We are given that
→a×→b=→c⇒→a⊥→c and →b⊥→c→a×→c=→c⇒→a⊥→b and →c⊥→b}⇒→a⊥→b⊥→c....(i)Now,→a×→b=→c⇒|→a×→b|=|→c|⇒|→a||→b|sin π2=|→c|⇒|→a||→b|=|→c|...(ii) [by (i),→a⊥→bAnd,→a×→c=→b⇒|→a||→c|sin π2=|→b|⇒|→a||→c|=|→b|...(iii) [by (i),→a⊥→cBy (ii)÷(iii),we get:|→c|2=|→b|2⇒|→c|=|→b|.Substitute |→c|=|→b| in (ii) to obtain,|→a|=1.