`(bar a xx bar i)(bar b xx bar i)+(bar axx bar j)(bar b xx bar j)+(bar a xx bar k)(bar b xx bark)=`

Prove that [[bar a xx bar p, bar b xx bar p, bar cxx bar r]]+[[bar axx bar q, bar b xx bar r, bar c xx bar p]]+[[bar a xx bar r, bar b xx bar p,bar c xx bar q]]=0

if bar a+2 bar b+3bar c=bar 0 then bar axxbar b+bar b xx bar c+bar cxx bara = lambda (bar b xx bar c) ,where lambda=

If bar r=l(bar b xx bar c)+m(bar c xx bar a)+n(bar a xx bar b) and [bar a bar b bar c]=2 , then l+m+n is equal to

If bar a = 2 bar I + bar j - bar k, bar b = - bari + 2 bar j - 4 bar k and bar c = bar I + bar j + bar k , then find ( bar a xx bar b) cdot (bar b xx bar c) .

bar( r )xx bar(a)=bar(b)xx bar(a),bar( r )xx bar(b)=bar(a)xx bar(b),bar(a) ne bar(0),bar(b) ne bar(0)bar(a)ne lambda bar(b) . If bar(a).bar(b)=0 then bar( r ) = ………….

bar (a) = 2bar (i) + 3bar (j) -bar (k), bar (b) = bar (i) + 2bar (j) -4bar (k), bar (c) = bar (i) + bar (j) + bar (k), bar (d) = bar (i) -bar (j) -bar (k) then

bar(i)xx(bar(a)xxvec i)+bar(j)xx(bar(a)xxbar(j))+(bar(a)xx(bar(a)xxbar(j))+bar(k)(bar(a)xxbar(k))=