Question

If the vectors A and B represented in such a way that A+B=A-B, then the angle b/w A and B is ?

Answer 1

Any idea @oksuz_ ?

Answer 2

You can use Cosine's law for that...

Answer 3

How ? Can u solve it please ?

Answer 4

@AllTehMaffs Any idea ?

Answer 5

For A+B

A+B= A^2 + B^2 - 2*A*B*Cos(180-alpha)= A^2 + B^2 + 2*A*B*Cos(alpha)

For A-B

A-B = A^2 + B^2 - 2*A*B*Cos(alpha)

if you make them equal each other..

2*A*B*Cos(alpha) =- 2*A*B*Cos(alpha)

2AB vanishes both side

Cos(alpha) = - Cos(alpha)

2Cos(alpha) = 0

this equation valid if and only if Cos(alpha)= (n*pi)/2 where n=1,3,5...

so if we put n= 1 you get alpha= pi/2 that means A and B are perpendicular each other..

Answer 6

Means 90 degree C @oksuz_ ?

Answer 7

Sorry not C .. :P

Answer 8

yes 90 degrees but i am not sure whether it is C or not :)

Answer 9

No it is not ..mii mistake :P

Answer 10

@AllTehMaffs right ans is 90 But Could n't figure out how ... Thanks @oksuz_ and @AllTehMaffs :)

Answer 11

A+B= A^2 + B^2 - 2*A*B*Cos(180-alpha) . How Cos is (180 - alpha )? @oksuz_

Answer 12

Cos(180-alpha)= Cos(180)*Cos(alpha) + Sin(180)*Sin(alpha) sin's terms vanishes

and Cos(180)=-1

Answer 13

ok .Same thing for A-B ?

Answer 14

For A-B you get directly Cos(alpha)

Answer 15

ok ty so much :)

Answer 16

u are welcome :)