Simplify :

`( veca + vecb - vecc ). ( vecb + vecc )`

Let veca , vecb,vecc be three vectors such that veca bot ( vecb + vecc), vecb bot ( vecc + veca) and vecc bot ( veca + vecb) , " if " |veca| =1 , |vecb| =2 , |vecc| =3 , " then " | veca + vecb + vecc| is,

If veca,vecb,vecc are unity vectors such that vecd=lamdaveca+muvecb+gammavecc then gamma is equal to (A) ([veca vecb vecc])/([vecb veca vecc]) (B) ([vecb vecc vecd])/([vecb vecc veca]) (C) ([vecb vecd vecc])/([veca vecb vecc]) (D) ([vecc vecb vecd])/([veca vecb vecc])

If (veca xx vecb) xx vecc = veca xx (vecb xx vecc) where veca, vecb and vecc are any three vectors such that veca.vecb =0, vecb.vecc=0 then veca and vecc are:

If (veca xx vecb) xx vecc = vec a xx (vecb xx vecc) , where veca, vecb and vecc are any three vectors such that veca.vecb ne 0, vecb.vecc ne 0 , then veca and vecc are:

vecb and vecc are non- collinear if veca xx (vecb xx vecc) + (veca .vecb) vecb = ( 4-2x- sin y) vecb + ( x^(2) -1) vecc andd (vec. vecc) veca =veca then

If veca + 2 vecb + 3 vecc = vec0 " then " veca xx vecb + vecb xx vecc + vecc xx veca=

If veca, vecb and vecc be any three non coplanar vectors. Then the system of vectors veca\',vecb\' and vecc\' which satisfies veca.veca\'=vecb.vecb\'=vecc.vecc\'=1 veca.vecb\'=veca.veca\'=vecb.veca\'=vecb.vecc\'=vecc.veca\'=vecc.vecb\'=0 is called the reciprocal system to the vectors veca,vecb, and vecc . The value of [veca\' vecb\' vecc\']^-1 is (A) 2[veca vecb vecc] (B) [veca,vecb,vecc] (C) 3[veca vecb vecc] (D) 0

for any three vectors, veca, vecb and vecc , (veca-vecb) . (vecb -vecc) xx (vecc -veca) = 2 veca.vecb xx vecc .

If veca, vecb, vecc are three non-zero vectors such that veca + vecb + vecc=0 and m = veca.vecb + vecb.vecc + vecc.veca , then:

if veca + vecb + vecc=0 , then show that veca xx vecb = vecb xx vecc = vecc xx veca .