Let θ be the angle between the vectors a→ and b→, where a→=4,b→=3 and θ∈(π)/4,(π)/3. Then a→-b→×a→+b→^2+4a→·b→^2 is equal to ______

Correct answer: 576. a→-b→×a→+b→^2+4a→·b→^2 =a→×a→+a→×b→−b→×a→−b→×b→^2+4a→⋅b→^2 =4a→×b→^2+a→·b→^2=4a→^2b→^2=4×9×16=576

Q52 / 90 · JEE Main 2022 (25 Jun Shift 1) · Mathematics

Let $θ$ be the angle between the vectors $\vec{a}$ and $\vec{b}$ , where $|\vec{a}| = 4, |\vec{b}| = 3$ and $θ \in (\frac{π}{4}, \frac{π}{3})$ . Then ${|(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})|}^{2} + 4 {(\vec{a} \cdot \vec{b})}^{2}$ is equal to ______

Correct answer: 576

${|(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})|}^{2} + 4 {(\vec{a} \cdot \vec{b})}^{2}$

$= {|\vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}|}^{2} + 4 {(\vec{a} \cdot \vec{b})}^{2}$

$= 4 [{|\vec{a} \times \vec{b}|}^{2} + {(\vec{a} \cdot \vec{b})}^{2}] = 4 {|\vec{a}|}^{2} {|\vec{b}|}^{2} = 4 \times 9 \times 16 = 576$

Hence, the correct answer is 576.