A vector a has components $3 \mathrm{p}$ and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, a has components $p+1$ and $\sqrt{10}$, then a value of $p$ is equal to:
Correct Option: , 4
$\overrightarrow{\mathrm{a}}$ old $=3 \mathrm{p} \hat{\mathrm{i}}+\hat{\mathrm{j}}$
$\overrightarrow{\mathrm{a}}_{\mathrm{New}}=(\mathrm{p}+1) \hat{\mathrm{i}}+\sqrt{10} \hat{\mathrm{j}}$
$\Rightarrow\left|\overrightarrow{\mathrm{a}}_{\mathrm{old}}\right|=\left|\overrightarrow{\mathrm{a}}_{\mathrm{New}}\right|$
$\Rightarrow \mathrm{ap}^{2}+1=\mathrm{p}^{2}+2 \mathrm{p}+1+10$
$8 \mathrm{p}^{2}-2 \mathrm{p}-10=0$
$4 p^{2}-p-5=0$
$(4 p-5)(p+1)=0 \rightarrow p=\frac{5}{4},-1$
