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B
c−c.ab.aa
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C
a−c.ab.ab
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D
b−c.ab.ab
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Solution
The correct option is Ac−c.ab.ab x×b=c×b⇒(x−c)×b=0⇒x−c is parallel to b ⇒x−c=λb for some scalar λ⇒x=c+λb x.a=0⇒(c+λb).a=0⇒c.a+λb.a=0⇒λ=−c.ab.a⇒x=c−c.ab.ab