Question 36 of 121

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$\dfrac{1}{2}a\left(\hat{k}-\hat{i}\right)$B$\dfrac{1}{2}a\left(\hat{i}-\hat{k}\right)$C$\dfrac{1}{2}a\left(\hat{j}-\hat{i}\right)$D$\dfrac{1}{2}a\left(\hat{j}-\hat{k}\right)$

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Answer: (C) $\dfrac{1}{2}a\left(\hat{j}-\hat{i}\right)$

$\vec{{r}_{BA}}={\vec{r}}_{B}-{\vec{r}}_{A}$

$=-\dfrac{a}{2}\hat{i}+\dfrac{a}{2}j$

JEE Main Physics practice question on Mathematics in Physics (Addition and Subtraction of Vectors), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.