A) \[abc\]
B) \[-1\]
C) 2
D) 0
Correct Answer: D
Solution :
Now, \[a\times b=b\times c\] \[\Rightarrow \] \[a\times b=-c\times b\] \[\Rightarrow \] \[(a+c)\times b=0\] \[\Rightarrow \] \[a+c||b\] \[\Rightarrow \]\[a+c=\lambda b\]for some scalar ... (i) Further, \[c\times (a+c)=\lambda (c\times b)\] \[\Rightarrow \] \[(c\times a)+(c\times c)=\lambda c\times b\] \[\Rightarrow \] \[(c\times a)=-\lambda b\times c\] \[\Rightarrow \] \[(c\times a)=-\lambda (c\times a)\Rightarrow \lambda =-1\] Then from Eq. (i), \[a+c=-b\] \[\Rightarrow \] \[a+b+c=0\]
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