Let `veca` and `vecb` be two non-collinear unit vectors. If `vecu=veca-(veca.vecb)vecb` and `vecv=vecaxxvecb`, then `|vecv|` is

Options

A. `|vecu|+|vecu.(vecaxxvecb)|`
B. `|vecu|+|vecu.veca|`
C. `|vecu|+|vecu.vecb|`
D. `|vecu|+vecu.(veca+vecb)`

Answer

C

Explanation

We have `|veca|=|vecb|=1`
Now,
`vecu=veca-(veca.vecb)vecb`
`impliesvecu=(vecb.vecb)veca-(vecb.veca)vecb`
`impliesvecu=vecbxx(vecaxxvecb)`………………i
`impliesvecu=vecbxxvec`
`implies |vecu|=|vecb||vecv|"sin"(pi)/2 [ :' vecv=vecaxxvecbimpliesvecv_|_vecb]`
`implies|vecu|=|vecv|`...........ii `[ :' |b|=1]`
So option a is not correct.
Again,
`vecu=vecbxx(vecaxxvecb)` [From i]
`impliesvecu` is perpendicular to `vecb`
`impliesvecu.vecb=0`

`:.|vecu|+|vecu.vecb|=|vecu|`
`implies |vecu|+|vecu.vecb|=|vecv|` [ Using ii]
Hence, option c is also correct.