If in triangle ABC, ` vec(AB) = vecu/|vecu|-vecv/|vecv| and vec(AC) = (2vecu)/|vecu| , " where " |vecu| ne |vecv|` , then `(a)1 + cos 2A + cos 2B + cos 2C=0` (b)`sin A = cos C` (c)projection of AC on BC is equal to BC (d) projection of AB on BC is equal to AB

Options

A. `1 + cos 2A + cos 2B + cos 2C=0`
B. `sin A = cos C`
C. projection of AC on BC is equal to BC
D. projection of AB on BC is equal to AB

Answer

a,b,c

Explanation

` vec(AB) + vec(BC) = vec(AC)`
`vec(BC) = (2vecu)/|vecu|- vecu/|vecu|+vecv/|vecv|= vecu/|vecu|+ vecv/|vecv|`
` vec(AB). vec(BC) = (vecu/|vecu|- vecv/|vecv|)(vecu/|vecu|+vecv/|vecv|)`
`(hatu - hatv . (hatu +hatv) = 1-1 =0`
v
`Rightarrow 1+ cos 2A + cos2B+ cos2C=0`