If a→×b→=c→,b→×c→=a→, where c→≠0→ then

a→×b→=c→,b→×c→=a→Taking cross with b→  in the first equation, we get     b→×(a→×b→)=b→×c→=a→ or      |b→|2a→−(a→⋅b→)b→=a→⇒     |b→|=1 and a→⋅b→=0 Also     |a→×b→|=|c→| or |a→||b→|sin⁡π2=|c→| or      |a→|=|c→|