A vector \(\overline{\mathrm{a}}\) has components 1 and \(2 \mathrm{p}\) with respect to a rectangular Cartesian system. This system is rotated through a certain angle about origin in the counter clock wise sense. If, with respect to the new system, \(\bar{a}\) has components 1 and \((p+1)\), then

\(
\begin{aligned}
\bar{a} & =1 \cdot \hat{i}+2 p \hat{j} \\
& =\hat{i}+2 p \hat{j}
\end{aligned}
\)
Let \(\bar{b}\) be the vector obtained on rotation with components 1 and \((p+1)\). Then,
\(
\begin{aligned}
& \overline{\mathrm{b}}=\hat{\mathrm{i}}+(\mathrm{p}+1) \hat{\mathrm{j}} \\
& |\overline{\mathrm{a}}|=|\overline{\mathrm{b}}|
\end{aligned}
\)
...[Magnitude remains unchanged after rotation]
\(
\begin{aligned}
& \Rightarrow|\vec{a}|^2=|\bar{b}|^2 \\
& \Rightarrow 1+(2 p)^2=1+(p+1)^2 \\
& \Rightarrow 4 p^2=p^2+2 p+1 \\
& \Rightarrow 3 p^2-2 p-1=0 \\
& \Rightarrow(3 p+1)(p-1)=0 \\
& \Rightarrow p=-\frac{1}{3} \text { or } p=1
\end{aligned}
\)