If `bara.i+bara.(i+j)+bara.(i+j+k)=1` then `bara=`

If [bara barb barc ] = 1 then show that (bara.bara)(barbxxbarc)+ (bara.barb)(barcxxbara)+(bara.barc)(baraxxbarb)=bara .

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

Let : bara = i - j, barb = j - k, barc = k - i If bard is a unit vector such that bara. bard = 0 = [barb barc bard] , then : bard =

Let bara=2overset(^)i+overset(^)j+overset(^)k,barb=overset(^)i+2overset(^)j-overset(^)k and a unit vectors barc be coplanar. If barc is prependicular to bara , then barc=

Let bara=2overset(^)i+overset(^)j+overset(^)k,barb=overset(^)i+2overset(^)j-overset(^)k and a unit vectors barc be coplanar. If barc is prependicular to bara , then barc=

If abs(bara+barb) = abs(bara-barb), " then " bara.barb =

If bara+barb+barc =bar0 , " then " bara.bara+barb.barb+2(bara.barb)-barc.barc =

If bara.i = 4, then bara.[j xx (2j - 3k)] =

If bara.i = 4, then bara.[j xx (2j - 3k)] =

If bara is any vector, then : (bara.i)(baraxx i)+(bara.j)(baraxxj)+(bara.k)(baraxxk)=