Answer:
Let \[{{\vec{b}}_{1}}=\hat{i}+2\hat{j}+\hat{k}\] and \[{{\vec{b}}_{2}}=-\,\hat{i}+3\hat{j}+4\hat{k}\] Then, \[{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 2 & 1 \\ -\,1 & 3 & 4 \\ \end{matrix} \right|\] \[=\hat{i}(5)-\hat{j}(5)+\hat{k}(5)=\vec{c}\,(say)\] \[\Rightarrow \] \[|{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}|\,\,=\sqrt{{{(5)}^{2}}+{{(5)}^{2}}+{{(5)}^{2}}}=5\sqrt{3}\] Now, \[\hat{c}=\frac{{{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}}{|{{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}|}=\frac{{\hat{i}}}{\sqrt{3}}-\frac{{\hat{j}}}{\sqrt{3}}+\frac{{\hat{k}}}{\sqrt{3}}\] Hence, required vector \[=\pm \,(10\sqrt{3}\hat{c})\] \[=\pm \,(10\hat{i}-10\hat{j}+10\hat{k})\]
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