A) \[l=m=n=1\]
B) \[l+m+n=1\]
C) \[l=m=n=0\]
D) \[l\ne 0,\,\,m\ne 0,\,\,n\ne 0\]
Correct Answer: C
Solution :
\[l\mathbf{a}+m\mathbf{b}+n\mathbf{c}=0\] Squaring both sides, we get \[{{a}^{2}}{{l}^{2}}+{{m}^{2}}{{b}^{2}}+{{n}^{2}}{{c}^{2}}+2l\,m\,\mathbf{a}\,.\,\mathbf{b}+2l\,n\,\mathbf{a}\,.\,\mathbf{c}+2m\,n\,\mathbf{b}\,.\,\mathbf{c}=0\] But \[\mathbf{a},\,\mathbf{b},\,\mathbf{c}\] are mutually perpendicular So, \[\mathbf{a}\,.\,\mathbf{b},\] \[\mathbf{b}\,.\,\mathbf{c}\] and \[\mathbf{c}\,.\,\mathbf{a}\] are equal to zero. Therefore, \[{{a}^{2}}{{l}^{2}}+{{m}^{2}}{{b}^{2}}+{{n}^{2}}{{c}^{2}}=0\,\,i.e.,\,\,l,\,\,m,\,\,n\]are equal to zero because \[{{a}^{2}},\,\,{{b}^{2}}\] and \[{{c}^{2}}\] cannot be equal to zero.
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