`veca, vecb and vecc` are unit vecrtors such that `|veca + vecb+ 3vecc|=4` Angle between `veca and vecb` is ` theta_(1)` , between `vecb and vecc` is `theta_(2)` and between `veca and vecc` varies `[pi//6, 2pi//3]` . Then the maximum value of `cos theta_(1)+3cos theta_(2)` is

Options

A. A. 3
B. B. 4
C. C. `2sqrt2`
D. D. 6

Answer

b

Explanation

`|veca + vecb + 3 vecc |^(2) = 16`
`Rightarrow |veca|^(2) + |vecb|^(2) + 9 |vecc|^(2) + 2cos theta_(1) + 6 cos theta_(2)`
` + 6 cos theta_(3) = 16, theta_(3) ne [ pi//6, 2pi//3]`
`or 2 cos theta_(1)+ 6 cos theta_(2) = 5-6 cos theta_(3)`
` or ( cos theta_(1) + 3 costheta_(2))_(max) = 4`