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 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 :
[b] |
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 \[2\left( 0,\,\frac{a}{2},\,\,\frac{a}{2} \right)\] |
\[{{\overrightarrow{r}}_{2}}\,=\,\frac{a}{2}\widehat{j}\,\,+\,\,\frac{a}{2}\,\widehat{k}\,\] |
\[{{\overrightarrow{r}}_{2}}\,={{\overrightarrow{r}}_{1}}\,=\,\,\frac{a}{2}\widehat{j}\,\,-\,\,\frac{a}{2}\,\widehat{i}\,\] |
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