A) \[\left( \frac{\hat{j}-\hat{k}}{\sqrt{2}} \right)\]
B) \[\left( \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \right)\]
C) \[\left( \frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}} \right)\]
D) \[\left( \frac{\hat{i}+2\hat{j}+\hat{k}}{\sqrt{6}} \right)\]
E) \[\left( \frac{-\hat{j}+2\hat{k}}{\sqrt{5}} \right)\]
Correct Answer: A
Solution :
Let unit vector is \[a\hat{i}+b\hat{j}+c\hat{k}\] \[\because \]\[a\hat{i}+b\hat{j}+c\hat{k}\]is\[\bot \]to\[\hat{i}+\hat{j}+\hat{k},\] then \[a+b+c=0\] ...(i) and \[a\hat{i}+b\hat{j}+c\hat{k},(\hat{i}+2\hat{j}+\hat{k})\] and \[(\hat{i}+\hat{j}+2\hat{k})\]are coplanar \[\therefore \] \[\left| \begin{matrix} a & b & c \\ 1 & 1 & 2 \\ 1 & 2 & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[-3a+b+c=0\] ...(ii) From Eqs. (i) and (ii), we get \[a=0\]and\[c=-b\] \[\because \]\[a\hat{i}+b\hat{j}+c\hat{k}\]is a unit vector, then \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1\] \[\Rightarrow \] \[0+{{b}^{2}}+{{b}^{2}}=1\] \[\Rightarrow \] \[b=\frac{1}{\sqrt{2}}\] \[\therefore \] \[a\hat{i}+b\hat{j}+c\hat{k}=\frac{1}{\sqrt{2}}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}\] \[=\frac{\hat{j}-\hat{k}}{\sqrt{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
