Physics - Vectors Question with Solution | TestHub

$\left|{\overrightarrow{\mathrm{A}}}_1+{\overrightarrow{\mathrm{A}}}_2\right|=5$
${\left|{\overrightarrow{\mathrm{A}}}_1+{\overrightarrow{\mathrm{A}}}_2\right|}^2=25$
${\left|{\overrightarrow{\mathrm{A}}}_1\right|}^2+{\left|{\overrightarrow{\mathrm{A}}}_2\right|}^2+2{\overrightarrow{\mathrm{A}}}_1{\overrightarrow{\mathrm{A}}}_2=25$
$9+25+2{\overrightarrow{\mathrm{A}}}_1\bullet {\overrightarrow{\mathrm{A}}}_2=25$
${\overrightarrow{\mathrm{A}}}_1.{\overrightarrow{\mathrm{A}}}_2=-\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
Now,$\left(2{\overrightarrow{\mathrm{A}}}_1+3{\overrightarrow{\mathrm{A}}}_2\right)\bullet \left(3{\overrightarrow{\mathrm{A}}}_1-2{\overrightarrow{\mathrm{A}}}_2\right)=6{\left|{\overrightarrow{\mathrm{A}}}_1\right|}^2-4{\overrightarrow{\mathrm{A}}}_1{\overrightarrow{\mathrm{A}}}_2+9{\overrightarrow{\mathrm{A}}}_1{\overrightarrow{\mathrm{A}}}_2-6{\left|{\overrightarrow{\mathrm{A}}}_2\right|}^2$
$=6.{\left|{\overrightarrow{\mathrm{A}}}_1\right|}^2+5{\overrightarrow{\mathrm{A}}}_1{\overrightarrow{\mathrm{A}}}_2-6{\left|{\overrightarrow{\mathrm{A}}}_2\right|}^2$
$=6\times 9+5\left(-\frac{9}{2}\right)-6\times 25$
$=54-\frac{45}{2}-150$
$= -118.5$