JEE PYQ: Vector Algebra - Question ID 04efa6dd4d2b (JEE Main 2019)

ID: 04efa6dd4d2bJEE Main 2019Single Correct MCQ
Let A1=3\left| {\mathop {{A_1}}\limits^ \to } \right| = 3, A2=5\left| {\mathop {{A_2}}\limits^ \to } \right| = 5 and A1+A2=5\left| {\mathop {{A_1}}\limits^ \to + \mathop {{A_2}}\limits^ \to } \right| = 5. The value of (2A1+3A2)(3A12A2)\left( {2\mathop {{A_1}}\limits^ \to + 3\mathop {{A_2}}\limits^ \to } \right)\left( {3\mathop {{A_1}}\limits^ \to - \mathop {2{A_2}}\limits^ \to } \right) is :-

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Step-by-step Explanation

A1=3,A2=5andA1+A2=5\left| {\overrightarrow {{A_1}} } \right| = 3,\left| {\overrightarrow {{A_2}} } \right| = 5\,and\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = 5

A1+A2=A12+A22+2A1A2cosθ\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = {\left| {\overrightarrow {{A_1}} } \right|^2} + {\left| {\overrightarrow {{A_2}} } \right|^2} + 2\left| {\overrightarrow {{A_1}} } \right|\left| {\overrightarrow {{A_2}} } \right|\cos \theta

cosθ=310\cos \theta = - {3 \over {10}}

(2A1+3A2).(3A12A2)\left( {2\overrightarrow {{A_1}} + 3\overrightarrow {{A_2}} } \right).\left( {3\overrightarrow {{A_1}} - 2\overrightarrow {{A_2}} } \right)

= 6A12+9A1.A24A1.A26A226{\left| {\overrightarrow {{A_1}} } \right|^2} + 9\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 4\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 6{\left| {\overrightarrow {{A_2}} } \right|^2}

= - 118.5