If the vectors `bara+barb+barc`, `bara+lambdabarb+2barc` and `-bara+barb+barc` are linearly dependent then `lambda` is

[bara barb barc]+[bara barc barb] =

If bara. barb = barb. barc = barc . bara = 0 then [bara barb barc] =

Show that [2bara +barb +barc bara +2barb +barc bara +barb +2barc] = 4[bara barb barc]

For any three vectors bara, barb, barc prove : [bara-barb, barb-barc, barc-bara]=0

The volume of parallelopiped with vectors bara+2barb-barc,bara-barb,bara-barb-barc as coterminous edges is K[bara barb barc] then abs(K) =

If bara,barb and barc be three non-zero vectors, no two of which are collinear. If the vectors bara+2barb is collinear with barc and barb+3barc is collinear with bara , then ( lambda being some non-zero scalar) bara+2barb+6barc is equal to: a) lambdabara b) lambdabarb c) lambdabarc d)0

bara.(barb+barc)xx(bara+barb+barc)=

Prove that [bara+barb,barb+barc,barc+bara]=2[bara barb barc]

If bara,barb,barc are non-coplanar vectors, prove that points with position vectors 2bara+3barb-barc,bara-2barb+3barc,3bara+4barb-2barc and bara - 6barb + 6barc are coplanar.

If bara, barb, barc are three vectors such that |bara|=|barc|=1, |barb|=4, |barb xx barc|=2 and 2barb=barc +lambda bara , then lambda=