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)

Reply 1

mqb2766

21

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.

Reply 2

shannonrhamilton

OP

2

It keeps saying error uploading, try again later :frown:

I’ve tried restarting etc but I don’t know how to fix it

Reply 3

mqb2766

21

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 ...?

Reply 4

shannonrhamilton

OP

2

Original post by mqb2766

Can you upload to a different site and link or is the question available on line

I really hope this link works
https://www.bartleby.com/questions-and-answers/figure-5-in-figure-5-oa-a-and-ob-b-the-point-c-divides-ob-in-the-ratio-13-the-point-d-is-the-midpoin/8016bf97-5534-472b-b055-8ef1674fe045

Reply 5

mqb2766

21

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.

Reply 6

shannonrhamilton

OP

2

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

Reply 7

mqb2766

21

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)

Reply 8

flo23323

4

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!

Reply 9

mqb2766

21

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

mqb2766

21

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!