Vector Addition

[faran]

Hi folk, I need help.
I encountered a question in my high school exam that was

" The resultant of two anti-parallel vectors A and B is:
1) A+B
2) A-B "

What is the correct answer?
I was told that it is A-B, but how could it be A-B, when resultant it self means that it is bascially the addition of two vectors.
If B vector is anti to A, then it should be -B as convention, but what if we say that -B=C
Then it becomes A+C, means we have to add them both to get answer.
I'm stuck please help me out.

[Fu-Kwun Hwang]

two anti-parallel vectors and means

The sum of two vectors and is

[faran]

Thanks for the response.
But what if the two vectors are of different magnitude and are anti parallel.
Then infact we'll have to subtract their magnitudes that will be |A-B| , but what if we want to write them in vector form, what would we write them if we want to get resultant?

A-B
or
A+B

Some people told me that as they are anti-parallel, then their resultant will be ultimately
A+(-B),
i.e A-B

One of my friend argued that the resultant should be A+B because

Let two vectors A and B,
A= |A| (k )
B= |B| (-k)

Now
Resultant:

A + (-B)

[|A| (k)] + [|B| (-k)]

As |B|(-k)= B

so

A + B
----
Now if it is A+B, then it means that they would be added to each other, and their magnitutude should also be added?

Please do try to understand what I've written, and help me.
Waiting for the reply

[faran]

[Fu-Kwun Hwang]

The sum of two vectors is always .
 
For example: if
  (1.0-0.2)
 
if , two vector are parallel.
  (1.0+0.2)

[THERITESHBABA]

if the angle between two vectors is a & b is @ and angle between vectors a+b and a is O
then
  

tan@=(bsinO )/(a+bcosO )

now

tan² @=(bsinO)²/(a+bcosO)²

 now

tan²@= sec²@ -1

and

sec²@ =1/cos²@

{a² sec²O + b² + 2absecO}

so cos²@=

____________________________________________________________________

{a²sec²O + b²sec²O + 2absecO}

if @ =0

b²=b²cos²O

O=0

if @ =90

a=-bcosO

if O=90
  

-b²

tan²@=

____

a²+b²

HERE BOLD CHARACTERS REPRESENT VECTORS.

[THERITESHBABA]

if the angle between two vectors is a & b is @ and angle between vectors a+b and a is O
then
 

tan@=(bsinO )/(a+bcosO )

now

tan2 @=(bsinO)2/(a+bcosO)2


 now

tan2@= sec2@ -1

and

sec2@ =1/cos2@

 

 

{a2 sec2O + b2 + 2absecO}

so cos2@=

 

{a2sec2O + b2sec2O + 2absecO}





 


if @ =0

 

b2=b2cos2O

 

O=0

 

if @ =90

 

a=-bcosO

 

if O=90


 

-b2

 

tan2@=  _________________

 

 

a2+b2

 

 

[nateuer]

-*-
I have just seen it,nice I love that.

[koclup1580]

thanks