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Question : 33 of 160
Marks:
+1,
-0
Solution:
a=,b= (r×a)=(b×a)⇒(r×a)−(b×a)=0 (r−b)×a=0⇒r−b=λa r=b+λa=λ+ ....(i)
r×b=a×b ⇒(r−a)×b=0 r=a+µb=+µ ....(ii)
Comparing Eqs. (i) and (ii), we get
λ+=+µ⇒λ=1,µ=1 r=+
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