`veca,vecb, vecc` arwe threee unit vectors and every two are two inclined to each at an angle `cos^(-1)(3//5)` . If `vecaxxvecb=pveca+qvecb+rvecc`, where p,q,r are scalars, then find the value of q.

Explanation

`veca xx vecb = pveca =qvecb + rvecc`
taking dot product with `veca, vecb and vecc`, we get
`0 = P + 3/5 q = 3/5 r`
`0 = 3/5 p+q+ 3/5 r`
`[veca vecb vecc] = 3/5p + 3/5 q + r `
`Also, [veca vecb vecc] ^(2)=|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc):}|`
`=[{:(1,3//5,3//5),(3//5, 1, 3//5),(3//5,3//5,1):}]= 44/125` solving (i), (ii) and (iii) for q, we get
`5/11 [veca vecb vecc] = - 2/3 q `
` 5/11 xx sqrt(44/125) = - 2/3 q`
` q = - 3 /sqrt55`