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Q. Let $\left|\vec{A}_{1}\right| =3, \left|\vec{A}_{2}\right| = 5 $ and $\left|\vec{A}_{1} +\vec{A_{2}}\right| = 5$
The value of $ \left(2 \vec{A}_{1} + 3\vec{A}_{2}\right). \left(3\vec{A}_{1} - 2\vec{A}_{2}\right) $ is : -

JEE MainJEE Main 2019Motion in a Plane

Solution:

$\left|\vec{A}_{1}\right| =3, \left|\vec{A}_{2}\right| = 5 \left|\vec{A}_{1} +\vec{A_{2}}\right| = 5 $
$ \left|\vec{A}_{1} + \vec{A}_{2}\right| = \sqrt{\left|\vec{A}_{1}\right|^{2} + \left|\vec{A_{2}}\right|^{2} +2\left|\vec{A}_{1}\right|\left|\vec{A_{2}}\right|\cos\theta} $
$5 = \sqrt{9+25+2\times3\times5\cos\theta} $
$\cos\theta = - \frac{9}{2\times3\times5} = - \frac{3}{10} $
$ \left( 2 \vec{A}_{1} + 3\vec{A}_{2}\right). \left(3\vec{A}_{1} - 2\vec{A}_{2}\right) $
$ = 6\left|\vec{A_{1}}\right|^{2} +9 \vec{A_{1} }. \vec{A}_{2} - 4 \vec{A_{1}} \vec{A_{2}} - 6 \left|\vec{A}_{2}\right|^{2} $
$ 54 + 5 \times3 \times5 \left(- \frac{3}{10}\right) - 6 \times25 $
$= 54 - 150 - \frac{45}{2} = -118.5 $