Question

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

Answer 1

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

Answer 2

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

Answer 3

@Saeeddiscover ?

Answer 4

@.Sam. ?
@experimentX ? :D

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

No I mean
Is

\(\vec{A}\) = 3i - 4j + 12k ?

If yes then HOW IS ITS MAGNITUDE 120?

Answer 11

130 sry.

Answer 12

@Saeeddiscover ?

Answer 13

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 ...

Answer 14

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

Answer 15

?

Answer 16

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.....

Answer 17

And is it the same for PQ?

Answer 18

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

Answer 19

@Saeeddiscover ?

Answer 20

@thomaster ?

Answer 21

@Yahoo!

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

Answer 22

@Preetha ?
@terenzreignz ?
@thomaster ?

Answer 23

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

Answer 24

@AravindG ?

Answer 25

wat u really want here ?

Answer 26

The vector sum of the 2 vectors! :/

Answer 27

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

Answer 28

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 :(

Answer 29

It is fine.
Thanks Stephen Hawking! :D

Answer 30

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

Answer 31

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

Answer 32

wat shuld be ur answer?

Answer 33

i mean the net force

Answer 34

sqrt 234.

Answer 35

the resultant vector b/w two vectors at an angle theta is

V= sqrt(A^2 + B^2 +2AB cos theta)

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

first find the vectors and just add it up

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

But then you are dividing also right?

Answer 47

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

Answer 48

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

Answer 49

No ...

Answer 50

Then?

Answer 51

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)

Answer 52

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

Answer 53

(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.

Answer 54

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

Answer 55

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

Answer 56

?

Answer 57

yes

Answer 58

What is (1,1,1) then?

Answer 59

this is just given direction.

Answer 60

But it is just a POINT.

Answer 61

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

Answer 62

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

Answer 63

anyway you got the point right.

Answer 64

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

Answer 65

no not quite.

Answer 66

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

Answer 67

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

Answer 68

I got that... :D
Thanks btw for helping.

Should I be knowing 3D Geometry to solve this question?

Answer 69

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

Answer 70

Yeah okay.
Thanks! :)

Answer 71

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.

Answer 72

Okay.

Answer 73

Have you been satisfied?

Answer 74

98%

Answer 75

What's still been unsolved?

Answer 76

Sorry I was away at that time.

Answer 77

It is totally alright! :)