If a¯×b¯=c¯ and b¯×c¯=a¯ then

a¯×b¯=c¯⇒c¯ is perpendicular to a¯ and b¯b¯×c¯=a¯⇒a¯ is perpendicular to b¯ and c¯⇒a¯,b¯,c¯ are mutually perpendicular  Again a¯×b¯=c¯⇒|a¯×b¯|=|c→|⇒|a¯||b¯|=|c→|→(1)b¯×c¯=a¯⇒|b¯×c¯|=|a¯|⇒|b¯||c¯|=|a¯|→(2)∣∴ From (1)&(2)|c¯|=|a¯|&|b¯|=1