JEE PYQ: Mathematics In Physics Question 19
Question 19 - 2019 (08 Apr Shift 2)
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:
(1) $-106.5$
(2) $-99.5$
(3) $-112.5$
(4) $-118.5$
Show Answer
Answer: (4)
Solution
Using, $R^2 = A_1^2 + A_2^2 + 2A_1 A_2\cos\theta$ $5^2 = 3^2 + 5^2 + 2 \times 3 \times 5\cos\theta$ or $\cos\theta = -0.3$ $(2\vec{A_1} + 3\vec{A_2}) \cdot (3\vec{A_1} - 2\vec{A_2}) = 2A_1 \times 3A_1$ $+ (3A_1)(3A_2)\cos\theta - (2A_1)(2A_2)\cos\theta - 3A_2 \times 2A_2$ $= 6A_1^2 + 9A_1A_2\cos\theta - 4A_1A_2\cos\theta - 6A_2^2$ $= 6A_1^2 - 6A_2^2 + 5A_1A_2\cos\theta$ $= 6 \times 3^2 - 6 \times 5^2 + 5 \times 3 \times 5 \times (-0.3)$ $= -118.5$
BETA
