JEE Mains · Physics · STD 11 - Vectors
- JEE Mains 2019
- Hard
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(-118.5\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\left| {{{\vec A}_1}} \right| = 3,\left| {{{\vec A}_2}} \right| = 5,\,and\,\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = 5.\\
\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = {\left| {{{\vec A}_1}} \right|^2} + {\left| {{{\vec A}_2}} \right|^2} + 2\left| {{{\vec A}_1}} \right|\left| {{{\vec A}_2}} \right|\,\cos \,\theta \\
\cos \,\theta \, = - \frac{3}{{10}}\\
\left( {2{{\vec A}_1} + 3{{\vec A}_2}} \right).\left( {3{{\vec A}_1} - 2{{\vec A}_2}} \right)\\
= 6{\left| {{{\vec A}_1}} \right|^2} + 9{{\vec A}_1}.{{\vec A}_2} - 4{{\vec A}_1}.{{\vec A}_2} - 6{\left| {{{\vec A}_2}} \right|^2}\\
= - 118.5
\end{array}\)
\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}\)
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