A vector $\overline{\mathrm{a}}$ has components 1 and $2 \mathrm{p}$ with respect to a rectangular Cartesian…
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
$\mathrm{p}=1$ or $\mathrm{p}=\frac{1}{3}$
$\mathrm{p}=-1$ or $\mathrm{p}=\frac{-1}{3}$
$\mathrm{p}=\frac{-1}{3}$ or $\mathrm{p}=1$
$\mathrm{p}=\frac{1}{3}$ or $\mathrm{p}=-1$
Solution
$\begin{aligned}
\bar{a} & =1 \cdot \hat{i}+2 p \hat{j} \\
& =\hat{i}+2 p \hat{j}
\end{aligned}$
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]
$\begin{aligned}
& \Rightarrow|\vec{a}|^2=|\bar{b}|^2 \\
& \Rightarrow 1+(2 p)^2=1+(p+1)^2 \\
& \Rightarrow 4 p^2=p^2+2 p+1 \\
& \Rightarrow 3 p^2-2 p-1=0 \\
& \Rightarrow(3 p+1)(p-1)=0 \\
& \Rightarrow p=-\frac{1}{3} \text { or } p=1
\end{aligned}$