Help needed.
Find the vector sum of the forces \(\vec{A}\) and \(\vec{B}\). \(\vec{| A |}\) is 130N and is along \(\vec{OA}\) where A is (3,-4,12) and \(\vec{| B |}\) = 200N and is along vector \(\vec{PQ}\) where \(\vec{P}\) is (-2,4,7) and Q is (-2,8,10).
DO NOT SOLVE IT.
Only tell me the concept here.
:D

- AkashdeepDeb

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- anonymous

Draw a coordinate system and then put vectors on ...

- AkashdeepDeb

Yes I tried.
Can you show it to me once?

- AkashdeepDeb

@Saeeddiscover ?

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## More answers

- AkashdeepDeb

@.Sam. ?
@experimentX ? :D

- anonymous

To add vectors, you need to put them at a common point without changing their directions.The vector OA makes no difficulty, you may calculate the direction of vector PQ in space, and then catch it to put in the origin.

- AkashdeepDeb

Why have they specified their Magnitudes?
And what is O in OA?

- anonymous

O just refers to the point (0,0,0). you might calculate the directional angles to find its direction into space...

- AkashdeepDeb

And why are the magnitudes given?
I mean How is the magnitude of A = 120N ?

- anonymous

Each of the coordinate axes are written in terms of newton rather than meter.

- AkashdeepDeb

No I mean
Is
\(\vec{A}\) = 3i - 4j + 12k ?
If yes then HOW IS ITS MAGNITUDE 120?

- AkashdeepDeb

130 sry.

- AkashdeepDeb

@Saeeddiscover ?

- anonymous

The magnitude of any vector can be obtained using the relation
|A|=(a1^2+a2^2+...+an^2)1/2
But in this example, we obtain sth different. Maybe you need to scale the coordinates ...

- AkashdeepDeb

Yes man I know that!
It is just that after I calculated the displacement vector for PQ
I am not getting the magnitude as 200N
And even after scaling OA I did not get 130N,
What seems to be the problem here chief? :D

- AkashdeepDeb

?

- anonymous

For OA and PQ, we have:
|OA|=(3^2+4^2+12^2)^1/2=13
|PQ|=[(-2--2)^2+(8-4)^2+(10-7)^2]^1/2=5
If we multiply OA in 10 we get 130.....

- AkashdeepDeb

And is it the same for PQ?

- AkashdeepDeb

It is NOT!
So now what?
These has to be a reason they gave the magnitudes! :/

- AkashdeepDeb

@Saeeddiscover ?

- AkashdeepDeb

@thomaster ?

- AkashdeepDeb

@Yahoo!
HELP?
No help on YahooAnswers too! ^_^ :D
Haha..My corny puns! :D

- AkashdeepDeb

@Preetha ?
@terenzreignz ?
@thomaster ?

- AkashdeepDeb

@radar ?
@goformit100 ?
@somebodyhelpmeee.... :/

- AkashdeepDeb

@AravindG ?

- anonymous

wat u really want here ?

- AkashdeepDeb

The vector sum of the 2 vectors! :/

- AkashdeepDeb

Please tell me the question is wrong!! :/
I ahve been at it for a long time

- anonymous

No,I think that the question is correct :P ... but I can't solve "3D coordinate system" and i'm not very good at it,sorry :(

- AkashdeepDeb

It is fine.
Thanks Stephen Hawking! :D

- anonymous

i dont thik so hw will |B| = 200N?

- AkashdeepDeb

The book says it. :/
Do you think it should be 50N?

- anonymous

wat shuld be ur answer?

- anonymous

i mean the net force

- AkashdeepDeb

sqrt 234.

- anonymous

the resultant vector b/w two vectors at an angle theta is
V= sqrt(A^2 + B^2 +2AB cos theta)

- anonymous

i think u have to find cose theta form given expression and subs

- AkashdeepDeb

No actually I just add the 2 vectors!
And then I got
3i + 15k

- AkashdeepDeb

Sorry....What I mean to say is that...
I just added the two vectors with their vector components
OA + PQ and then I got the vector = 3i + 15k
So then I got the magnitude as sqrt 234!! :D

- AkashdeepDeb

@Yahoo!
I don't get it. :/

- experimentX

first find the vectors and just add it up

- AkashdeepDeb

I did.
Can you PLEASE just check if my answer is right?
THERE IS NO SOLUTION IN THIHS BOOK.

- AkashdeepDeb

The thing is how do I find OA?
Do I take O as 0?
Should O actually be Q? [Considering it is a typo?]

- experimentX

A = 130*(3,-4,12)/sqrt(3^2 + 4^2 + 12^2) = 130*A/|A|
B = 200 * (Q - P)/|Q-P|
then just add them up

- AkashdeepDeb

The question is why are you multiplying the given magnitudes by their ACtual magnitudes?

- experimentX

you are given direction ... to get vector multiply the magnitude by direction.

- AkashdeepDeb

But then you are dividing also right?

- experimentX

what is the vector along (1,1,1) whole magnitude is 12?

- AkashdeepDeb

Okay so the vector would be
i + j +k
But direction would be
1/12.1 + 1/12.j + 1/12.k
??

- experimentX

No ...

- AkashdeepDeb

Then?

- experimentX

the first thing that comes to your mind is
12*(i + j + k) = 12 i + 12 j + 12 k
but if you take its magnitude ... it turns out to be
sqrt(12^2 + 12^2 + 12^2) = 12 sqrt(3)

- experimentX

whne you are given direction first, you should normalize it.
that means you should find unit vector along that direction.

- experimentX

(1,1,1) is not unit vector. to find unit vector along (1,1,1) you need to divide it by it's magnitude.
(1,1,1)/|(1,1,1)| = 1/sqrt(3)(1,1,1) <- now this is unit vector.

- experimentX

then multiply it by 12, you get a vector whose magnitude is 12 and direction is along (1,1,1)

- AkashdeepDeb

Multiply this by 12 [1/sqrt(3)(1,1,1)]

- AkashdeepDeb

?

- experimentX

yes

- AkashdeepDeb

What is (1,1,1) then?

- experimentX

this is just given direction.

- AkashdeepDeb

But it is just a POINT.

- experimentX

Oh ... you are not familiar with tuple notation of vector.
(1,1,1) = i + j + k

- AkashdeepDeb

Not quite....
We have not studied 3D Geometry yet. :/

- experimentX

anyway you got the point right.

- AkashdeepDeb

Yes pretty much!
And If we are given just one point does that mean that
It starts from the origin?

- experimentX

no not quite.

- experimentX

it has to be specified that other end is at origin.

- experimentX

as in your question. I didn't do it becuase i am lazy to type.

- AkashdeepDeb

I got that... :D
Thanks btw for helping.
Should I be knowing 3D Geometry to solve this question?

- experimentX

no .. just use your i,j,k notation instead of (a,b,c)

- AkashdeepDeb

Yeah okay.
Thanks! :)

- experimentX

fist make the vector OA = 3i-4j+12k
then make it unit vector (unit OA)=OA/|OA|
then multiply it by magnitude = 130 *(unit OA)
then you are done, you can do what is specified.

- AkashdeepDeb

Okay.

- anonymous

Have you been satisfied?

- AkashdeepDeb

98%

- anonymous

What's still been unsolved?

- AravindG

Sorry I was away at that time.

- AkashdeepDeb

It is totally alright! :)

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