JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 18)
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 -
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$${1 \over 2}a\left( {\widehat k - \widehat i} \right)$$
$${1 \over 2}a\left( {\widehat j - \widehat i} \right)$$
$${1 \over 2}a\left( {\widehat j - \widehat k} \right)$$
$${1 \over 2}a\left( {\widehat i - \widehat k} \right)$$
Explanation
$$\overrightarrow {{r_g}} = {a \over 2}\widehat i + {a \over 2}\widehat k$$
$$\overrightarrow {{r_H}} = {a \over 2}\widehat i + {a \over 2}\widehat k$$
$$\overrightarrow {{r_H}} - \overrightarrow {{r_g}} = {a \over 2}\left( {\widehat j - \widehat i} \right)$$
$$\overrightarrow {{r_H}} = {a \over 2}\widehat i + {a \over 2}\widehat k$$
$$\overrightarrow {{r_H}} - \overrightarrow {{r_g}} = {a \over 2}\left( {\widehat j - \widehat i} \right)$$
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