Let `veca, vec b, vec c, vec d` be unit vectors such that `veca.vecb+veca.vec c+veca.vecd+vecb.vec c+vecb.vecd+vec c.vecd=-2` Then `[veca vecb vec c]+[veca vecb vecd] + [veca vec c vecd]+[vecb vec c vecd]=`

If [veca+vecb, vecb+vec c, vec c+veca]=8 then [veca, vecb, vec c] is

If veca, vecb,vec c are unit vectors such that veca+vecb+vec c=veco , then the value of veca*vecb+vecb*vec c+vec c*veca is -

[vecaxx vecb " " vecc xx vecd " " vecexx vecf] is equal to (a) [veca vecb vecd] [vecc vece vecf]-[veca vecb vecc] [vecd vece vecf] (b) [veca vecb vece ] [vecf vecc vecd] - [veca vecb vecf] [vece vecc vecd] (c) [vecc vecd veca] [vecb vece vecf] - [veca vecd vecb] [vecavece vecf] (d) [veca vecc vece] [vecb vecd vecf]

If vec a,vec b , vec c are unit vectors such that veca + vec b+ vecc = 0 and lambda = veca.vecb + vecb.vecc+vecc.veca and d = veca xx vecb + vecb xx vecc + vecc xx veca , then (lambda,vecd) is

If vec a,vec b , vec c are unit vectors such that veca + vec b+ vecc = 0 and lambda = veca.vecb + vecb.vecc+vecc.veca and d = veca xx vecb + vecb xx vecc + vecc xx veca , then (lambda,vecd) is

Let vec a, vec b, vec c are three vectors such that veca .veca=vecb . vecb = vecc . vecc = 3 and |veca-vecb|^2+|vecb-vecc|^2+|vecc-veca|^2=27, then

Let vec a, vec b, vec c are three vectors such that veca .veca=vecb . vecb = vecc . vecc = 3 and |veca-vecb|^2+|vecb-vecc|^2+|vecc-veca|^2=27, then

Show that [vecb, vec c, vecd]veca-[veca, vec c, vecd]vecb+[veca, vecb, vecd]vec c-[veca, vecb, vec c]vecd=0 .

If veca .vecb = veca.vec cand veca xx vecb = veca xx vec c,veca ne0 , then

Prove that : veca*(vecb+vec c)xx(veca+2vecb+3vec c)=[veca vecb vec c]