`bar i xx [(bar a xx bar b) xx bar i]+bar j xx [(bar a xx bar b) xx bar j]+bar k xx [(bar a xx bar b) xx bar k]=`

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 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) .

Compute bar(a) xx (bar(b) + bar(c )) + bar(b) xx (bar(c ) +bar(a)) + bar(c ) xx (bar(a)+bar(b))

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(a)=bar(i)-2bar(j)+bar(k), bar(b)=2bar(i)+bar(j)+bar(k), bar(c ) = bar(i) + 2bar(j)-bar(k) , then find bar(a) xx (bar(b) xx bar(c )) and |(bar(a)xxbar(b))xxbar(c )| .

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))=

If bar = hat i + hat j and bar b = 2 hat i - hat k then the intersection piont of the lines bar r xx bar a = bar b xx bar a and bar r xx bar b = bar a xx bar b is .............