If `vecmu` and `vecv` be unit vector. If `vecv` is a vector such that `vecv + (vecv xx vecu)=vecv`, then `vecu(vecv xx vecv)` will be equal to:

Options

A. `1-vecv.vecw`
B. `1-|vecw|^(2)`
C. `|vecw|^(2)-(vecv.vecw)^(2)`
D. all of these

Answer

D

Explanation

We have
`vecw+(vecwxxvecu)=vecv`……………..i
Taking scalar product with `vecv` we obtain
`{vecw+(vecwxxvecu)}.vecv=vecv.vecv`
`impliesvecw.vecv+(vecwxxvecu).vecv=|vecc|^(2)`
`impliesvecw.vecv+[(vecu, vecv, vecw)]=1 [ :' |vecv|=1]`
`implies[(vecu, vecv, vecw)]=1-vecv. vecw`................ii
So option a is correct.
Taking cross product of i with `vecu`, we obtain
`vecuxxvecw+vecuxx(vecwxxvecu)=vecuxxvecv`
`impliesvecuxxvecw+(vecu.vecu)vecw-(vecu.vecw)vecu=vecuxxvecv`
`impliesvecuxxvecw+vecw-(vecu.vecw)vecu=vecuxxvecv`
Taking dot product with `vecw` we get
`vecw.(vecuxxvecw)+vecw.vecw-(vecu-vecw)(vecw.vecu)=vecw.(vecuxxvecc)`
`implies|vecw|^(2)-(vecu.vecw)^(2)=[(vecu, vecv, vecw)]`
`implies|vecw|^(2)-(vecu.vecw)^(2)=[(vecu, vecv, vecw)]`..............iii
So option is correct
Taking scalar product of i with `vecw` we get
`vecw.vecw+vecw.(vecwxxvecu)=vecw.vecv`
`implies|vecw|^(2)+0=vecv.vecw`
`impliesvecv.vecw=|vecw|^(2)`
Substituting the value of `vecv.vecw` in ii, we get
`[(vecu, vecv, vecw)]=1-|vecw|^(2)`
So optio b is correct.