JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Vector

Reading the coordinates off the cube with the origin at $A$, the centre of face $ABOD$ sits at $(0,\frac{a}{2},\frac{a}{2})$ and the centre of face $BEFO$ at $(\frac{a}{2},\frac{a}{2},0)$. The displacement from the first to the second is the difference of these position vectors, $(\frac{a}{2},\frac{a}{2},0)-(0,\frac{a}{2},\frac{a}{2})=(\frac{a}{2},0,-\frac{a}{2})$. The $\hat{j}$ parts cancel because both centres share the same height along that axis, leaving $\frac{a}{2}(\hat{i}-\hat{k})$. The whole trick is to anchor one corner as origin and take each face centre as the average of its four corners.