If $\overrightarrow{A}=\left(2\hat{i}+3\hat{j}\right)\text{and}\overrightarrow{B}=\left(\hat{i}-\hat{j}\right)$ then component of $\overrightarrow{A}$  perpendicular to vector $\overrightarrow{B}$ and in the same plane is

Component of $\overrightarrow{A}$ perpendicular to $\overline{B}\text{be}\overrightarrow{r}$

$\overrightarrow{r}=x\hat{i}+y\hat{j}$

$\left|\overrightarrow{r}\right|=\text{A}\times \sin \theta =\sqrt{x^2+y^2}$

$\text{r}=\text{A}.\sqrt{1-{\cos }^2\theta }=\text{A}\sqrt{1-{\left(\frac{\overline{A}.\overline{B}}{\left|\overline{A}\right|\left|\overline{B}\right|}\right)}^2}$

$\sqrt{x^2+y^2}=\sqrt{13}\sqrt{1-{\left(\frac{-1}{\sqrt{13}\sqrt{2}}\right)}^2}$

$\sqrt{x^2+y^2}=\frac{5}{\sqrt{2}}\to \left(1\right)$

$\overrightarrow{r}$ Perpendicular to $\overline{B}$

$x-y=0$

$x=y\to \left(2\right)$

From (1) & (2)

$\sqrt{2x^2}=\frac{5}{\sqrt{2}};x=\frac{5}{2}=y$

$\therefore \overrightarrow{r}=\frac{5}{2}\left(\hat{i}+\hat{j}\right)$