A parallelogram is constructed on `3 vec a+ vec ba n d vec a-4 vec b ,w h e r e| vec a|=6a n d| vec b|=8,a n d vec aa n d vec b` are anti-parallel. Then the length of the longer diagonal is a .`40` b. `64` c. `32` d. `48`

Options

A. 40
B. 64
C. 32
D. 48

Answer

c

Explanation

`(3veca + vecb) . (veca - 4 vecb)`
`= 3 |veca|^(2)- 11 veca.vecb - 4|vecb|^(2)`
` 3 xx 36 - 11 xx 6xx 8 cos pi - 4 xx 64 gt 0 `
Therefore, the angle between `veca and vecb` is acute. The longer, diagonal is given by
`vecalpha = ( 3veca + vecb) + (veca -4vecb) = 4veca - 3vecb)`
Now, `|vecalpha|^(2) = |4veca - 3vecb|^(2)`
`16|veca|^(2)+ 9 |vecb|^(2)- 24veca.vecb`
`16 xx 36 + 9 xx 64 - 24 xx 6 xx 8 cos pi`
`16 xx 144`
` or |4veca - 3vecb|= 48`