Let `veca.vecb=0` where `veca and vecb` are unit vectors and the vector `vecc` is inclined an anlge `theta` to both `veca and vecb. If vecc=mveca+nvecb + p(veca xx vecb) , (m,n,p in R)` then

Options

A. `pi/4lethetalepi/4`
B. `pi/4lethetale(3pi)/4`
C. `0 le thetale pi/4`
D. `0 le thetale(3pi)/4`

Answer

a

Explanation

`vecc = mveca + nvecb + p(veca xx vecb)`
taking dot product with `veca and vecb` we have
`m =n= cos theta`
`Rightarrow |vecc|= |cos theta veca + cos theta veca + cos theta vecb + p (veca xx vecb)|=1`
squaring both sides, we get
`cos^(2)theta + cos ^(2) theta + p^(2) =1 `
`or cos theta= +-sqrt(1-p^(2))/sqrt2`
Now ` - 1/sqrt2 le cos theta ge 1/sqrt2` ( for real value of `theta`)
`pi/4 le cos theta le (3pi) / 4`