Solve.it.dont send link i.have already referred to link and its.not at all explainatory Q 42 . F o r t h r e e v e c t o r s a → , b → , a n d c → i f a → × b → = c → a n d a → × c → = b → , t h e n p r o v e t h a t a → , b → , a n d c → a r e m u t u a l l y p e r p e n d i c u l a r v e c t o r s , b → = c → a n d a → = 1 . Share with your friends Share 0 Aarushi Mishra answered this Dear studentHere I am assuming that a→, b→ and c→ are non-zero vectorsa→×b→=c→ __________1a→×c→=b→ __________2a→.a→×b→=a→.c→Since a→×b→ ⊥a→, therefore a→.a→×b→=0a→.c→=0a→ ⊥c→a→.a→×c→=a→.b→Since a→×c→ ⊥a→, therefore a→.a→×c→=0a→.b→=0a→ ⊥b→c→.a→×c→=c→.b→Since a→×c→ ⊥c→, therefore c→.a→×c→=0c→.b→=0b→ ⊥c→Therefore we havea→, b→ and c→ are mutually perpendiculara→×b→=c→a→×b→=c→a→b→sin 90°=c→ Since a→ and b→ are perpendiculara→b→=c→ __________3Alsoa→×c→=b→a→×c→=b→a→c→sin 90°=b→ Since a→ and c→ are perpendiculara→c→=b→ __________4Dividing equation 4 and 3a→c→a→b→=b→c→c→b→=b→c→c→2=b→2c→=b→Put in equation 4a→c→=c→a→=1 0 View Full Answer
