Let `vecx, vecy and vecz` be three vectors each of magnitude `sqrt2` and the angle between each pair of them is `pi/3 if veca` is a non-zero vector perpendicular to `vecx and vecy xx vecz and vecb` is a non-zero vector perpendicular to `vecy and vecz xx vecx`, then

Options

A. `vecb=(vecb.vecz)(vecz-vecx)`
B. `veca=(veca.vecy)(vecy-vecz)`
C. `veca.vecb=-(veca.vecy)(vecb.vecz)`
D. `veca=(veca.vecy)(vecz-vecy)`

Answer

A, B, C

Explanation

We have `|vecx|=|vecy|=|vecz|=sqrt(2)`
`vecx.vecy=|vecx||vecy|"cos"(pi)/3=1,vecy.vecz=1` and `vecz.vecx=1`
It is given that `veca` is perpendicular to `vecx` and `vecyxxvecz` and `vecb` is perpendicular to `vecy` and `veczxxvecx`. Therefore
`veca||vecx xx (vecyxxvecz)` and `vecb||vecyxx(veczxxvecx)`
`veca=lamda_(1){vecxxx(vecyxxvecz)}` and `vecb=lamda_(2){vecyxx(veczxx vecx)}`
for scalar `lamda_(1)` and `lamda_(2)`
`implies veca=lamda_(1){(vecx.vecz)vecy-(vecx.vecy)vecz}`
and `vecb=lamda_(2){(vecy.vecx)vecz-(vecy.zvecz)vecx}`
`impliesveca=lamda_(1)(vecy-vecz)` and `vecb=lamda_(2)(vecz-vecx)`
`impliesveca.vecy=lamda_(1)(vecy.vecy-vecy.vecz)` and `vecb.vecz=lamda_(2)(vecz.vecz-vecx.vecz)`
`impliesveca.vecy=lamda_(1)(2-1)` and `vecb.vecz=lamda_(2)(2-1)`............i
`implies lamda_(1)=veca.vecy` and `lamda_(2)=vecb.vecz`
Sunstituting the values of `lamda_(1)` and `lamda_(2)` in i we get
`impliesveca=(veca.vecy)(vecy-vecz)` and `vecb=(vecb.vecz)(vecz-vecx)`
Thus option b and c are correct.
Now `veca.vecb=(veca.vecy)(vecy-vecz).(vecb.vecz)(vecz-vecx)`
`impliesveca.vecb=(veca.vecy)(vecb.vecz){(vecy-vecz).(vecz-vecx)}`
`impliesveca.vecb=(veca.vecy(vecb.vecz)){(vecy.vecz-vecy.vecx-vecz.vecz+vecz.vecx)}`
`implies veca.vecb=(veca.vecy(vecb.vecz)(1-1-2+1)`
`implies veca.vecb=-(veca.vecy)(vecb.vecz)`
so option c is correct.