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JEE Main 201910 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 10 January
Shift: Morning Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2019 · 10 January · Morning Shift · Physics · Mathematics in Physics

In the cube of side 'a' shown in the figure, the vector from the...

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:

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

Options

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

Correct answer

C. $\dfrac{1}{2}a\left(\overset{^}{j} - \overset{^}{i}\right)$

Step-by-step solution

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

\overset{\to}{{r}_{BA}} = \overset{\to}{r}B - \overset{\to}{r}A

= - \dfrac{a}{2}\overset{^}{i} + \dfrac{a}{2}j

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