If `(bar a xx bar b)(bar c xx bar b)+(bar a .bar b)(bar b.bar c)=lambda(bar a.bar c)` then `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

bar (a) = bar (a) + bar (j) + bar (k), bar (b) = bar (i) + bar (j), bar (c) = bar (i) and (bar (a) xxbar (b)) xxbar (c) = lambdabar (a) + mubar (b), then lambda + mu =

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=

bar a, bar b, bar c are three vectros, then prove that: (bar a xx bar b) xx bar c = (bar a. bar c) bar b-(bar b. bar c) bar a.

If bar a=hat i +hat j+ hat k, bar b=hat i-hat j +hat k, bar c=hat i +2hat j-hat k then the value of |[bar a.bar a, bar a.bar b, bar a.bar c],[bar b.bar a,bar b.bar b, bar b.bar c],[bar c.bar a,bar c.bar b,bar c.barc]|

bar(d)=lambda(bar(a)xx bar(b))+mu(bar(b)xx bar( c ))+v(bar( c )xx bar(a)) and [bar(a)bar(b)bar( c )]=(1)/(8) then lambda+mu+v = …………..

If bar(a)=hati+hatj+hatk,bar(b)=hati+hatj,bar( c )=hati and (bar(a)xx bar(b))xx bar( c )=lambda bar(a)+mu bar(b) then lambda+mu = …………

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(b)-bar(c))*(bar(a)-bar(b))xx(bar(a)-bar(bar(c)))=

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) .