If `veca` and `vecb` are two vectors such that `|veca|=1, |vecb|=4, veca.vecb=2`. If `vecc=(2veca xx vecb)-3vecb`, then the angle between `veca` and `vecc` is

Options

A. `pi/3`
B. `pi/6`
C. `(3pi)/(4)`
D. `(5pi)/(6)`

Answer

D

Explanation

`|veca|=1,|vecb|=4,veca.vecb=2`
`vecc = (2veca xx vecb)-3vecb`
`rArr vecc +3vecb=2veca xx vecb`
`therefore veca.vecb=2`
`rArr |veca|.|vecb|costheta=2`
`rArr costheta=2/(|veca|.|vecb|)=2/4`
`rArr costheta=1/2`
`therefore theta=pi/3`
`rArr |vecc+3vecb|^(2)=|2veca xx vecb|^(2)`
`rArr |vecc|^(2)+9|vecb|^(2)+2vecc.3vecb=4|veca|^(2)|vecb|^(2)sin^(2)theta`
`rArr |vecc|^(2)+96+6(vecb.vecc)=0`.............(1)
`therefore vecc=2veca xx vecb-3vecb`
taking dot product with `vecb`
`rArr vecb.vecc=0-3 xx 16`
`rArr vecb.vecc=-48`
Putting value of `vecb.vecc` in equation (1)
`|vecc|^(2)+96-6 xx 48=0`
`rArr |vecc|^(2)=192`
Again, putting value of `|vecc|` in equation (1)
`192+96+6|vecb|.|vecc|cosalpha=0`
`rArr 6 xx 4 xx 8sqrt(3) cosalpha=-288`
`rArr cosalpha=(-288)/(6 xx 4 xx 8sqrt(3)) = -3/2sqrt(3)`
`rArr cosalpha=-sqrt(3)/2`
`therefore alpha=(5pi)/(6)`