A parallelogram is constructed on the vectors `veca=3vecalpha-vecbeta, vecb=vecalpha+3vecbeta. If |vecalpha|=|vecbeta|=2` and angle between `vecalpha and vecbeta is pi/3` then the length of a diagonal of the parallelogram is

Options

A. `4sqrt5`
B. `4sqrt3`
C. `4sqrt7`
D. none of these

Answer

b,c

Explanation

`vec(AC)=veca+vecb`
`|vec(AC)|=|veca +vecb|`
`|vec(AC)|^(2)=|veca|^(2)+|vecb|^(2)+2veca.vecb`
` |3vecalpha-vecbeta|^(2)+ |vecalpha + 3vecbeta|^(2)`
`+2(3vecalpha-vecbeta).(vecalpha+3beta)`
`=16 |vecalpha|^(2) + 4|vecbeta|^(2)+ 16vecalpha.vecbeta`
80 + 16 (2) (2) (1/2)
=112
`|vec(AC)|= 4sqrt7`
`|vec(BD)|=|veca -vecb|`
`|vec(BD)|^(2) = |veca|^(2) + |vecb|^(2) - 2veca. vecb`
`|3vecalpha -vecbeta|^(2) + |vecalpha + 3vecbeta|^(2)`
`4|vecalpha|^(2) + 4|vecbeta|^(2) - 16 vecalpha.vecbeta`
80 -16(2) (2)(1/2)
= 48
`|vec(BD)|= 4sqrt3`