Q 11-03-159JEE MainJEE Main 2019 (8 Apr, Shift 2)Easy
Let $|\vec A_1| = 3$, $|\vec A_2| = 5$ and $|\vec A_1 + \vec A_2| = 5$. The value of $(2\vec A_1 + 3\vec A_2)\cdot(3\vec A_1 - 2\vec A_2)$ is
Answer: (B) $-118.5$
From $|\vec A_1 + \vec A_2|^2 = A_1^2 + A_2^2 + 2\vec A_1\cdot\vec A_2$:
$$25 = 9 + 25 + 2\vec A_1\cdot\vec A_2 \Rightarrow \vec A_1\cdot\vec A_2 = -4.5$$
Expanding:
$$(2\vec A_1 + 3\vec A_2)\cdot(3\vec A_1 - 2\vec A_2) = 6A_1^2 + 5\,\vec A_1\cdot\vec A_2 - 6A_2^2 = 54 - 22.5 - 150 = -118.5$$
Solution by Sreeraj P, M.Sc Physics