If `veca,vecb,vecc` are non-coplanar vectors and `vecu` and `vecv` are any two vectors. Prove that `vecuxxvecv=(1)/([vecavecbvecc])|{:(vecu.veca,vecv.veca,veca),(vecu.vecb,vecv.vecb,vecb),(vecu.vecc,vecv.vecc,vecc):}|`

Options

A. `-cos^(-1)(19/(5sqrt43))`
B. `cos^(-1)(19/(5sqrt43))`
C. `picos^(-1)(19/(5sqrt43))`
D. cannot of these

Answer

b

Explanation

we have
`vecp.vecq = 0 `
`Rightarrow ( 5veca - 3vecb) . (-veca - 2vecb) =0`
`Rightarrow 6|vecb|^(2) - 5|veca|^(2) = 7 veca. Vecb =0`
Also, `vecr. Vecs=0`
`Rightarrow ( -4veca -vecb) (-veca + vecb) =0`
`Rightarrow 4|veca|^(2) - |vecb|^(2) - 3 veca. vecb= 0`
Now, `vecx,= 1/3 (vecp + vecr +vecs)`
` =1/3 (5 veca - 3vecb -4veca - vecb - veca + vecb)=- vecb`
` and vecy = 1/5 (vecr + vecs) = 1/5 (-5veca) = -veca`
Angle between `vecx and vecy` i.e.
` cos theta = (vecx. vecy)/(|vecx||vecy|) = (veca.vecb)/(|veca||vecb|)`
from (i) and (ii)
`|veca|= sqrt(25/19) sqrt(veca. vecb) and |vecb| = sqrt(43/19) sqrt(veca. vecb)`
`|veca||vecb|= (sqrt(25 xx 43))/ 19 . veca . vecb`
` theta = cos^(-1) (19/ (5 sqrt(43)))`