A) \[\frac{\sqrt{3}}{2},\,\,\frac{1}{2},\,0\]
B) \[0,\,\,\frac{\sqrt{3}}{2},\,\,\frac{1}{2}\]
C) \[\frac{\sqrt{3}}{2},\,\,0,\,\,\frac{1}{2}\]
D) \[0,\,\,\frac{1}{2},\,\frac{\sqrt{3}}{2}\]
Correct Answer: B
Solution :
Let unit vector be \[y\mathbf{i}+z\mathbf{k},\] then \[\sqrt{{{y}^{2}}+{{z}^{2}}}=1\] ?..(i) Since given that \[\cos 30{}^\circ =\frac{(y\mathbf{j}+z\mathbf{k})\,.\,(y\mathbf{j})}{|y\mathbf{j}+z\mathbf{k}|\,\,|y\mathbf{j}|}\] \[\Rightarrow \frac{{{y}^{2}}}{\left( \sqrt{{{y}^{2}}+{{z}^{2}}} \right)\,y}=\frac{\sqrt{3}}{2}\Rightarrow y=\frac{\sqrt{3}}{2}\], \[(\because \,\sqrt{{{y}^{2}}+{{z}^{2}}}=1\]by (i)) Similarly, \[\cos 60{}^\circ =\frac{(y\mathbf{j}+z\mathbf{k})\,.\,z\mathbf{k}}{|y\mathbf{j}+z\mathbf{k}|\,\,|z\mathbf{k}|}\Rightarrow z=\frac{1}{2}\] Hence the components of unit vector are \[0,\,\,\frac{\sqrt{3}}{2},\,\,\frac{1}{2}.\] Trick : Since the vector lies in \[yz-\]plane, so it will be either \[0\mathbf{i}+\frac{\sqrt{3}}{2}\mathbf{j}+\frac{1}{2}\mathbf{k}\] or \[0\mathbf{i}+\frac{1}{2}\mathbf{j}+\frac{\sqrt{3}}{2}\mathbf{k}.\] But the vector \[\frac{\sqrt{3}}{2}\mathbf{j}+\frac{1}{2}\mathbf{k}\] makes angle \[30{}^\circ \]with \[y-\]axis and that of \[60{}^\circ \] with z-axis.
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