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QJEEM22Ma95ed: JEE Main 2022 Physics - Class 11 JEEM22Ma95ed Solution: Let be the angle between the vectors and , where and . Then is…

Let θ\theta be the angle between the vectors a→\overrightarrow a and b→\overrightarrow b , where ∣a→∣=4,|\overrightarrow a | = 4, ∣b→∣=3|\overrightarrow b | = 3 and θ∈(π4,π3)\theta \in \left( {{\pi \over 4},{\pi \over 3}} \right) . Then ∣(a→−b→)×(a→+b→)∣2+4(a→ . b→)2{\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2} is equal to __________.

Step-by-Step Verified Solution

∣(a→−b→)×(a→+b→)∣2+4(a→ . b→)2{\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}

⇒∣a→×a→+a→×b→−b→×a→−b→×b→∣2+4(a→ . b→)2\Rightarrow {\left| {\overrightarrow a \times \overrightarrow a + \overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}

⇒∣2(a→×b→)∣2+4(a→ . b→)2\Rightarrow {\left| {2\left( {\overrightarrow a \times \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}

⇒4(a→×b→)2+(a→ . b→)2\Rightarrow 4{\left( {\overrightarrow a \times \overrightarrow b } \right)^2} + {\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}

⇒4∣a→∣2∣b→∣2=4 . 16 . 9=576\Rightarrow 4{\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2} = 4\,.\,16\,.\,9 = 576

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