Question

Let $\left|{\vec{A}}_1\right|=3,\left|{\vec{A}}_2\right|=5$ and $\left|{\vec{A}}_1+\overrightarrow{A_2}\right|=5$
The value of $\left(2{\vec{A}}_1+3{\vec{A}}_2\right).\left(3{\vec{A}}_1-2{\vec{A}}_2\right)$ is : -

• A. -112.5
• B. -106.5
• C. -118.5
• D. -99.5

Answer 1

Answer: (C)

$\left|{\vec{A}}_1\right|=3,\left|{\vec{A}}_2\right|=5\left|{\vec{A}}_1+\overrightarrow{A_2}\right|=5$
$\left|{\vec{A}}_1+{\vec{A}}_2\right|=\sqrt{{\left|{\vec{A}}_1\right|}^2+{\left|\overrightarrow{A_2}\right|}^2+2\left|{\vec{A}}_1\right|\left|\overrightarrow{A_2}\right|\cos \theta }$
$5=\sqrt{9+25+2\times 3\times 5\cos \theta }$
$\cos \theta =-\frac{9}{2\times 3\times 5}=-\frac{3}{10}$
$\left(2{\vec{A}}_1+3{\vec{A}}_2\right).\left(3{\vec{A}}_1-2{\vec{A}}_2\right)$
$=6{\left|\overrightarrow{A_1}\right|}^2+9\overrightarrow{A_1}.{\vec{A}}_2-4\overrightarrow{A_1}\overrightarrow{A_2}-6{\left|{\vec{A}}_2\right|}^2$
$54+5\times 3\times 5\left(-\frac{3}{10}\right)-6\times 25$
$=54-150-\frac{45}{2}=-118.5$