If `veca bot vecb` then vector `vecv` in terms of `veca and vecb` satisfying the equations `vecv. veca = 0` and `vecv. vecb = 1` and `[ vecv.(vecaxx vecb)] =1` is

Options

A. `(vecb)/(|vecb|^(2))+ (vecaxx vecb)/(|vecaxxvecb|^(2))`
B. `(vecb)/(|vecb|)+ (vecaxx vecb)/(|vecaxxvecb|^(2))`
C. `(vecb)/(|vecb|)+ (vecaxx vecb)/(|vecaxxvecb|)`
D. none of these

Answer

a

Explanation

Let `vecc = xveca + yvecb +z veca xx vecb`
Given : `veca. Vecb = 0, veca. Vecb = 1 , [ vecv veca vecb] =1 `
`Rightarrow vecv.veca=xveca.veca = x|veca|^(2)`
` ( veca. Vecb =0, veca.veca xx vecb =0)`
` Rightarrow x =0`
Again, ` vecv. (veca xx vecb) = z (veca xx vecb)^(2)`
` Rightarrow 1=z (veca xx vecb)^(2) or z= 1/(|veca xx vecb|^(2))`
Hence, `vecv= 1/|vecb|^(2) vecb+ 1 / (|veca xx vecb|^(2)) veca xx vecb`