A) \[\frac{1}{4}[(\sqrt{6}+\sqrt{2})i-(\sqrt{6}-\sqrt{2})j]\]
B) \[\frac{1}{4}[(\sqrt{6}-\sqrt{2})i+(\sqrt{6}+\sqrt{2})j]\]
C) \[\frac{1}{4}[(\sqrt{6}-\sqrt{2})i+(\sqrt{6}+\sqrt{2})j]\]
D) \[\frac{1}{3}[(\sqrt{6}+\sqrt{2})i+(\sqrt{2}-\sqrt{6})j]\]
E) \[\frac{1}{4}[(\sqrt{6}+\sqrt{2})i+(\sqrt{6}-\sqrt{2})j]\]
Correct Answer: E
Solution :
Let a unit vector in XOY-plane is, \[\hat{a}=\frac{a}{|a|}\]. \[a={{a}_{1}}i+{{a}_{2}}j,|a|=\sqrt{a_{1}^{2}+a_{2}^{2}}\] Given vectors let \[b=i+j\] and \[c=i-j\] From question; \[\hat{a}.b=|\hat{a}|.|b|\cos 30{}^\circ \] \[=1.\sqrt{2}.\frac{\sqrt{3}}{2}\Rightarrow \frac{\sqrt{6}}{2}\] \[\Rightarrow \] \[\frac{a.b}{|a|}=\frac{\sqrt{6}}{2}\] \[\Rightarrow \] \[{{a}_{1}}+{{a}_{2}}=\frac{\sqrt{6}}{2}|a|\] ...(i) and \[\hat{a}.c=|\hat{a}||c|.\cos 60{}^\circ \] \[=1.\sqrt{2}.\frac{1}{2}\] \[\frac{a.c}{|a|}=\frac{\sqrt{2}}{2}\] \[\Rightarrow \] \[{{a}_{1}}-{{a}_{2}}=\frac{\sqrt{2}}{2}|a|\] ...(ii) From Eqs. (i) and (ii), \[2{{a}_{1}}=(\sqrt{6}+\sqrt{2})\frac{|a|}{2}\] \[{{a}_{1}}=\frac{1}{4}(\sqrt{6}+\sqrt{2})|a|\] \[\Rightarrow \] \[\frac{a}{|a|}=\frac{1}{4}(\sqrt{6}+\sqrt{2})\] and \[{{a}_{2}}=\frac{1}{4}(\sqrt{6}-\sqrt{2})|a|\] \[\Rightarrow \] \[\frac{{{a}_{2}}}{|a|}=\frac{1}{4}(\sqrt{6}-\sqrt{2})\] So, the unit vector is \[\hat{a}=\frac{1}{|a|}({{a}_{1}}i+{{a}_{2}}j)\] \[\hat{a}=\frac{1}{4}\{(\sqrt{6}+\sqrt{2})i+(\sqrt{6}-\sqrt{2})j\}\]
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