10. Use the sin area formula with a common angle and the fact you know the two side ratios
Reply 11
Original post by mqb2766
You posted this 4 months ago in #9 and I replied in #10. Use the sin area formula with a common angle and the fact you know the two side ratios

I literally have no clue how I reposted that comment. Sorry.

Quick Reply