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If `vec(a)` and `vec(b)` are unit vectors inclined at an angle `theta` then prove that(i)`sin(theta/2)=1/2|vec(a-vec(b)|` (ii) `tan(theta/2)=(|vec(a)-vec(b)|)/(|vec(a)-vec(b)|)` |
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Answer» Here, `veca` and `vecb` are unit vectors. `:. |hata| = |hatb| = 1` (i)`|hata-hatb|^2 = (hata)^2+(hatb)^2-2hata*hatb` `=|hata|^2+|hatb|^2-2|hata||hatb|cos theta` `=1^2+1^2-2(1)(1)costheta` `=>|hata-hatb|^2=2(1-costheta)` `=>|hata-hatb|^2/2 = 1-costheta` `=>|hata-hatb|^2/2 = 2sin^2(theta/2)` `=>|hata-hatb|^2/4 = sin^2(theta/2)` `=>sin (theta/2) =1/2 |hata-hatb| ->(1)` (ii)`|hata+hatb|^2 = (hata)^2+(hatb)^2+2hata*hatb` `=|hata|^2+|hatb|^2+2|hata||hatb|cos theta` `=1^2+1^2+2(1)(1)costheta` `=>|hata+hatb|^2=2(1+costheta)` `=>|hata+hatb|^2/2 = 1+costheta` `=>|hata+hatb|^2/2 = 2cos^2(theta/2)` `=>|hata+hatb|^2/4 = cos^2(theta/2)` `=>cos (theta/2) =1/2 |hata+hatb| ->(2)` Now, dividing (1) by (2), `sin (theta/2)/cos (theta/2) = |hata-hatb|/|hata+hatb|` `=> tan (theta/2) = |hata-hatb|/|hata+hatb|` |
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