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The set of vectors constituting an orthogonal basis in $\mathbb{R} ^3$ is

  1. $\begin{Bmatrix} \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, & \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, & \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \end{Bmatrix}$
  2. $\begin{Bmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{pmatrix}, & \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, & \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \end{Bmatrix}$
  3. $\begin{Bmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{pmatrix}, & \begin{pmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{pmatrix} \end{Bmatrix}$
  4. None of these
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For being Orthogonal Basis the set of vectors must be orthogonal to one another(or perpendicular to one another hence dot product will be zero) and must be linearly independent. 

Here in Option (A) the set of vectors are linearly independent as the matrix formed from these vectors has a non zero determinant :-  $\begin{vmatrix} 1 & 1 & 0\\ -1 & 1 & 0\\ 0& 0& 1 \end{vmatrix}=2$

Further if we do the dot product of any of the vectors with another it is equal to $0$  

For Example : ${\begin{pmatrix} 1 & -1 & 0 \end{pmatrix}} \cdot {\begin{pmatrix} 1 & 1 & 0 \end{pmatrix}}=0$

Hence Option (A) forms a orthogonal basis in  $\mathbb{R}^3$

In Option (B) the vectors are not orthogonal to each other.

In Option (C) the matrix formed with the given vectors has rank 2, hence it cannot span $\mathbb{R}^3$

 

 

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