If `a +b+c = alphad, b+c+d=beta a and a, b, c` are non-coplanar, then the sum of `a +b+c+d =`

Options

A. 0
B. `alpha a`
C. `betab`
D. `(alpha+beta)c`

Answer

A

Explanation

We have, `a+b+c =alphad`
and b+c+d=`beta a`
`therefore a+b+c+d=(alpha+1)d`
and `a+b+c+d=(beta+1)a`
`implies (alpha+1)d=(beta+1)a`
if `alphane-1`, then `(alpha+1)d=(beta+1)a`
`implies d=(beta+1)/(alpha+1)a`
`implies a+b+c=alphad`
`implies a+b+c=alpha((beta+1)/(alpha+1))a`
`implies[1-(alpha(beta+1))/(alpha+1)]a+b+c=0`.
a,b and c are coplanar which is contradiction to the given condition.
`therefore alpha=-1`
and so a+b+c+d=0.