If #|hata+hatb| =√3# then #|hata -hat b|# is?

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Answer 1 · 2017-05-25T11:20:32

#|hata+hatb| =√3#

#=>sqrt(abshata^2+abshatb^2+2abshataabshatbcostheta)=sqrt3#

#=>sqrt(1^2+1^2+2xx1xx1costheta)=sqrt3#

[ #abshata=1 and abs hatb=1# as they are unit vector]

#=>1^2+1^2+2xx1xx1costheta=3#

#=>costheta=(3-2)/2=1/2=cos60^@#

#=>theta=60^@#

Now

#|hata -hat b|#

#=sqrt(abshata^2+abshatb^2+2abshataabshatbcos(pi-theta))#

#=>sqrt(1^2+1^2-2xx1xx1costheta)#

#=>sqrt(1^2+1^2-2xx1xx1cos60^@#

#=>sqrt(1^2+1^2-2xx1xx1xx1/2)=1#