QJEEM21Madac2: JEE Main 2021 Physics - Class 11 JEEM21Madac2 Solution: For p > 0, a vector is obtained by rotating the vector by an…

Question

Verified Solution • 2025–26 Session moderate

For p > 0, a vector

{\overrightarrow v _2} = 2\widehat i + (p + 1)\widehat j

is obtained by rotating the vector

{\overrightarrow v _1} = \sqrt 3 p\widehat i + \widehat j

by an angle

\theta

about origin in counter clockwise direction. If

\tan \theta = {{\left( {\alpha \sqrt 3 - 2} \right)} \over {\left( {4\sqrt 3 + 3} \right)}}

, then the value of

\alpha

is equal to _____________.

Accepted Answer

\left| {\overrightarrow {{V_1}} } \right| = \left| {\overrightarrow {{V_2}} } \right|

3{P^2} + 1 = 4 + {(P + 1)^2}

2{P^2} - 2P - 4 = 0 \Rightarrow {P^2} - P - 2 = 0

P = 2, - 1

(rejected)

\cos \theta = {{\overrightarrow {{V_1}} .\overrightarrow {{V_2}} } \over {\left| {\overrightarrow {{V_1}} } \right|\left| {\overrightarrow {{V_2}} } \right|}} = {{2\sqrt 3 P + (P + 1)} \over {\sqrt {{{(P + 1)}^2} + 4} \sqrt {3{P^2} + 1} }}

\cos \theta = {{4\sqrt 3 + 3} \over {\sqrt {13} \sqrt {13} }} = {{4\sqrt 3 + 3} \over {13}}

\tan \theta = {{\sqrt {112 - 24\sqrt 3 } } \over {4\sqrt 3 + 3}} = {{6\sqrt 3 - 2} \over {4\sqrt 3 + 3}} = {{\alpha \sqrt 3 - 2} \over {4\sqrt 3 + 3}}

\Rightarrow \alpha = 6

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