Q 11-03-152JEE MainJEE Main 2019 (10 Jan, Shift 1)Medium
In the cube of side $a$ shown in the figure, the vector from the central point of the face $ABOD$ to the central point of the face $BEFO$ will be:
Answer: (C) $\tfrac12 a\left(\hat j - \hat i\right)$
Face $ABOD$ lies in the $xz$-plane ($y = 0$), so its centre is $G\left(\tfrac a2, 0, \tfrac a2\right)$.
Face $BEFO$ lies in the $yz$-plane ($x = 0$), so its centre is $H\left(0, \tfrac a2, \tfrac a2\right)$.
$$\vec{GH} = \left(0-\tfrac a2\right)\hat i + \left(\tfrac a2 - 0\right)\hat j + 0\,\hat k = \tfrac12 a\left(\hat j - \hat i\right)$$
Solution by Sreeraj P, M.Sc Physics