दिया है left|overrightarrow{ A }_{1}right|=3,left|overrightarrow{ A }

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3-1.Vectors

hard

दिया है $\left|\overrightarrow{ A }_{1}\right|=3,\left|\overrightarrow{ A }_{2}\right|=5$ तथा $\left|\overrightarrow{ A }_{1}+\overrightarrow{ A }_{2}\right|=5$ तो $\left(2 \overrightarrow{ A }_{1}+3 \overrightarrow{ A }_{2}\right) \cdot\left(3 \overrightarrow{ A }_{1}-2 \overrightarrow{ A }_{2}\right)$ का मान होगा ?

$-106.5$

$-112.5$

$-118.5$

$-99.5$

(JEE MAIN-2019)

Solution

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