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

Options

A. `veca + 4vecc`
B. `veca-4vecc`
C. `4vecc-veca`
D. `2vecc-veca`

Answer

A, C

Explanation

`veca xx vecb = 4(veca xx vecc)`
`veca xx (vecb-4vecc)=vec0`
`rArr veca = t(vecb-4vecc)`, where t is scalar.
`therefore |veca|^(2)=t^(2){|b^(2)|+16|c|^(2)-8(vecb.vecc)}`
`therefore 1=t^(2){1+16.1/16-8.1.1/4.1/2}`
`therefore 1=t^(2)[1]`
`therefore t=+-1`
`t=+-1`
`therefore veca=vecb-4vecc` or `veca=vecb+4vecc`
`rArr vecb=veca+4vec`or `vecb=4vecc-veca`