If vector A& B are such that |A+B|=|A|=|B|, then |A-B| may be equated to √3|A|. proof??

1 Answer
Jun 7, 2018

See below

Explanation:

From scalar product:

  • #bb X * bb Y = abs bb X abs bb Y cos alpha_("xy")#

#abs (bb A + bb B)^2#

#= abs (bb A + bb B) abs (bb A + bb B) cos 0 #

#= (bb A + bb B )* (bb A + bb B)#

#= underbrace(A^2 + B^2)\_("these are equal") + underbrace(bb A * bb B + bb B * bb A)\_ ( "these commute")#

# = 2A^2 + 2 bb A * bb B #

And:

#abs (bb A + bb B)^2 = A^2 #

#implies bbA * bb B = - 1/2 A^2#

Similarly:

#abs ( bb A - bb B)^2 #

#= 2A^2 - 2 bb A * bb B#

#= 2A^2 + A^2 = 3 A^2#

So:

#abs ( bb A - bb B) = sqrt3 abs bbA#