question_answer

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 -   [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

A) \[\frac{1}{2}a\,(\widehat{k}-\widehat{i})\]                                   

B) \[\frac{1}{2}a\,(\widehat{j}-\widehat{i})\]

C) \[\frac{1}{2}a\,(\widehat{j}-\widehat{k})\]                       

D) \[\frac{1}{2}a\,(\widehat{i}-\widehat{k})\]

Correct Answer: B

Solution :

Let side of cube is a coordinates of point \[1\left( \frac{a}{2},\,0,\,\frac{a}{2} \right)\] \[\overrightarrow{{{r}_{1}}}\,\,=\,\,\frac{a}{2}\widehat{i}\,\,+\,\,\frac{a}{2}\widehat{k}\] Coordinates of point \[2\left( 0,\,\frac{a}{2},\,\,\frac{a}{2} \right)\] \[{{\overrightarrow{r}}_{2}}\,=\,\frac{a}{2}\widehat{j}\,\,+\,\,\frac{a}{2}\,\widehat{k}\,\] \[{{\overrightarrow{r}}_{2}}\,={{\overrightarrow{r}}_{1}}\,=\,\,\frac{a}{2}\widehat{j}\,\,-\,\,\frac{a}{2}\,\widehat{i}\,\]