A) \[\frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}}\]
B) \[\frac{\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{2}}\]
C) \[\frac{\mathbf{i}+\mathbf{j}+2\mathbf{k}}{\sqrt{6}}\]
D) \[\frac{-\mathbf{j}+2\mathbf{k}}{\sqrt{5}}\]
Correct Answer: A
Solution :
Let\[\mathbf{a}=t(\mathbf{i}+\mathbf{j}+2\mathbf{k})+s(\mathbf{i}+2\mathbf{j}+\mathbf{k})\] \[=(t+3s)\mathbf{i}+(t+2s)\mathbf{j}+(2t+s)\mathbf{k}\] given,\[\mathbf{a}\cdot (\mathbf{i}+\mathbf{j}+\mathbf{k})=0\] \[\therefore \]\[t+s+t+2s+2t+s=0\] \[\Rightarrow \] \[4(t+s)=0\] \[\Rightarrow \] \[t+s=0\] \[\therefore \]\[\mathbf{a}=t(\mathbf{i}+\mathbf{j}+2\mathbf{k})-t(\mathbf{i}+2\mathbf{j}+\mathbf{k})\] \[=t(-\mathbf{j}+\mathbf{k})\] But\[|\mathbf{a}|\,\,=1\] \[\Rightarrow \,\,\,t=\pm \,\frac{1}{\sqrt{2}}\] \[\therefore \,\,\,\,\,a=\pm \,\left( \frac{k-j}{\sqrt{2}} \right)\].
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