Three landmarks of Dehradun are joined by straight roads. Clock tower of Dehradun is considered as…

Clock Tower is the origin O $(0, 0)$. Position vector of IMA (I) relative to O: $\vec{OI} = 3\hat{i} + 9\hat{j}$. Position vector of Rajpur (R): $\vec{OR} = \vec{OI} + (5\hat{i} - 5\hat{j}) = (3+5)\hat{i} + (9-5)\hat{j} = 8\hat{i} + 4\hat{j}$. Since the bus stop S is two thirds of the way along the road from Clock Tower to IMA: $\vec{OS} = \frac{2}{3}\vec{OI} = \frac{2}{3}(3\hat{i} + 9\hat{j}) = 2\hat{i} + 6\hat{j}$.

The vector from Rajpur (R) to Bus Stop (S) is: $\vec{RS} = \vec{OS} - \vec{OR} = (2\hat{i} + 6\hat{j}) - (8\hat{i} + 4\hat{j}) = -6\hat{i} + 2\hat{j}$. The line from Clock Tower to IMA is in the direction of $\vec{OI} = 3\hat{i} + 9\hat{j}$ (or $\vec{OS} = 2\hat{i} + 6\hat{j}$). Taking the dot product of $\vec{RS}$ with $\vec{OI}$: $\vec{RS} \cdot \vec{OI} = (-6)(3) + (2)(9) = -18 + 18 = 0$. Since $\vec{RS}$ is perpendicular to the line joining Clock Tower and IMA, S is the foot of the perpendicular from R to that line. Therefore, S is the closest point to Rajpur on the Clock Tower to IMA road.