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→)(3A1→−2A2→)\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→∣=5 and ∣A1→+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→∣=∣A1→∣2+∣A2→∣2+2∣A1→∣∣A2→∣cos⁡θ\,\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

cos⁡θ=−310\cos \theta = - {3 \over {10}}

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

= 6∣A1→∣2+9A1→.A2→−4A1→.A2→−6∣A2→∣26{\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