A) \[-\hat{i}+\hat{j}-2\hat{k}\]
B) \[2\hat{i}-\hat{j}+2\hat{k}\]
C) \[\hat{i}-\hat{j}-2\hat{k}\]
D) \[\hat{i}+\hat{j}-2\hat{k}\]
Correct Answer: A
Solution :
Let \[b={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\] given \[\overrightarrow{a}.\text{ }\overrightarrow{b}=3\] \[{{b}_{2}}{{b}_{3}}=3\] (1) and \[\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{c}=0\] \[\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{c}\] \[\left| \begin{matrix} i & j & k \\ 0 & 1 & -1 \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ \end{matrix} \right|=-\hat{i}+\hat{j}+\hat{k}\] \[{{b}_{3}}+{{b}_{2}}=1\] (2) \[-{{b}_{1}}=1\] ----- (3) \[-{{b}_{1}}=1\] ------(4) \[{{b}_{1}}=1\] from (1) and (2) \[{{b}_{2}}=1\] \[{{b}_{3}}=2\] \[\overrightarrow{b}=-\hat{i}+\hat{j}-2\hat{k}\]
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