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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 $\left( {2{{\vec A}_1} + 3{{\vec A}_2}} \right)\cdot \left( {3{{\vec A}_1} - 2{{\vec A}_2}} \right)$ is
$-106.5$
$-112.5$
$-118.5$
$-99.5$
Solution
$\begin{array}{l}
\left| {{{\vec A}_1}} \right| = 3,\left| {{{\vec A}_2}} \right| = 5,\,and\,\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = 5.\\
\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = {\left| {{{\vec A}_1}} \right|^2} + {\left| {{{\vec A}_2}} \right|^2} + 2\left| {{{\vec A}_1}} \right|\left| {{{\vec A}_2}} \right|\,\cos \,\theta \\
\cos \,\theta \, = – \frac{3}{{10}}\\
\left( {2{{\vec A}_1} + 3{{\vec A}_2}} \right).\left( {3{{\vec A}_1} – 2{{\vec A}_2}} \right)\\
= 6{\left| {{{\vec A}_1}} \right|^2} + 9{{\vec A}_1}.{{\vec A}_2} – 4{{\vec A}_1}.{{\vec A}_2} – 6{\left| {{{\vec A}_2}} \right|^2}\\
= – 118.5
\end{array}$
