A) \[\left[ a\,b\,c \right]=-\frac{8}{3}and\lambda =-\frac{1}{3}\]
B) \[\left[ a\,b\,c \right]=\frac{8}{3}and\lambda =-\frac{1}{3}\]
C) \[\left[ a\,b\,c \right]=-\frac{8}{3}and\lambda =-\frac{2}{3}\]
D) \[\left[ a\,b\,c \right]=-\frac{8}{3}and\lambda =\frac{2}{3}\]
Correct Answer: B
Solution :
| We have, \[a=\overset{\hat{\ }}{\mathop{i}}\,+\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,,b=\overset{\hat{\ }}{\mathop{i-}}\,\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,\]\[\Rightarrow \,\,\,\,\,\,\,\,\,\,a\times b=b+\lambda a\]\[\Rightarrow \,\,\,\,\,\,\,\,\,\,a.\left( a\times c \right)=a.b+\lambda |a{{|}^{2}}\] |
| \[0=1+3\lambda \]\[\Rightarrow \lambda =-\frac{1}{3}\]\[\Rightarrow a\times (b\times c)=a\times b+\lambda (a\times a)\]\[\Rightarrow \,\,\,|a{{|}^{2}}c-(a.c)a=b\times a\]\[\Rightarrow c=\frac{b\times a+(a.c)a}{|a{{|}^{2}}}\] |
| \[\Rightarrow \left[ a\,b\,c \right]=\frac{|a\times b{{|}^{2}}}{|a{{|}^{2}}}=\frac{8}{3}\] |
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