\(\vec{r}=\frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}\)

\(\vec{v} \cdot \vec{r}=0\) as velocity is always tangential to the path.

\(\left(v_x \hat{i}+v_y \hat{j}\right) \cdot \frac{1}{2}(\hat{i}+\hat{j})=0\)

\(v_x+v_y=0\)

\(\Rightarrow v_x=-v_y\)

\(\text { or }\)

\(v_y=-v_x\)

\(v=\sqrt{v_x^2+v_y^2} \quad \Rightarrow \sqrt{2} v_x=v \Rightarrow v_x=\frac{v}{\sqrt{2}}\)

\(v_y=-\frac{v}{\sqrt{2}}\)

\(\text { or } v_x=-\frac{v}{\sqrt{2}}, v_y=\frac{v}{\sqrt{2}}\)

So, possible value of \(v \Rightarrow v_x \hat{i}+v_y \hat{j} \Rightarrow \frac{v}{\sqrt{2}} \hat{i}-\frac{v}{\sqrt{2}} \hat{j}\) or \(\frac{-v}{\sqrt{2}} \hat{i}+\frac{v}{\sqrt{2}} \hat{j}\)