A) \[(2i-j-k)\sqrt{6}\]
B) \[\frac{(2i-j-k)}{\sqrt{6}}\]
C) \[2i+j+k\]
D) \[\frac{3i+j-2k}{\sqrt{6}}\]
Correct Answer: B
Solution :
Given vectors \[b=i+j+k,c=2i+j+3k\] Let the unit vector \[a=\frac{a}{|a|}\] ... (i) then \[a\text{ }\text{.}\,\,b=0\] \[\Rightarrow \] \[\frac{({{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k)}{|a|}.(i+j+k)=0\] \[\Rightarrow \] \[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}=0\] ?....(ii) and \[a\text{ }\text{. }c=0\] \[\Rightarrow \] \[\frac{({{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k)}{|a|}.(2i+j+3k)=0\] \[\Rightarrow \] \[2{{a}_{1}}+{{a}_{2}}+3{{a}_{3}}=0\] ??.(iii) On solving Eqs. (ii) and (iii) by cross multiplication method \[\frac{{{a}_{1}}}{3-1}=\frac{{{a}_{2}}}{2-3}=\frac{{{a}_{3}}}{1-2}\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{2}=\frac{{{a}_{2}}}{-1}=\frac{{{a}_{3}}}{-1}\] From Eq. (i), we get \[a=\frac{1i-j-k}{\sqrt{4+1+1}}=\frac{(2i-j-k)}{\sqrt{6}}\]
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