A) \[\frac{1}{\sqrt{3}}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{\sqrt{5}}\]
D) \[\frac{1}{\sqrt{6}}\]
Correct Answer: B
Solution :
Here,\[A=\left( \begin{matrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \\ \end{matrix} \right)\] \[\therefore \]\[{{A}^{T}}=\left( \begin{matrix} 0 & p & p \\ 2q & q & -q \\ r & -r & r \\ \end{matrix} \right)\] \[\therefore \]\[A{{A}^{T}}=\left( \begin{matrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \\ \end{matrix} \right)\left( \begin{matrix} 0 & p & p \\ 2q & q & -q \\ r & -r & r \\ \end{matrix} \right)\] \[=\left( \begin{matrix} 4{{q}^{2}}+{{r}^{2}} & 2{{q}^{2}}-{{r}^{2}} & -2{{q}^{2}}+{{r}^{2}} \\ 2{{q}^{2}}-{{r}^{2}} & {{p}^{2}}+{{q}^{2}}+{{r}^{2}} & {{p}^{2}}-{{q}^{2}}-{{r}^{2}} \\ -2{{q}^{2}}+{{r}^{2}} & {{p}^{2}}-{{q}^{2}}-{{r}^{2}} & {{p}^{2}}+{{q}^{2}}+{{r}^{2}} \\ \end{matrix} \right)\] Also, \[A{{A}^{T}}={{I}_{3}}\] [Given] \[\Rightarrow \]\[2{{q}^{2}}={{r}^{2}},{{p}^{2}}={{q}^{2}}+{{r}^{2}}\]and\[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}=1\] \[\therefore \]\[|p|=\frac{1}{\sqrt{2}}\]
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