Solve this
Question:

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

1. (1) $-106.5$

2. (2) $-99.5$

3. (3) $-112.5$

4. (4) $-118.5$

Correct Option: , 4

Solution:

(4) Using,

$R^{2}=A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \theta$

$5^{2}=3^{2}+5^{2}+2 \times 3 \times 5 \cos \theta$

or $\cos \theta=-0.3$

$\left(2 \overrightarrow{A_{1}}+3 \overrightarrow{A_{2}}\right) \cdot\left(3 \overrightarrow{A_{1}}-2 \overrightarrow{A_{2}}\right)=2 A_{1} \times 3 A_{1}$

$+\left(3 A_{2}\right)\left(3 A_{1}\right) \cos \theta-\left(2 A_{1}\right)\left(2 A_{2}\right) \cos \theta-3 A_{2} \times 2 A_{2}$

$=6 A_{1}^{2}+9 A_{1} A_{2} \cos \theta-4 A_{1} A_{2} \cos \theta-6 A_{2}^{2}$

$=6 A_{1}^{2} 6 A_{2}^{2}+5 A_{1} A_{2} \cos \theta$

$=6 \times 3^{2}-6 \times 5^{2}+5 \times 3 \times 5(-0.3)$

$=-118.5$