A) (1, 0, 0)
B) (0, 0, 1)
C) (0, ?1, 0)
D) None of these
Correct Answer: B
Solution :
Let \[\mathbf{b}={{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{b}_{3}}\mathbf{k}\] But \[(\mathbf{i}-\mathbf{j}+\mathbf{k}).({{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{b}_{3}}\mathbf{k})=1\]\[\Rightarrow {{b}_{1}}-{{b}_{2}}+{{b}_{3}}=1\] ......(i) and \[\mathbf{a}\times \mathbf{b}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ \end{matrix} \right|\] \[=-\mathbf{i}({{b}_{2}}+{{b}_{3}})+\mathbf{j}({{b}_{1}}-{{b}_{3}})+\mathbf{k}({{b}_{2}}+{{b}_{1}})\] \[\Rightarrow \mathbf{a}\times \mathbf{b}=\mathbf{c}\] Comparing the coefficients of \[\mathbf{i},\,\,\mathbf{j}\] and \[\mathbf{k}\] respectively, we get \[{{b}_{2}}+{{b}_{3}}=1\] ?..(ii) \[{{b}_{1}}-{{b}_{3}}=-1\] ?..(iii) \[{{b}_{2}}+{{b}_{1}}=0\] ?..(iv) By solving the equations (i), (ii), (iii) and (iv), we get \[{{b}_{1}}=0,\] \[{{b}_{2}}=0\] and \[{{b}_{3}}=1\].
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