In the cube of side a shown in the figure,the | vedclass

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(C) The origin $O$ is at $(0, 0, 0)$. The cube has side length $a$. Face $ABOD$ lies in the $xz$-plane. Its vertices are $O(0,0,0)$,$D(a,0,0)$,$A(a,0,

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$\frac{1}{2} a (\hat{j} - \hat{i})$

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(C) The origin $O$ is at $(0, 0, 0)$. The cube has side length $a$. Face $ABOD$ lies in the $xz$-plane. Its vertices are $O(0,0,0)$,$D(a,0,0)$,$A(a,0,a)$,and $B(0,0,a)$. The center $G$ of face $ABOD$ is the average of its vertices: $G = (\frac{a+0}{2}, 0, \frac{a+0}{2}) = (\frac{a}{2}, 0, \frac{a}{2})$. Thus,$\vec{r}_G = \frac{a}{2}\hat{i} + \frac{a}{2}\hat{k}$. Face $BEFO$ lies in the $yz$-plane. Its vertices are $O(0,0,0)$,$F(0,a,0)$,$E(0,a,a)$,and $B(0,0,a)$. The center $H$ of face $BEFO$ is the average of its vertices: $H = (0, \frac{a+0}{2}, \frac{a+0}{2}) = (0, \frac{a}{2}, \frac{a}{2})$. Thus,$\vec{r}_H = \frac{a}{2}\hat{j} + \frac{a}{2}\hat{k}$. The vector from $G$ to $H$ is $\vec{GH} = \vec{r}_H - \vec{r}_G = (0\hat{i} + \frac{a}{2}\hat{j} + \frac{a}{2}\hat{k}) - (\frac{a}{2}\hat{i} + 0\hat{j} + \frac{a}{2}\hat{k}) = -\frac{a}{2}\hat{i} + \frac{a}{2}\hat{j} = \frac{a}{2}(\hat{j} - \hat{i})$.

In the cube of side $a$ shown in the figure,the vector from the central point $G$ of the face $ABOD$ to the central point $H$ of the face $BEFO$ is:

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