`(axxb)xxc=axx(bxxc)`

Prove that axx(bxxc)+bxx(cxxa)+cxx(axxb)=0

If alpha(axxb)+beta(bxxc)+gamma(cxxa)=0 , then

If |a|=2a n d|b|=3 and adotb=0,t h e n(axx(axx(axx(axxb)))) is equal to 48 hat b b. -48 hat b c. 48 hat a d. -48 hat a

(a/bxxc/d) = (axxc)/(bxxd)

Let a= 2hat(i) -2hat(k) , b=hat(i) +hat(j) and c be a vectors such that |c-a| =3, |(axxb)xx c|=3 and the angle between c and axx b" is "30^(@) . Then a. c is equal to

If a,b and c are three mutually perpendicular vectors, then the projection of the vectors l(a)/(|a|)+m(b)/(|b|)+n((axxb))/(|axxb|) along the angle bisector of the vectors a and b is

If A={1,4},B={2,3,6} and C={2,3,7} , then verify that Axx(BcapC)=(AxxB)cap(AxxC)

Given A={1,2},B={3},C={4,5} , test whether the following is true: Axx(BcapC)=(AxxB)cap(AxxC)

Axx(B-C)=(AxxB)-(BxxC)

If [axxb bxxc c xxa]=lambda[abc]^(2) , then lambda is equal to