If ` vecalpha+ vecbeta+ vecgamma=a vecdeltaa n d vecbeta+ vecgamma+ vecdelta=b vecalpha, vecalphaa n d vecdelta` are non-colliner, then ` vecalpha+ vecbeta+ vecgamma+ vecdelta` equals a. `a vecalpha` b. `b vecdelta` c. `0` d. `(a+b) vecgamma`

Options

A. `a vecalpha `
B. `b vecdelta`
C. `0`
D. `(a+ b)vecgamma`

Answer

C

Explanation

Given `vecalpha + vecbeta + vecgamma = a vecdelta" "` (i)
`" "vecbeta + vecgamma + vecdelta = b vecalpha" "` (ii)
From (i), ` vecalpha + vecbeta + vecdelta = (a+1) vecdelta" "` (iii)
From (ii), `vecalpha + vecbeta + vecgamma + vecdelta = (b+1)vecalpha " "` (iv)
From (iii) and (iv), we get
`" "(a+1) vecdelta = (b+1) vecalpha " "` (v)
Since `vecalpha` is not parallel to `vecdelta`,
From (v), `a=1 =0 and b+1 =0`
From (iii), `vecalpha + vecbeta + vecgamma + vecdelta =0`