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For three non-zero vectors vec(a),\vec(b...

For three non-zero vectors `vec(a),\vec(b) " and"vec(c )`, prove that `[(vec(a)-vec(b))\ \ (vec(b)-vec(c))\ \ (vec(c )-vec(a))]=0`

Text Solution

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The correct Answer is:
`=(veca-vecb).{(vecb-vecc)xx(vecc-veca)}`
`=(veca-vecb).{vecbxxvecc-vecbxxveca-veccxxvecc+veccxxveca)}`
`=(veca-vecb).{vecbxxvecc-vecbxxveca+vecxxveca}........(veccxxvecc=0)`
`=(veca-vecb).{vecbxxvecc+vecaxxvecb+vecxxveca}`
` =veca.(vecbxxvecc)+veca.(vecaxxvecb)+veca.(veccxxveca)-vecb.(vecbxxvecc)-vecb.(vecaxxvecb)-vecb.(vecxxveca)`
`=veca. (vecbxxvecc)+0+0-0-0-vecb.(veccxxveca)`
`=veca. (vecbxxvecc)-vecb.(veccxxveca)`
`=0`
(STP remains same if vectors `veca,vecb,vec` are changed in cyclic order)
`=(veca-vecb).{(vecb-vecc)xx(vecc-veca)}`
`=(veca-vecb).{vecbxxvecc-vecbxxveca-veccxxvecc+veccxxveca)}`
`=(veca-vecb).{vecbxxvecc-vecbxxveca+vecxxveca}........(veccxxvecc=0)`
`=(veca-vecb).{vecbxxvecc+vecaxxvecb+vecxxveca}`
` =veca.(vecbxxvecc)+veca.(vecaxxvecb)+veca.(veccxxveca)-vecb.(vecbxxvecc)-vecb.(vecaxxvecb)-vecb.(vecxxveca)`
`=veca. (vecbxxvecc)+0+0-0-0-vecb.(veccxxveca)`
`=veca. (vecbxxvecc)-vecb.(veccxxveca)`
`=0`
(STP remains same if vectors `veca,vecb,vec` are changed in cyclic order)
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