A) \[+\frac{1}{\sqrt{2}}\,(i+k)\]
B) \[+\frac{1}{\sqrt{5}}\,(2i+j)\]
C) \[+\frac{1}{\sqrt{2}}\,(i+j)\]
D) \[+\frac{1}{\sqrt{2}}\,(2i+3j)\]
Correct Answer: C
Solution :
Let given two vectors are \[a=i-j\] and \[b=i+2j\] Again let third unit vector is c. \[\because \] c is coplanar with a, b \[\therefore \] \[c=xa+yb\] \[=x(i-j)+y(i+2j)\] \[\Rightarrow \] \[c=(a+y)i+(-x+2y)j\] ?..(i) Also, c is perpendicular to a \[\therefore \] \[a.c=0\] \[\Rightarrow \] \[(i-j).\{(x+y)i+(-x+2y)j\}=0\] \[\Rightarrow \] \[(x+y)-(-x+2y)=0\] \[\Rightarrow \] \[x+y+x-2y=0\] \[\Rightarrow \] \[2x-y=0\Rightarrow y=2x\] On putting this value of y in Eq. (i), we get \[c=(x+2x)i+(-x+4x)j\] \[\Rightarrow \] \[c=3xi+3xj\] ?.(ii) But c is a unit vector. So, \[|c|=1\] \[\Rightarrow \] \[{{(3x)}^{2}}+{{(3x)}^{2}}=1\,\Rightarrow 9{{x}^{2}}+9{{x}^{2}}=1\] \[\Rightarrow \] \[{{x}^{2}}=\frac{1}{18}\,\,\,\Rightarrow \,\,\,x=\frac{1}{3\sqrt{2}}\] On putting this value of x in Eq. (ii), we get \[c=3.\frac{1}{3\sqrt{2}}i+3.\frac{1}{3\sqrt{2}}j\,\,\Rightarrow c=\frac{1}{\sqrt{2}}\,(i+j)\]
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