Given that `bara` is a perpendicular to `barb` and p is a non zero scalar, solve for `barr` when `p barr+(barr.barb)bara=barc`.

If bara is perpendicular to barb and barc , then : abs(baraxx(barbxxbarc))=

Given that bara and barb are not mutually perpendicular. If barc and bard are two vectors such that barb xx barc = barb xx bard and bara.bard =0, then : bard=

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

bara,barb,barc,bard are four distinct vectors satisfying the conditions bar a times barb =barc times bard and bara times barc=barb times bard then show that bara.barb+bard.barc ne bard.barb+bara.barc.

[[bara-barb, barb-bara, barc-bara]]=

If bara, barb,barc are non-coplanar vectors, then (bara+barb+barc).(bara+barb)xx(bara+barc)=

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

If bara is perpendicular to barb and barc , abs(bara)=2,abs(barb)=3,abs(barc)=4 and mangle(barb,barc)=(2pi)/3, then : [barabarb barc] =

IF bara times barb = barc and barb times barc=bara then