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

a,,b., c. According to the question
`vecx.vecz=vecx.vecy=vecy.vecz=sqrt(2).sqrt(2). cos ""(pi)/(3)=1`
Given `veca` is perpendicular to `vecx and vecyxxvecz`
`therefore veca =lamda_(1)(vecx xx(vecyxx vecz))`
`implies veca=lamda_(1) ((vecx.vecz)vecY-(vecx . vecy)vecz)`
`implies veca=lamda_(1) (vecy-vecz)`
Now` veca. vecy = lamda _(1) ( vecy. vecy- vecy.vecz) = lamda_(1) (2-1) `
` implies lamda_(1) = veca . vecy`
From (1) and (2) , `veca= ( veca. vecy) ( vecy- vexz)`
Similarly , ` vecb = ( vecb. vecz)( vecz - vecx) `
Now , `veca. vecb =(vecb. vecz) [ ( vecy-vecz).( vecz- vecx)]`
`= ( veca . vecy) (vecb. vecz) [1-1-2+1]`
` =- ( veca. vecy) ( vecb. vecz) `