The condition for equations r→×a→=b→ and r→×c→=d→ to be consistent is
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a
b→⋅c→=a→⋅d→
b
a→⋅b→=c→⋅d→
c
b→⋅c→+a→⋅d→=0
d
a→⋅b→+c→⋅d→=0
answer is C.
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Detailed Solution
r→×a→=b→ or d→×(r→×a→)=d→×b→or (a→⋅d→)r→−(d→⋅r→)a→=d→×b→----ir→×c→=d→or b→×(r→×c→)=b→×d→or (b→⋅c→)r→−(b→⋅r→)c→=b→×d---ii→Adding (i) and (ii), we get (a→⋅d→+b→⋅c→)r→−(d→⋅r→)a→−(b→⋅r→)c→=0→ Now r→⋅d→=0 and b→⋅r→=0 as d→ and r→ as well as b→ and r→ are mutually perpendicular. Hence (b→⋅c→+a→⋅d→)r→=0→