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Let veda,vecb,vecc be three noncolanar v...

Let `veda,vecb,vecc` be three noncolanar vectors and `vecp,vecq,vecr` are vectors defined by the relations vecp= (vecbxxvecc)/([veca vecb vecc]), vecq= (veccxxvecca)/([veca vecb vecc]), vecr= (vecaxxvecb)/([veca vecb vecc])` then the value of the expression `(veca+vecb).vecp+(vecb+vecc).vecq+(vecc+veca).vecr`. is equal to (A) 0 (B) 1 (C) 2 (D) 3

A

0

B

1

C

2

D

3

Text Solution

Verified by Experts

The correct Answer is:
d

Given that `veca, vecb, vecc` are non - coplanar, therefore,
`'[veca vecb vecc]ne0`
Also `vecp= (vecb xx vecc)/([veca vecb vecc]). vecq= (vecc xx veca)/([veca vecb vecc]),vecr= (veca xx vecb)/([veca vecb vecc])`
Now, `(veca +vecb).vecp=(vecb + vecc). vecq+(vecc +veca).vecr `
`=(veca+vecb).(vecb xx vecc)/([veca vecb vecc])+(vecb +vecc).(vecc xx veca)/([veca vecb vecc])+ (vecc+veca).(veca xx vecb)/([veca vecb vecc])`
`(veca .vecb xx vecc)/([veca vecb vecc])+(vecb.veccxx veca)/([veca vecb vecc])+(vecc.vecaxxvecb)/([veca vecb vecc])`
`[because vecb.vecb xx vecc=vecc.veccxxveca=veca.vecaxxvecb=0]`
`([veca vecb vecc])/([veca vecb vecc])+([veca vecb vecc])/([veca vecbvecc])+([veca vecbvecc])/([veca vecb vecc])`
1+1+1
=3
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