A) \[\frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}\]
B) \[\pm \,\left( \frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}} \right)\]
C) \[\frac{\mathbf{k}-\mathbf{i}}{\sqrt{2}}\]
D) \[\frac{\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}}\]
Correct Answer: B
Solution :
Let the vector be given as \[a\mathbf{i}+b\mathbf{j}+c\mathbf{k}.\] For this vector to be coplanar with \[\mathbf{i}+\mathbf{j}+2\mathbf{k}\] and \[\mathbf{i}+2\mathbf{j}+\mathbf{k},\] we will have \[a\mathbf{i}+b\mathbf{j}+c\mathbf{k}=p(\mathbf{i}+\mathbf{j}+2\mathbf{k})+r(\mathbf{i}+2\mathbf{j}+\mathbf{k})\] This gives, \[a=p+r\] .....(i) \[b=p+2r\] .....(ii) \[c=2p+r\] .....(iii) For the vector \[a\mathbf{i}+b\mathbf{j}+c\mathbf{k}\] to be perpendicular to \[\mathbf{i}+\mathbf{j}+\mathbf{k},\] we will have \[(a\mathbf{i}+b\mathbf{j}+c\mathbf{k})\,.\,(\mathbf{i}+\mathbf{j}+\mathbf{k})=0\] \[\Rightarrow a+b+c=0\] ......(iv) Adding equation (i) to (iii), we get \[4p+4r=a+b+c\] \[\Rightarrow 4(p+r)=0\Rightarrow p=-r\] Now with the help of (i), (ii) and (iii), we get \[a=0,\] \[b=r,\] \[c=p=-r\] Hence the required vector is \[r(\mathbf{j}-\mathbf{k})\] To be its unit vector \[{{r}^{2}}+{{r}^{2}}=1\Rightarrow r=\pm \frac{1}{\sqrt{2}}\] Hence the required unit vector is, \[\pm \frac{1}{\sqrt{2}}(\mathbf{j}-\mathbf{k})\]. Trick : Check for option \[\text{(a)},\,\,\frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}\] is a unit vector and perpendicular to \[\mathbf{i}+\mathbf{j}+\mathbf{k}\]. But \[\left| \begin{matrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0 \\ 1 & 1 & 2 \\ 1 & 2 & 1 \\ \end{matrix} \right|=-\frac{4}{\sqrt{2}}\ne 0\]. So it is not coplanar with the given vector. Check for option \[\text{(b}),\,\,\pm \left( \frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}} \right)\] is a unit vector and also perpendicular to \[\mathbf{i}+\mathbf{j}+\mathbf{k},\] \[\left| \,\begin{matrix} 0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ 1 & 1 & 2 \\ 1 & 2 & 1 \\ \end{matrix}\, \right|=0\]. So, it is also coplanar with the given vectors.
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