Let `vecu, vecv and vecw` be three unit vectors such that `vecu + vecv + vecw = veca, vecuxx (vecv xx vecw)= vecb, (vecu xx vecv) xx vecw= vecc, vec a.vecu=3//2, veca.vecv=7//4 and |veca|=2`
Vector `vecu` is

Options

A. `2/3 (2vecc -vecb)`
B. `1/3(veca -vecb -vecc)`
C. `1/3 veca - 2/3 vecb -2vecc`
D. `4/3 (vecc -vecb)`

Answer

d

Explanation

taking dot product of `vecu + vecv + vecw =veca "with" vecu,` we have
`1+vecu.vecv+vecu.vecw=veca.vecu=3/2or vecu.vecv+vecu.vecw=1/2 (i)`
similarly,l taking dot product with `vecv`.we have
`vecu .vecv + vecw.vecv = 3/4`
Also , `veca. vecu+ veca.vecv +veca.vecw=veca.veca =4 `
`Rightarrow veca.vecw =4 - (3/2 + 7/4) = 3/4`
Again, taking dot product with `vecw`, we have
`vecu.vecw+vecu.vecw=3/4-1=-1/4`
Adding (i), (ii) and (iii) , we have
`2(vecu.vecv+vecw+vecv.vecw)=1`
`or vecu.vecv+vecu.vecw+vecv.vecw=1/2`
Subtracting (i), (ii) and (iii) form (iv) , we have
`vecv.vecw=0,vecu.vecw -1/4 and vecu.vecv=3/4`
Now , the equations `vecu xx (vecv xx vecw) = vecb and (vecuxx vecv) xx vecw =vecc` can be written as
`(vecu.vecw)vecv- (vecu.vecv) vecw=vecb`
`and (vecu.vecw)vecv- (vecv.vecw)vecu=vecc`
`Rightarrow -1/4vecv-3/4vecw=vecb,-1/4vecv=vecc,i.e, vecv= -4vecc`
`Rightarrow vecc-3/4vecw=vecb Rightarrow vecw=4/3 (vecc-vecb)`
`and vecu=veca-vecv-vecw=veca+4vecc-4/3vecc+4/3vecb`
`veca+4/3vecb+ 8/3vecc`