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Let vecx, vecy and vecz be three vectors...

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

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)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
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) `
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Knowledge Check

  • Let vec(x), vec(y) and vec(z) be three vectors each of magnitude sqrt2 and the angle between each pair of them is (pi)/(3) . If vec(a) is a non-zero vector perpendicular to vec(x) and vec(y) xx vec(z) and vec(b) is a non-zero vector perpendicular to vec(y) and vec(z) xx vec(x) , then

    A
    `vec(b) = (vec(b).vec(z)) (vec(z) - vec(x))`
    B
    `vec(a) = (vec(a).vec(y)) (vec(y) - vec(z))`
    C
    `vec(a).vec(b) = - (vec(a).vec(y)) (vec(b).vec(z))`
    D
    `vec(a) = (vec(a).vec(y)) (vec(z) - vec(y))`

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