QJEEM21Ph93e4: JEE Main 2021 Physics - Class 11 JEEM21Ph93e4 Solution: The magnitude of vectors , and in the given figure are equal.…

Question

Verified Solution • 2025–26 Session moderate

The magnitude of vectors

\overrightarrow {OA}

,

\overrightarrow {OB}

and

\overrightarrow {OC}

in the given figure are equal. The direction of

\overrightarrow {OA}

+

\overrightarrow {OB}

-

\overrightarrow {OC}

with x-axis will be :

JEE Main 2021 (Online) 26th August Morning Shift Physics - Vector Algebra Question 23 English

Accepted Answer

Let magnitude be equal to

\lambda

\overrightarrow {OA} = \lambda \left[ {\cos 30^\circ \widehat i + \sin 30\widehat j} \right] = \lambda \left[ {{{\sqrt 3 } \over 2}\widehat i + {1 \over 2}\widehat j} \right]

\overrightarrow {OB} = \lambda \left[ {\cos 60^\circ \widehat i - \sin 60\widehat j} \right] = \lambda \left[ {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right]

\overrightarrow {OC} = \lambda \left[ {\cos 45^\circ ( - \widehat i) + \sin 45\widehat j} \right] = \lambda \left[ { - {1 \over {\sqrt 2 }}\widehat i + {1 \over {\sqrt 2 }}\widehat j} \right]

\therefore

\overrightarrow {OA} + \overrightarrow {OB} - \overrightarrow {OC}

= \lambda \left[ {\left( {{{\sqrt 3 + 1} \over 2} + {1 \over {\sqrt 2 }}} \right)\widehat i + \left( {{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \right)\widehat j} \right]

\therefore

Angle with x-axis

{\tan ^{ - 1}}\left[ {{{{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \over {{{\sqrt 3 } \over 2} + {1 \over 2} + {1 \over {\sqrt 2 }}}}} \right] = {\tan ^{ - 1}}\left[ {{{\sqrt 2 - \sqrt 6 - 2} \over {\sqrt 6 + \sqrt 2 + 2}}} \right]

= {\tan ^{ - 1}}\left[ {{{1 - \sqrt 3 - \sqrt 2 } \over {\sqrt 3 + 1 + \sqrt 2 }}} \right]

Hence option (a).

Step-by-Step Verified Solution