A) \[\frac{6\mathbf{i}-5\mathbf{k}}{\sqrt{61}}\]
B) \[\frac{3\mathbf{j}-\mathbf{k}}{\sqrt{10}}\]
C) \[\frac{2\mathbf{i}-5\mathbf{j}}{\sqrt{29}}\]
D) \[\frac{2\mathbf{i}+\mathbf{j}-2\mathbf{k}}{3}\]
Correct Answer: B
Solution :
Let a unit vector in the plane of \[2\mathbf{i}+\mathbf{j}+\mathbf{k}\] and \[\mathbf{i}-\mathbf{j}+\mathbf{k}\] be \[\mathbf{\hat{a}}=\alpha (2\mathbf{i}+\mathbf{j}+\mathbf{k})+\beta (\mathbf{i}-\mathbf{j}+\mathbf{k})\] \[\mathbf{\hat{a}}=(2\alpha +\beta )\mathbf{i}+(\alpha -\beta )\mathbf{j}+(\alpha +\beta )\mathbf{k}\] As \[\mathbf{\hat{a}}\]is unit vector, we have Þ \[{{(2\alpha +\beta )}^{2}}+{{(\alpha -\beta )}^{2}}+{{(\alpha +\beta )}^{2}}=1\] Þ \[6{{\alpha }^{2}}+4\alpha \beta +3{{\beta }^{2}}=1\] ?..(i) As \[\mathbf{\hat{a}}\] is orthogonal to \[5\mathbf{i}+2\mathbf{j}+6\mathbf{k}\], we get \[5(2\alpha +\beta )+2(\alpha -\beta )+6(\alpha +\beta )=0\] Þ \[18\alpha +9\beta =0\Rightarrow \beta =-2\alpha \] From (i), we get \[6{{\alpha }^{2}}-8{{\alpha }^{2}}+12{{\alpha }^{2}}=1\] Þ\[\alpha =\pm \frac{1}{\sqrt{10}}\Rightarrow \beta =\mp \frac{2}{\sqrt{10}}\].Thus \[\mathbf{\hat{a}}=\pm \left( \frac{3}{\sqrt{10}}\mathbf{j}-\frac{1}{\sqrt{10}}\mathbf{k} \right)\].
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