Solved: Vector Problem "Dividing A Line Segment (AB) in 3:4 and AX in 1:2

[manmachine]

Homework Statement

X divides a line segment (AB) in a ratio of 3:4. Y divides line segment AX in a ratio of 1:2.
If K is any point of the line containing A,B,X and Y; Please show that AY= 1/7OB - 1/7OA

Homework Equations

Not sure if any. Just adding and subtracting vectors.

The Attempt at a Solution

Alright so I know that 1/4 AB +BO would equal XO, but i don't know what do next. Thanks for the help.

 

[tiny-tim]

hi manmachine!

X divides a line segment (AB) in a ratio of 3:4. Y divides line segment AX in a ratio of 1:2.
If K is any point of the line containing A,B,X and Y; Please show that AY= 1/7OB - 1/7OA

do you mean "If O is any point of the line containing A,B,X and Y; Please show that AY= 1/7OB - 1/7OA" ?

hint: what is OB - OA ? ​

 

[106267]

This has nothing to do with the question but how do i ask question on this forum?

 

[tiny-tim]

welcome to pf!

hi 106267! welcome to pf!

… how do i ask question on this forum?

here's something I've been drafting … (some of the opinions are my own)​

How do I ask a question?

You start a new thread.

Do not send a Private Message, do not send a Visitors Message, do not create a Library entry, do not add it to someone else's thread!

Find the appropriate forum index page, and click the "NEW TOPIC" button at the top of the page.

It does not have to be a new topic for the forum … it is almost always better to start your own thread than to add to someone else's.

Homework questions:

Homework questions and homework-style questions always belong in the homework forums: https://www.physicsforums.com/forumdisplay.php?f=152

In the homework forums, the following "template" will always appear:

Homework Statement

Homework Equations

The Attempt at a Solution

DO NOT DELETE IT! DO NOT IGNORE IT! ​

It is there to help you, and to help others to help you … in particular, you should use it to make it clear what you do know, since we usually have no idea how advanced you are.

btw, there is a software glitch which sometimes results in more than one copy of the template appearing … sorry, but we can't stop that … just delete all except one!

Images:

Please type your work, rather than post an image of it.

Your handwriting is not as easy to read as your typing.

If you want other members to spend their time on helping you, please save some of it by spending some of yours on typing!

And you're more likely to get an answer if we don't have to keep toggling between different windows to understand you!

Images are useful in some problems, particularly in mechanics and electric circuits.

Do not post images that are wider than the usual page width, that is extremely annoying!

(If you can't post an image, then just write the link to it.)

Tags

Do not add tags, they will not help other members to find your question and answer it.

(If the answers are useful, then it may be helpful to add a tag)

Whole question?

When in doubt, post more of the original question rather than less.

It can waste a lot of time if you leave out some information which turns out to be useful.

Also, if you know the answer, it helps to post that too!

Editing:

If you notice a mistake after posting your question (except in the title), you can edit it … guess which button that is!

Do not make a reply to correct it … that takes your question off the "No replies" list, and makes it less likely that you will be answered!

When can I expect an answer?

You may "bump" the thread after about 24 hours.

Do not expect an early reply if you post when most people in America and Europe are asleep! :zzz:

Can I expect expect a full answer?

Noooo …

we do not do your work for you, we help you to do it yourself …

that makes sure that you understand it, and that you will be able to do it in the exam!

It also prevents cheating.

"SOLVED":

We do not have a "SOLVED" tag. Your last post will usually make it clear that you no longer need further help.

You can of course always re-open a thread, if you change your mind!

If you solve the question yourself, please say so, even if nobody has replied!

 

[manmachine]

It is AB? I don't know what to do next!

 

[tiny-tim]

… show that AY= 1/7OB - 1/7OA

hint: what is OB - OA ? ​

It is AB?

yes!

ok, so now all you need to prove is AY = 1/7 AB

 

[manmachine]

And then what next?? I know, I know, I know I ask so many questions. Shame on me. Had surgery so couldn't come to many lectures.

 

[tiny-tim]

X divides a line segment (AB) in a ratio of 3:4. Y divides line segment AX in a ratio of 1:2.

write that in vector equations …

what do you get? ​

 

[manmachine]

Ab= ax+ xy?

 

[tiny-tim]

uhh?

there's two equations

try this one first …

X divides a line segment (AB) in a ratio of 3:4​

 

[manmachine]

So AX+XB=AB?

 

[tiny-tim]

So AX+XB=AB?

well, yes, but that is true wherever X is, isn't it?

you need an equation with 3 and 4 (or 7) in !​

 

[manmachine]

How can I do that? What is it? I am so bad at this :P

 

[HallsofIvy]

First, in your original post you said "If K is any point of the line containing A,B,X and Y; Please show that AY= 1/7OB - 1/7OA" which makes no sense. tiny-tim, in his first response, asked if you meant "O" rather than "K" but you never answered. tiny-tim proceeded to assume you meant "O".

Here's how I would do this problem. Let "A" be "0" on a number line. Since 3+ 4= 7, take "B" to be "7" on that same number line. That choice makes "X divides a line segment (AB) in a ratio of 3:4" easy: "X" is at the number "3" on that number line.- AX= 3 and XY= 4, a ratio of 3:4. And "Y divides line segment AX in a ratio of 1:2" is also easy: "Y" is at the number "1"- AY= 1 and YX= 2 so the ratio is 1:2.

Let "O" be any point on that number line, at, say "z". Then OB is |z- 7| and OA is |z|. We have already seen that AY= 1. Do the algebra.

 

[tiny-tim]

"X divides a line segment (AB) in a ratio of 3:4"​

manmachine, what does that mean in ordinary english?