` vec aa n d vec b` are two unit vectors that are mutually perpendicular. A unit vector that is equally inclined to ` vec a , vec ba n d vec axx vec b` is `1/(sqrt(2))( vec a+ vec b+ vec axx vec b)` b. `1/2( vec axx vec b+ vec a+ vec b)` c. `1/(sqrt(3))( vec a+ vec b+ vec axx vec b)` d. `1/3( vec a+ vec b+ vec axx vec b)`

Options

A. `1/sqrt2(veca+vecb+vecaxxvecb)`
B. `1/2(vecaxxvecb+veca+vecb)`
C. `1/sqrt3(veca+vecb+vecaxxvecb)`
D. `1/3(veca+vecb+vecaxxvecb)`

Answer

a

Explanation

Let the required vector `vecr` be such that
` vecr =x_(1) veca + x_(2) vecb = x_(3) veca xx vecb`
we must have `vecr. Veca = vecr .vecb = vecr . (veca xx vecb) `
` vecr,veca, vecb and veca xx vecb` are unit vectors and `vecr` is equally inclined to `veca ,vecb and veca xx vecb)`
Now ` vecr. veca = x_(1) , vecr. vecb = x_(2) vecr. (veca xx vecb) = x_(3)`
`Rightarrow vecr = lambda ( veca + vecb + (veca xx vecb))`
Also , `vecr .vecr=1`
` Rightarrow lambda^(2) (veca + vecb + veca xx vecb) . (veca + vecb + (veca xx vecb)) =1 `
`or lambda^(2) (|veca|^(2) + |vecb|^(2) + |veca xx vecb|^(2) ) =1`
` or lambda^(2) = 1/3`
`or lambda = +- 1/sqrt3`
` Rightarrow vecr = +- 1/sqrt3 (veca = vecb + vecaxx vecb) `