In the cube of side 'a' shown in the figure, the vector from the central point of the face A B O D…

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:

1. $\frac{1}{2}a \left(\widehat{k}-\widehat{i}\right)$
2. $\frac{1}{2}a\left(\widehat{i}-\widehat{k}\right)$
3. $\frac{1}{2}a\left(\widehat{j}-\widehat{i}\right)$
4. $\frac{1}{2}a\left(\widehat{j}-\widehat{k}\right)$

Solution

$\overrightarrow{r_{BA}}={\overrightarrow{r}}_B-{\overrightarrow{r}}_A$

$=-\frac{a}{2}\widehat{i}+\frac{a}{2}j$

Asked in: JEE Main 2019 (10 Jan Shift 1)

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