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← JEE Main 2022 · 25 Jun · Shift 1
MathematicsInteger+4 / -1Question 23 of 90

JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 23

Let θ\theta be the angle between the vectors a⃗\vec a and b⃗\vec b, where ∣a⃗∣=4|\vec a|=4, ∣b⃗∣=3|\vec b|=3 and θ∈(π4,π3)\theta\in\left(\dfrac\pi4,\dfrac\pi3\right). Then
∣(a⃗−b⃗)×(a⃗+b⃗)∣2+4(a⃗⋅b⃗)2\left|\left(\vec a-\vec b\right)\times\left(\vec a+\vec b\right)\right|^2+4\left(\vec a\cdot\vec b\right)^2
is equal to ______.

Answer and worked solution

Official answer: 576

Cross product. (a⃗−b⃗)×(a⃗+b⃗)=2 a⃗×b⃗(\vec a-\vec b)\times(\vec a+\vec b)=2\,\vec a\times\vec b, so the first term is 4∣a⃗∣2∣b⃗∣2sin⁡2θ4|\vec a|^2|\vec b|^2\sin^2\theta.

Total. 4∣a⃗∣2∣b⃗∣2(sin⁡2θ+cos⁡2θ)=4×16×9=5764|\vec a|^2|\vec b|^2(\sin^2\theta+\cos^2\theta)=4\times16\times9=576.

Answer: 576