Question

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

• A. $\begin{array}{l}\frac{1}{2}a(\hat{k}-\hat{i})\end{array}$
• B. $\begin{array}{l}\frac{1}{2}a(\hat{i}-\hat{k})\end{array}$
• C. $\begin{array}{l}\frac{1}{2}a(\hat{j}-\hat{i})\end{array}$
• D. $\begin{array}{l}\frac{1}{2}a(\hat{j}-\hat{k})\end{array}$

Answer 1

The correct option is C

$\begin{array}{l}\frac{1}{2}a(\hat{j}-\hat{i})\end{array}$

Explanation for the correct option:

Finding the vector from the central point of the face$ABOD$:

Let $\begin{array}{l}\hat{j},\hat{i},\end{array}\begin{array}{l}\hat{k}\end{array}$be the unit vectors along $X,Y\&Z$ axis respectively

Considering the coordinates of $O$ as $(0,0)$

Then coordinates of central points $G$ and $H$ are $\left(\frac{a}{2}\hat{i},\frac{a}{2}\begin{array}{l}\hat{k}\end{array}\right)$ and $\left(\frac{a}{2}\hat{j},\frac{a}{2}\begin{array}{l}\hat{k}\end{array}\right)$ respectively

Then position vectors

$$\begin{gathered} \overrightarrow{OG}=\frac{a}{2}\hat{i}+\frac{a}{2}\begin{array}{l}\hat{k}\end{array} \\ \overrightarrow{OH}=\frac{a}{2}\hat{j}+\frac{a}{2}\begin{array}{l}\hat{k}\end{array} \end{gathered}$$

From figure

$$\begin{gathered} \overrightarrow{GH}=\overrightarrow{OH}-\overrightarrow{OG} \\ \overrightarrow{GH}=\left(\frac{a}{2}\hat{j}+\frac{a}{2}\begin{array}{l}\hat{k}\end{array}\right)-\left(\frac{a}{2}\hat{i}+\frac{a}{2}\begin{array}{l}\hat{k}\end{array}\right) \\ \overrightarrow{GH}=\frac{a}{2}\left(\hat{j}-\hat{i}\right) \end{gathered}$$

Hence, option C is correct.