If ` vec aa n d vec b` are any two vectors of magnitudes 1 and 2, respectively, and `(1-3 vec a . vec b)^2+|2 vec a+ vec b+3( vec axx vec b)|^2=47 ,` then the angel between ` vec aa n d vec b` is `pi//3` b. `pi-cos^(-1)(1//4)` c. `(2pi)/3` d. `cos^(-1)(1//4)`

Options

A. `pi//3`
B. `pi-cos^(-1) (1//4)`
C. `(2pi)/3`
D. `cos^(-1) (1//4)`

Answer

c

Explanation

` 1 + 9 (veca .vecb)^(2) - 6( veca .vecb) + 4 |veca|^(2)`
`+ |vecb|^(2) + 9 |veca xx vecb|^(2) + 4veca.vecb = 47`
`or 1 + 4+4 +36 - 4 cos theta = 47`
`or cos theta = -1/2`
Hence, the angle between `veca and vecb is (2pi)/3`