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Discovery consists of seeing what everybody has seen and thinking what nobody has thought. ..."Albert von Szent-Gyorgyi(1893-1986, 1937 Nobel Prize for Medicine, Lived to 93)"

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 Author Topic: Vector Addition  (Read 317224 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
faran
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 « Embed this message Reply #30 on: July 06, 2011, 03:34:59 am » posted from:-,-,PAKISTAN

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 "

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.
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Fu-Kwun Hwang
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 « Embed this message Reply #31 on: July 06, 2011, 06:28:00 pm » posted from:Taipei,T'ai-pei,Taiwan

two anti-parallel vectors $\vec{A}$ and $\vec{B}$ means $\vec{B}=-\vec{A}$

The sum of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{A}+\vec{B}= \vec{A}+(-\vec{A})=\vec{0}$
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faran
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 « Embed this message Reply #32 on: July 07, 2011, 03:05:59 am » posted from:-,-,PAKISTAN

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.
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faran
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 « Embed this message Reply #33 on: July 08, 2011, 02:57:45 am » posted from:-,-,PAKISTAN

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Fu-Kwun Hwang
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 « Embed this message Reply #34 on: July 08, 2011, 04:32:35 pm » posted from:Taipei,T'ai-pei,Taiwan

The sum of two vectors is always $\vec{A}+\vec{B}$.

For example: if $\vec{B}=-0.2 \vec{A}$
$\vec{A}+\vec{B}=\vec{A}+ (-0.2\vec{A})=0.8 \vec{A}$ (1.0-0.2)

if$\vec{B}= 0.2 \vec{A}$ , two vector are parallel.
$\vec{A}+\vec{B}=\vec{A}+ (0.2\vec{A})=1.2 \vec{A}$ (1.0+0.2)
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THERITESHBABA
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 « Embed this message Reply #35 on: January 02, 2012, 12:35:36 pm » posted from:Kolkata,West Bengal,India

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.
 « Last Edit: January 02, 2012, 01:45:26 pm by THERITESHBABA » Logged
THERITESHBABA
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 « Embed this message Reply #36 on: January 02, 2012, 01:48:13 pm » posted from:Kolkata,West Bengal,India

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

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nateuer
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 « Embed this message Reply #37 on: November 17, 2012, 09:42:01 pm » posted from:Dhaka,Dhaka,Bangladesh

-*-
I have just seen it,nice I love that.
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koclup1580
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 « Embed this message Reply #38 on: December 29, 2012, 01:13:07 pm » posted from:,,Satellite Provider

thanks
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Discovery consists of seeing what everybody has seen and thinking what nobody has thought. ..."Albert von Szent-Gyorgyi(1893-1986, 1937 Nobel Prize for Medicine, Lived to 93)"