If `veca and vecb` are two non collinear vectors `and `vecuveca0(veca.vecb)vecb and vecv=vecaxxvecb` then `|vecv|` is (A) `|vecu|` (B) `|vecu|+|vecu.vecb|` (C) `|vecu|+|vecu.veca|` (D) none of these

Options

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

Answer

a,c

Explanation

we have
`vecv= vecaxxvecb= |veca||vecb| sin theta hatn = sin theta hatn`
where `veca and vecb` are unit vectors. Therefore,
`|vecv|= sin theta`
Now, `vecu = veca - (veca.vecb)vecb`
`= veca -vecb cos theta ( " where " veca. Vecb = cos theta)`
`|vecu|^(2) = | veca-vecb cos theta|^(2)`
` 1 + cos^(2) theta -2 cos theta . cos theta`
` =1 - cos^(2) theta = sin^(2) theta = |v|^(2)`
` Rightarrow |vecu|= |vecv|`
Also , `vecu . vecb = veca. vecb - (veca.vecb) (vecb.vecb)`
` = veca.vecb-veca.vecb=0`
`|vecu.vecb|=0`
`|vecv|=|vecu|+ |vecu.vecb|` is also correct.