QJEEM19Phbb54: JEE Main 2019 Physics - Class 11 JEEM19Phbb54 Solution: Let , and . The value of is :-

Question

Verified Solution • 2025–26 Session moderate

Let

\left| {\mathop {{A_1}}\limits^ \to } \right| = 3

,

\left| {\mathop {{A_2}}\limits^ \to } \right| = 5

and

\left| {\mathop {{A_1}}\limits^ \to + \mathop {{A_2}}\limits^ \to } \right| = 5

. The value of

\left( {2\mathop {{A_1}}\limits^ \to + 3\mathop {{A_2}}\limits^ \to } \right)\left( {3\mathop {{A_1}}\limits^ \to - \mathop {2{A_2}}\limits^ \to } \right)

is :-

Accepted Answer

\left| {\overrightarrow {{A_1}} } \right| = 3,\left| {\overrightarrow {{A_2}} } \right| = 5\,and\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = 5

\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = {\left| {\overrightarrow {{A_1}} } \right|^2} + {\left| {\overrightarrow {{A_2}} } \right|^2} + 2\left| {\overrightarrow {{A_1}} } \right|\left| {\overrightarrow {{A_2}} } \right|\cos \theta

\cos \theta = - {3 \over {10}}

\left( {2\overrightarrow {{A_1}} + 3\overrightarrow {{A_2}} } \right).\left( {3\overrightarrow {{A_1}} - 2\overrightarrow {{A_2}} } \right)

=

6{\left| {\overrightarrow {{A_1}} } \right|^2} + 9\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 4\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 6{\left| {\overrightarrow {{A_2}} } \right|^2}

= - 118.5

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